General Physics 1 PDF Module - First Quarter

E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG1  INTRODUCTION TO PHYSICS Lesson Competencies: 1. Solve measurement problems involving conversion of units, expression of measurements in scientific notation (STEM_GP12EU-Ia-1) 2. Differentiate accuracy from pre...

E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG1  INTRODUCTION TO PHYSICS Lesson Competencies: 1. Solve measurement problems involving conversion of units, expression of measurements in scientific notation (STEM_GP12EU-Ia-1) 2. Differentiate accuracy from precision (STEM_GP12EU-Ia-2) 3. Differentiate random errors from systematic errors (STEM_GP12EU-Ia-3) 4. Estimate errors from multiple measurements of a physical quantity using variance (STEM_GP12EU-Ia5) Physics is about the world and everything in it. Physics describes that world and the kinds of things that take place in it. Sometimes, however, physics seems like an imposition from outside — a requirement you have to get through. That’s a shame, because it’s your world that physics describes. Under the burden of physics problems, though, things can get tough. That’s where this module comes in, because it’s designed to let you tackle those problems with ease. Kirchhoff’s laws? No problem. Carnot engines? No worries. Everything’s here in this module. After you’re done reading, you’ll be a problem-solving pro. What is Physics? It is the science of matter, it’s motion, as well as space and time. Physics can be thought of as humanity’s attempts to describe and explain our universe. Physics has many different branches within it: Mechanics: Statics, Kinematics, Dynamics: Waves, Light, Heat, Electricity, Mechanical Energy, Nuclear Physics. Measuring the Universe Any number that is used to describe a physical phenomenon quantitatively is called a physical quantity. The distinction between these two measurements is in their uncertainty. o Accuracy of a measured value—that is, how close it is likely to be to the true value. o Precision the closeness of each measurements from each other. Errors in Measurements Systematic Error: occurs as a result of a flaw in the experimental design or apparatus Random Error: caused by unpredictable changes in the experiment The accuracy of a measurement can be indicated by the number of significant figures or by a stated uncertainty. The result of a calculation usually has no more significant figure E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG2  than the input data. When only crude estimates are available for input data, we can often make useful order-of-magnitude estimates. Try this! Converting speed units. The world land speed record is 763.0 mi h, set on October 15, 1997, by Andy Green in the jet-engine car Thrust SSC. Express this speed in meters per second. Putting Scientific Notation to Work Significant figures (sig. figs) are those digits in a number or measurement that are not being used and considered as place-values. Zeroes are not significant if they are used only to indicate the position of the decimal point. For example, if the length of a computer desk, as measured by a ruler graduated in millimetres, was found to be 1564.3mm, the measurement has five significant figures. Here are the Rules for Significant Figures which will help you to understand them better. a. All non-zero figures are significant: 25.4 has three significant figures. b. All zeros between non-zeros are significant: 30.08 has four significant figures. c. Zeros to the right of a non-zero figure but to the left of the decimal point are not significant (unless specified with a bar): 109 000 has three significant figures. d. Zeros to the right of a decimal point but to the left of a non-zero figure are not significant: 0.050, only the last zero is significant; the first zero merely calls attention to the decimal point. e. Zeros to the right of the decimal point and following a non-zero figure are significant: 304.50 have five significant figures. Scientific notation or standard index notation is a way of writing any number between 1 and 10 multiplied by an appropriate power of 10 notations. It is a shorthand method of writing numbers that are very large or very small. Scientific notation involves writing the number in the form M x 10 n, where M is a number between 1 and 10 but not 10, and n is an integer. NOTE: Integer is a positive and negative whole number. For Example: 1. The distance from the earth to the nearest star is about 39 900 000 000 000 000m. In scientific notation it is written as 3.99 x 1016m. The exponent tells you how many times to multiply by 10. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG3 2. The mass of hydrogen atom is 0.000 000 000 000 000 000 000 000 001 7 kilograms. In scientific notation it is written as 1.7 x 10-27kg. In this case, the exponent tells you how many times to divide by 10. Metric Units of Measurement: Prefix Symbol Value Prefix Symbol Value Computer value deci d 10-1 = 0.1 Deka da, D 101 = 10 centi c 10-2 = 0.01 Hecto h 102 = 100 milli m 10-3 = 0.001 Kilo k,K 103 = 1000 210 =1024 micro µ 10-6= Mega M 106 = 1,000,000 220 =(1024)2 0.000001 nano n 10-9 Giga G 109 230 =(1024)3 pico P 10-12 Tera T 1012 240 =(1024)4 femto f 10-15 Peta P 1015 250 =(1024)5 atto a 10-18 Exa E 1018 260 =(1024)6 zepto z 10-21 Zetta Z 1021 270 =(1024)7 yocto y 10-24 Yotta Y 1024 280 =(1024)8 xenno x 10-27 Xenna X 1027 290 =(1024)9 weko w 10-30 Weka W 1030 2100 =(1024)10 vendeko v 10-33 Vendeka V 1033 2110 =(1024)11 (unofficial) u 10-36 (unofficial) U 1036 Other Powers of Ten: Examples: Googol (Go) = 10100 1 millimeter = 1 mm = 10-3 m Undecillon (U) = 1036 1 picogram = 1 pg = 10-12 g Myria (my) = 104 1 megasecond = 1 Ms = 106 s Powers of Two and Computer Terms: Powers of two (2) are used primarily in computer memory sizes: 210 = K, kilo = 1024 or kilobinary (proposed IEC prefix of kibi or Ki) 220 = M, mega = (1024)2 mebi or Mi 230 = G, giga = (1024)3 gibi or Gi 240 = T, tera = (1024)4 tebi or Ti 250 = P, peta = (1024)5 pebi or Pi 2 60 = E, exa = (1024)6 exbi or Ei Therefore 1 kilobyte = 1 kB = 1024 bytes ǂ 1000 bytes Try this! Read and answer each question. All working out must be shown where necessary. 1. Convert the following numbers into scientific notation: (a) 27 000 000 = __________________________________________ (b) 0.000 007 12 = __________________________________________ (c) 821 = __________________________________________ (d) 0.000 101 = __________________________________________ (e) 81 250 000 000 = __________________________________________ 2. Fill in the table below. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG4 3. Convert the following: (a) 35cm into m __________m (b) 562g into kg __________kg (c) 1ml into cm3 __________cm3 (d) 1.5 hours into seconds __________seconds 4. What is the mass of 5m3 of cement of density 3000kg/m3? 5. When a golf ball is put in a measuring cylinder of water, the water level rises by 30cm3 when the ball is completely submerged. If the ball weighs 33g in the air, find its density. References: Science TG pages 1-26; MELC General Physics 1 page 537, PIVOT BOW R4QUBE pages 134- 135; K to 12 Science Curriculum Guide University Physics with Modern Physics, Author: Young and Freedom, 3 rd edition pp.1-10 IGSE Physics 0625 notes for topic 1, Gen.Physics Fig. Name: Units, measurements Physics (Principle with Application), Author: Douglas Giancoli pp. 50-74 Conceptual Physics 12th edition, Global edition, Author: Paul G. Hewitt pp.55-77 “Everything happens for a reason, and the reason is usually PHYSICS. “ E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG5 E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG6  VECTORS: KNOWING WHERE YOURE HEADED Lesson Competencies: 1. Differentiate vector and scalar quantities (STEM_GP12V-Ia-8) 2. Perform addition of vectors (STEM_GP12V-Ia-9) 3. Rewrite a vector in component form (STEM_GP12V-Ia10) Vector quantity has both a magnitude and a direction in space. A vector is a combination of exactly two values: a magnitude (like the speed of an object in motion) and a direction (such as the direction of an object in motion). All kinds of things can be described with vectors, including constant motion, acceleration, displacement, magnetic fields, electric fields, and many more. Physical quantity is described by a single number, we call it a scalar quantity. If two vectors have the same direction, they are parallel. If they have the same magnitude and the same direction, they are equal. When the vectors are antiparallel the magnitude of equals the difference of the magnitudes. Vector notation There are many ways of writing the symbol of a vector. Vectors are denoted by symbols with an arrow pointing the direction above it. Representation of vectors vectors are represented by drawing arrows. The length represents magnitude and arrowhead indicates direction. vectors are added geometrically by placing the tail of one vector on the head of another. The resultant is the vector that begins at the tail of the first vector and ends at the arrow head of the final vector. vectors can be subtracted by adding the negative vector. FINDING A VECTOR’S COMPONENT Unit Vectors: A unit vector is a vector that has a magnitude of 1, with no units. Method of components: simple but general method for adding vectors E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG7  To convert this vector into the coordinate way of looking at vectors, you have to use the trigonometry shown in the figure. The x coordinate equals v cos θ, and the y coordinate equals v sin θ: vx = v cos θ vy = v sin θ Components are not vectors. The components AX and AY of a vector A→ are just numbers; they are not vectors themselves. This is why we print the symbols for components in light italic type with no arrow on top instead of in boldface italic with an arrow, which is reserved for vectors. WORK ON THIS! 1.a. Suppose a person walks 3.0 meters to the right of the origin. What is his displacement vector in terms of coordinates? 1.b. Suppose that you’ve walked away from the origin so that you’re now at 5.0 kilometers from the origin, at an angle of 45°. Resolve that into vector coordinates. 2. Find the component vectors of the following: x- y- component component a. 3.0 m, θ=15˚ b. 9.0 m, θ=35˚ c. 6.0m, θ=125˚ d. 4.0m, θ=255˚ VECTOR’S MAGNITUDE AND DIRECTION If you’re given the coordinates of a vector, such as (4 , 3), you can convert it easily to the magnitude/angle way of expressing vectors using trigonometry. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG8  1. You know that: vx = v cos θ 4 vy = v sin θ 2. In other words: 3 3. Which means that: 4. You can calculate the inverse sine (sin–1) or inverse cosine (cos–1) on your calculator. (Look for the sin–1 and cos–1 button.) 5. In the figure above, you’re given x and y, the coordinates, but not v, the magnitude. Dividing the expressions for y and x above gives you: 6. Where tan θ is the tangent of the angle. This means that: 7. Suppose that the coordinates of the vector are (4, 3). You can find the angle θ as the tan– 1(3/4) = 36.87°. And you can use the Pythagorean theorem to find the hypotenuse — the magnitude, v — of the triangle formed by x, y, and v: 8. Plug in the numbers for this example to get: 9. So if you have a vector given by the coordinates (4, 3), its magnitude is 5, and its angle is 36.87°. VECTOR ALGEBRA Vectors are added graphically. Arrows representing two or more vectors can be combined to produce a single resultant or outcome vector. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG9  Finding v isn’t so hard because you can use the Pythagorean theorem: Y=20 Plugging in the numbers to get: X=20 So the convention is 28.3 miles away. What about the angle θ? You know that: In this case, you can find the angle θ like so: And that’s it — you now know that the convention is 28.3 units away at an angle of 45°. LET’S TRY! Example 1: An Air Niugini aeroplane is travelling east at a velocity of 115km/h. The wind also blows east at 20km/h. The resultant can be found by drawing two vectors to the same scale and adding head to tail. The resultant of 135km/h has the same direction as the component velocities. Example 2: The velocity of an aeroplane is 125km/h and its course is eastward. The wind blows towards the west at 25km/h. The headwind will slow the plane down. The resultant velocity of the aeroplane is found by adding the vectors head to tail again. The resultant velocity of the aeroplane will be 100km/h east. (The resultant, has the same direction as the greater velocity). Example 3: A pilot in Air Niugini aeroplane flying eastward at 121km/h encounters a strong wind of 89km/h blowing to the north. To find the resultant velocity, we draw a scale diagram and use a trigonometry principle – Pythagoras Theorem. In this case, the vectors are added head to tail as shown in the diagram above. The diagram is found by joining the starting point to the finishing point. The resultant is the diagonal of the triangle constructed by using the component velocity vectors x and y as sides. The resultant is 150km/h (since the scale is 1cm = 30N there are 5cm in the resultant 5 x 30 = 150N from the sphere 3-4-5) in a direction E36.9°N can be calculated by E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG10 Pythagoras’s theorem or measured from the scale diagram that is tan θ = 90/120 = 0.75, then θ = 36.9° In a right triangle (the triangle that forms 90° angle), the longest side is called the hypotenuse. Y axis is the side that is opposite the right angle. We say that YZ is the side opposite angle X while XY is the side adjacent to angle at X. Example 4: Add the two vectors in Figure shown. One has a magnitude 5.0 and angle 45°, and the other has a magnitude 7.0 and angle 35°. Resolve the two vectors into their components. For the first vector, apply the equation vx = v cos θ to find the x coordinate. That’s 5.0 cos 45° = 3.5. Apply the equation vy = v sin θ to find the y coordinate of the first vector. That’s 5.0 sin 45°, or 3.5. So the first vector is (3.5, 3.5) in coordinate form. For the second vector, apply the equation vx = v cos θ to find the x coordinate. That’s 7.0 cos 35° = 5.7. Apply the equation vy = v sin θ to find the y coordinate of the second vector. That’s 7.0 sin 35° = 4.0. So the second vector is (5.7, 4.0) in coordinate form. To add the two vectors, add them in coordinate form: (3.5, 3.5) + (5.7, 4.0) = (9.2, 7.5). Convert (9.2, 7.5) into magnitude/angle form. Apply the equation θ = tan–1(y/x) to find the angle, which is tan–1(.82) = 39°. Apply the equation 𝑣 = √𝑥 2 + 𝑦 2 to find the magnitude, which is 𝑣 = √9.22 + 7.52 = 11.9. Converting to two significant digits gives you 12. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG11  WORK ON THIS! Solve for the following problems, supplement your answers with necessary solutions. 1. Two forces act on a block as shown below. Calculate the resultant force. 2. Add a vector whose magnitude is 13.0 and angle is 27° to one whose magnitude is 11.0 and angle is 45°. 3. Add a vector whose magnitude is 16.0 and angle is 56° to one whose magnitude is 10.0 and angle is 25°. 4. Add two vectors: Vector one has a magnitude 22.0 and angle of 19°, and vector two has a magnitude 19.0 and an angle of 48°. 5. Add a vector whose magnitude is 10.0 and angle is 257° to one whose magnitude is 11.0 and angle is 105°. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG12 References: Science TG pages 1-26; MELC General Physics 1 page 537, PIVOT BOW R4QUBE pages 134- 135; K to 12 Science Curriculum Guide University Physics with Modern Physics, Author: Young and Freedom, 3rd edition pp.1-10 IGSE Physics 0625 notes for topic 1, Gen.Physics Fig. Name: Units, measurements Physics (Principle with Application), Author: Douglas Giancoli pp. 50-74 Conceptual Physics 12th edition, Global edition, Author: Paul G. Hewitt pp.55-77 https://study.com/academy/lesson/what-is-a-vector-definition-types.html https://www.physics-prep.com/index.php/vectors-activity I’d be unstoppable if not for LAW ENFORECEMENT aSnd PHYSICS! E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG13 E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG14  THE BIG THREE: Acceleration, Distance and Time Kinematics: Motion along Straight Line Lesson Competencies: 1. Convert a verbal description of a physical situation involving uniform acceleration in one dimension into a mathematical description (STEM_GP12Kin-Ib12) 2. Interpret displacement and velocity, respectively, as areas under velocity vs. time and acceleration vs. time curves (STEM_GP12KINIb-14) 3. Interpret velocity and acceleration, respectively, as slopes of position vs. time and velocity vs. time curves (STEM_GP12KINIb-15) 4. Construct velocity vs. time and acceleration vs. time graphs, respectively, corresponding to a given position vs. time-graph and velocity vs. time graph and vice versa (STEM_GP12KINIb-16) 5. Solve for unknown quantities in equations involving one-dimensional uniformly accelerated motion, including free fall motion (STEM_GP12KINIb-17) 6. Solve problems involving one-dimensional motion with constant acceleration in contexts such as, but not limited to, the “tail-gating phenomenon”, pursuit, rocket launch, and freefall problems (STEM_GP12KINIb-19) From Point A to B: Displacement Displacement occurs when something moves from here to there. For example, suppose that you have a ball at the zero position, as in the Figure. Now suppose that the ball rolls over to a new point, 3 meters to the right, as you see in the second Figure. The ball is at a new location, so there’s been displacement. In this case, the displacement is just 3 meters to the right. In physics terms, you’ll often see displacement referred to as the variable s. In this case, s = +3 meters. LET’S TRY! You’ve taken the pioneers’ advice to “Go West.” You started in New York City and went west 10 miles the first day, 14 miles the next day, and then back east 9 miles on the third day. What is your displacement from New York City after three days? 1. You first went west 10 miles, so at the end of the first day, your displacement was 10 miles west. 2. Next, you went west 14 days, putting your displacement at 10 + 14 miles = 24 miles west of New York City. 3. Finally, you traveled 9 miles east, leaving you at 24 – 9 = 15 miles west of New York City. So s = 15 miles west of New York City How Fast I’ll Go: Speed Before the time of Galileo, people described moving things as simply “slow” or “fast.” Such descriptions were vague. Galileo is credited with being the first to measure speed by considering the distance covered and the time it takes. He defined speed as the distance covered per unit of time. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG15 For example, if you went a displacement s in a time t, then your speed, v, is determined as follows: Technically speaking, speed is the change in position divided by the change in time, so you also can represent it like this if, for example, you’re moving along the x axis: We use the symbol vav-x for average x-velocity (the subscript “av” signifies average value and the subscript x indicates that this is the x-component): The meaning of Δx. Note that Δx is not the product of Δ and x; it is a single symbol that means “the change in the quantity x.” We always use the Greek capital letter Δ (delta) to represent a change in a quantity, equal to the final value of the quantity minus the initial value—never the reverse. Likewise, the time interval from t1 to t2 is Δt the change in the quantity t: Δt=t2-t1 (final time minus initial time). Choice of the positive x-direction. You might be tempted to conclude that positive average x- velocity must mean motion to the right, and that negative average x-velocity must mean motion to the left. But that’s correct only if the positive x-direction is to the right, as we chose it to be in. In most problems the direction of the coordinate axis will be yours to choose. Once you’ve made your choice, you must take it into account when interpreting the signs of and other quantities that describe motion! Instantaneous Speed Things in motion often have variations in speed. A car, for example, may travel along a street at 50 km/h, slow to 0 km/h at a red light, and speed up to only 30 km/h because of traffic. You can tell the speed of the car at any instant by looking at its speedometer. The speed at any instant is the instantaneous speed. A car traveling at 50 km/h usually goes at that speed for less than 1 hour. If it did go at that speed for a full hour, it would cover 50 km. If it continued at that speed for half an hour, it would cover half that distance: 25 km. If it continued for only 1 minute, it would cover less than 1 km. In the language of calculus, the limit of Δx/Δt as Δt approaches zero is called the derivative of x with respect to t and is written dx/dt. The instantaneous velocity is the limit of the average velocity as the time interval approaches zero; it equals the instantaneous rate of change of position with time. We use the symbol with no “av” subscript, for the instantaneous velocity along the x- axis, or the instantaneous x-velocity: The terms “velocity” and “speed” are used interchangeably in everyday language, but they have distinct definitions in physics. We use the term speed to denote distance traveled divided by time, on either an average or an instantaneous basis. Instantaneous speed, for which we use the symbol v with no subscripts, measures how fast a particle is moving; instantaneous velocity measures how fast and in what E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG16 direction it’s moving. Instantaneous speed is the magnitude of instantaneous velocity and so can never be negative. For example, a particle with instantaneous velocity vx = 25m/s and a second particle with vx = -25m/s are moving in opposite directions at the same instantaneous speed. Average Speed In planning a trip by car, the driver often wants to know the time of travel. The driver is concerned with the average speed for the trip. Average speed can be calculated rather easily. For example, if we travel a distance of 80 kilometers in a time of 1 hour, we say our average speed is 80 kilometers per hour. Likewise, if we travel 320 kilometers in 4 hours, We see that, when a distance in kilometers (km) is divided by a time in hours (h), the answer is in kilometers per hour (km/h). Since average speed is the distance traveled divided by the total time of travel, it doesn’t indicate the different speeds and variations that may have taken place during shorter time intervals. On most trips, we experience a variety of speeds, so the average speed is often quite different from the instantaneous speed at any particular moment. LET’S TRY! Average and Instantaneous Velocity A cheetah is crouched 20 m to the east of an observer. At time t=0 the cheetah begins to run due east toward an antelope that is 50 m to the east of the observer. During the first 2.0 s of the attack, the cheetah’s coordinate x varies with time according to the equation x = 20m + (5.0m/s2) t2. (a) Find the cheetah’s displacement between t 1 = 1.0s and t2 = 2.0s (b) Find its average velocity during that interval. (c) Find its instantaneous velocity at t1 = 1.0s by taking Δt1 = 0.1s then 0.01s then 0.001s (d) Derive an expression for the cheetah’s instantaneous velocity as a function of time, and use it to find vx at t1 = 1.0s and t2 = 2.0s. A cheetah attacking an antelope from ambush. The animals are not drawn to the same scale as the axis. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG17  Finding Velocity on an x-t Graph Using an x-t graph to go from (a), (b) average x-velocity to (c) instantaneous x-velocity vx. In (c) we find the slope of the tangent to the x-t curve by dividing any vertical interval (with distance units) along the tangent by the corresponding horizontal interval (with time units). The Figure below depicts the motion of a particle in two ways: as (a) an x-t graph and (b) a motion diagram that shows the particle’s position at various instants (like frames from a video of the particle’s motion) as well as arrows to represent the particle’s velocity at each instant. We will use both x-t graphs and motion diagrams in this chapter to help you understand motion. You will find it worth your while to draw both an x-t graph and a motion diagram as part of solving any problem involving motion. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG18 (a) The x-t graph of the motion of a particular particle. The slope of the tangent at any point equals the velocity at that point. (b) A motion diagram showing the position and velocity of the particle at each of the times labeled on the x-t graph. WORK ON THIS! Compute for the following unknowns, supplement you answer/s with necessary solutions. 1. What is the average speed of a cheetah that sprints 100 meters in 4 seconds? if it sprints 50 m in 2 s? 2. If a car moves with an average speed of 60 km/h for an hour, it will travel a distance of 60 km. a. how far would it travel if it moved at this rate for 4 h? b. For 10 h? 3. A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t) = bt2 – ct3 where b = 2.40 m/s2 and c = 0.120 m/s3. (a) Calculate the average velocity of the car for the time interval t = 0 to t = 10.0s. (b) Calculate the instantaneous velocity of the car at t = 0, t = 5.0s and t = 10.0s (c) How long after starting from rest is the car again at rest? Putting Pedal to Metal: Acceleration In physics terms, acceleration is the amount by which your speed changes in a given amount of time. In terms of equations, it works like this: Given initial and final velocities, v o and vf, and initial and final times over which your speed changed, t o and tf, you can also write the equation like this: To get the units of acceleration, you divide speed by time as follows: E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG19  Suppose we are driving and, in 1 second, we steadily increase our velocity from 30 kilometers per hour to 35 kilometers per hour, and then to 40 kilometers per hour in the next second, to 45 in the next second, and so on. We change our velocity by 5 kilometers per hour each second. This change in velocity is what we mean by acceleration. In this case, the acceleration is 5 kilometers per hour second (abbreviated as 5 km/h·s), in the forward direction. Note that a unit for time enters twice: once for the unit of velocity and again for the time interval in which the velocity is changing. Also note that acceleration is not just the total change in velocity; it is the time rate of change, or change per second, in velocity. Average Acceleration Let’s consider a particle moving along the x-axis. Suppose that at time t1 the particle is at point P1 and has x-component of (instantaneous) velocity v1x and at a later time t2 it is at point P2 and has x-component of velocity So the x-component of velocity v2x changes by an amount Δvx = v2x - v1x during the time interval Δt = t2 - t1. We define the average acceleration of the particle as it moves from P1 to P2 to be a vector quantity whose x-component aav-x (called the average x-acceleration) equals Δvx, the change in the x-component of velocity, divided by the time interval Δt. Instantaneous Acceleration We can now define instantaneous acceleration following the same procedure that we used to define instantaneous velocity. As an example, suppose a race car driver is driving along a straightaway as shown in Figure below. To define the instantaneous acceleration at point P1 we take the second point P2 in Figure below to be closer and closer to P1 so that the average acceleration is computed over shorter and shorter time intervals. The instantaneous acceleration is the limit of the average acceleration as the time interval approaches zero. In the language of calculus, instantaneous acceleration equals the derivative of velocity with time. Thus Note that in Equation is really the x-component of the acceleration vector, or the E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG20 instantaneous x-acceleration; in straight-line motion, all other components of this vector are zero. From now on, when we use the term “acceleration,” we will always mean instantaneous acceleration, not average acceleration. LET’S TRY! Average and Instantaneous Acceleration Suppose the x-velocity vx of the car in Figure at any time t is given by the equation vx = 60 m/s + (0.50 m/s3) t2 (a) Find the change in x-velocity of the car in the time interval t1 = 1.0s to t2 = 3.0s (b) Find the average x-acceleration in this time interval. (c) Find the instantaneous x-acceleration at time t1 = 1.0 s by taking Δt to be first 0.1 s, then 0.01 s, then 0.001 s. (d) Derive an expression for the instantaneous x-acceleration as a function of time, and use it to find ax at t1 = 1.0s and t2 = 3.0s Execute: (a) Before we can apply the equation, we must find the x-velocity at each time from the given equation. At t1 = 1.0s and t2 = 3.0s , the velocities are: E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG21  Finding Acceleration on a vx-t Graph or an x-t Graph On a graph of x-velocity as a function of time, the instantaneous x-acceleration at any point is equal to the slope of the tangent to the curve at that point. Tangents drawn at different points along the curve in Figure have different slopes, so the instantaneous x-acceleration varies with time. The signs of x-acceleration and x-velocity By itself, the algebraic sign of the x-acceleration does not tell you whether a body is speeding up or slowing down. You must compare the signs of the x-velocity and the x-acceleration. When and have the same sign, the body is speeding up. If both are positive, the body is moving in the positive direction with increasing speed. If both are negative, the body is moving in the negative direction with an x-velocity that is becoming more and more negative, and again the speed is increasing. When and have opposite signs, the body is slowing down. If vx is positive and ax is negative, the body is moving in the positive direction with decreasing speed; if is negative and is positive, the body is moving in the negative direction with an x-velocity that is becoming less negative, and again the body is slowing down. (a) A vx-t graph of the motion of a different particle from that shown in the figure. The slope of the tangent at any point equals the x-acceleration at that point. (b) A motion diagram showing the position, velocity, and acceleration of the particle at each of the times labeled on the graph. The positions are consistent with the vx-t graph; for instance, from to the velocity is negative, so at tb the particle is at a more negative value of x than at tA. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG22 (a) The same x-t graph as shown. The x-velocity is equal to the slope of the graph, and the acceleration is given by the concavity or curvature of the graph. (b) A motion diagram showing the position, velocity, and acceleration of the particle at each of the times labeled on the x-t graph. An acceleration-time graph (ax-t) for straight- A velocity-time graph (vx-t) for straight-line line motion with constant positive x- motion with constant positive x- acceleration ax. acceleration ax. The initial x-velocity vox is also positive in this case. WORK ON THIS! Compute for the following unknowns, supplement you answer/s with necessary solutions. 1. Suppose that you’re driving at 75 miles an hour and suddenly see red flashing lights in the rearview mirror. “Great,” you think, and you pull over, taking 20 seconds to come to a stop. You could calculate how quickly you decelerated as you were pulled over (information about your law-abiding tendencies that, no doubt, would impress the officer). So just how fast did you decelerate, in cm/sec 2? E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG23  2. A car’s velocity as a function of time is given by vx(t) = α + βt2, where α = 3.00 m/s and β = 0.100 m/s. (a) Calculate the average acceleration for the time interval t = 0 to t = 5.0s. (b) Calculate the instantaneous acceleration for t = 0 and t = 5.0s. (c) Draw and graphs for the car’s motion between t = 0 and t = 5.0s. Connecting Acceleration, Time, and Displacement When the x-acceleration is constant, four equations relate the position x and the x-velocity vx at any time t to the initial position xo the initial x-velocity vox (both measured at time t = 0 ) and the x- acceleration ax. LET’S TRY! A motorcyclist heading east through a small town accelerates at a constant 4.0 m/s after he leaves the city limits. At time he is 2 5.0 m east of the city-limits signpost, moving east at 15 m/s. (a) Find his position and velocity at t = 2.0s (b) Where is he when his velocity is 25 m/s? Straight-Line Motion with Varying Acceleration When the acceleration is not constant but is a known function of time, we can find the velocity and position as functions of time by integrating the acceleration function. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG24  LET’S TRY! Sally is driving along a straight highway in her 1965 Mustang. At t = 0 when she is moving at 10m/s in the positive x-direction, she passes a signpost at x= 50m. Her x-acceleration as a function of time is: (a) Find her x-velocity and position x as functions of time. (b) When is her x-velocity greatest? (c) What is that maximum x-velocity? (d) Where is the car when it reaches that maximum x-velocity? (a). Integrating the function: Now we use the distance equation to find x as a function of t: (b). We set the expression for ax equal to zero and solve for t: The position, velocity, and acceleration of the car in Example as functions of time. Can you show that if this motion continues, the car will stop at t = 44.5 s? E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG25  WORK ON THIS! Compute for the following unknowns, supplement you answer/s with necessary solutions. 1. A trailer breaks loose from its truck on a steep incline. If the truck was moving uphill at 20 meters per second when the trailer broke loose, and the trailer accelerates down the hill at 10.0 meters per second2, how far downhill does the trailer go after 10 seconds? 2. A minivan puts on the brakes and comes to a stop in 12 seconds. If it took 200 meters to stop, and decelerates at 10 meters per second 2, how fast was it originally going, in meters per second? 3. An antelope moving with constant acceleration covers the distance between two points 70.0 m apart in 7.00 s. Its speed as it passes the second point is (a) What is its speed at the first point? (b) What is its acceleration? 4. The fastest measured pitched baseball left the pitcher’s hand at a speed of 45.0 m/s. If the pitcher was in contact with the ball over a distance of 1.50 m and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it? E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG26  Love: Free Falling Bodies Free fall is a case of motion with constant acceleration. The constant acceleration of a freely falling body is called the acceleration due to gravity, and we denote its magnitude with the letter g. We will frequently use the approximate value of g at or near the earth’s surface: LET’S TRY! Falling from a tower. A one-euro coin is dropped from the Leaning Tower of Pisa and falls freely from rest. What are its position and velocity after 1.0 s, 2.0 s, and 3.0 s? At a time t after the coin is dropped, its position and y-velocity are: WORK ON THIS! Compute for the following unknowns, supplement you answer/s with necessary solutions. 1. A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 13.0 m/s. The rock misses the edge of the cliff as it falls back to earth. Calculate the position and velocity of the rock 1.00 s, 2.00 s, and 3.00 s after it is thrown, neglecting the effects of air resistance. 2. You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at a point even with the roof railing with an upward speed of 15 m/s; the ball is E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG27  then in free fall. On its way back down, it just misses the railing. Find (a) the ball’s position and velocity 1.00 s and 4.00 s after leaving your hand; (b) the ball’s velocity when it is 5.00 m above the railing; (c) the maximum height reached; (d) the ball’s acceleration when it is at its maximum height. REFERENCES: Science TG pages 1-26; MELC General Physics 1 page 537, PIVOT BOW R4QUBE pages 134-135; K to 12 Science Curriculum Guide University Physics with Modern Physics, Author: Young and Freedom, 3rd edition pp.35- 59 Physics (Principle with Application), Author: Douglas Giancoli pp. 50-74 Conceptual Physics 12th edition, Global edition, Author: Paul G. Hewitt pp.55-77 Student Workbook for College Physics: A Strategic Approach Volume 1. Pearson New International Edition. Knight et.al. pp 1-8 http://www.mrmont.com/games/carequationchallenge.html https://ophysics.com/k7.html PHYSICS “The branch of science concerned with using extremely long and complicated formula to describe a coin falling from a tower” E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG28 E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG29  Round and Round: Circular Motion Kinematics: Motion along 2D or 3D Lesson Competencies: Describe motion using the concept of relative velocities in 1D and 2D Deduce the consequences of the independence of vertical and horizontal components of projectile motion Calculate range, time of flight, and maximum heights of projectiles Infer quantities associated with circular motion such as tangential velocity, centripetal acceleration, tangential acceleration, radius of curvature Solve problems involving two dimensional motion in contexts such as, but not limited to ledge jumping, movie stunts, basketball, safe locations during firework displays, and Ferris wheels Projectile Motion A projectile is any body that is given an initial velocity and then follows a path determined entirely by the effects of gravitational acceleration and air resistance. A batted baseball, a thrown football, a package dropped from an airplane, and a bullet shot from a rifle are all projectiles. The path followed by a projectile is called its trajectory. In projectile motion with no air resistance, ax = 0 and ay = -g. The coordinates and velocity components are simple functions of time, and the shape of the path is always a parabola. We usually choose the origin to be at the initial position of the projectile. Figure 1: Projectile Motion List of Equations 1: Projectile Motion Figure 2: Trajectory of Projectile Motion E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG30  LET’S TRY! A body projected horizontally. A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 9.0m/s. Find the motorcycle’s position, distance from the edge of the cliff, and velocity 0.50 s after it leaves the edge of the cliff. A. The motorcycle’s position, distance from the edge of the cliff: B. Velocity 0.50 s after it leaves the edge of the cliff: The velocity is 29˚ below the horizontal. WORK ON THIS! Compute for the following unknowns, supplement you answer/s with necessary solutions. 1. Height and range of a projectile I: A batted baseball A batter hits a baseball so that it leaves the bat at speed vo = 37.0 m/s at an angle αo = 53.1˚. (a) Find the position of the ball and its velocity (magnitude and direction) at t = 2.00s. (b) Find the time when the ball reaches the highest point of its flight, and its height h at this time. (c) Find the horizontal range R—that is, the horizontal distance from the starting point to where the ball hits the ground. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG31  Converting between Angles The first step in working with angular motion is to know about the way of measuring that motion. You use radians, not meters, and you have to know what that means. There are 2π radians in a full, 360° circle. That means that in order to convert 45° from degrees to radians, you multiply by 2π / 360°, like so: Period and Frequency When describing the way things go in circles, you don’t just use radians; you also can specify the time it takes. The time it takes for an object to complete an orbit is referred to as its period. For example, if the object is traveling at speed v, then the time it takes to go around the circle — the distance it travels in the circle’s circumference, 2πr — will be: Equation 1 : Period Note the symbol of the radius of a circle: r. That’s half the circle’s diameter, which is d. So r = d / 2. Note also the symbol for the period: T. With this equation, given an orbiting object’s speed and the radius of the circle, you can calculate the object’s period. Another time measurement you’ll see in physics problems is frequency. Whereas the period is the time an object takes to go around in a circle, the frequency is the number of circles the object makes per second. The Equation 2: Frequency frequency, f, is connected to the period like this: Uniform and nonuniform circular motion When a particle moves in a circular path of radius R with constant speed (uniform circular motion), its acceleration a is directed toward the center of the circle and perpendicular to v. The magnitude arad of the acceleration can be expressed in terms of v and R or in terms of R and the period T (the time for one revolution), where v = 2πR/T. If the speed is not constant in circular motion (nonuniform circular motion), there is still a radial component of a given by the equations below, but there is also a component of parallel (tangential) to the path. This tangential component is equal to the rate of change of speed, dv/dt. List of Equations 2: Uniform Circular Motion E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG32  LET’S TRY! 1. Centripetal acceleration on a curved road: An Aston Martin V8 Vantage sports car has a “lateral acceleration” of 0.96g = (0.96)(9.8 m/s2) = 9.4 m/s2. This is the maximum centripetal acceleration the car can sustain without skidding out of a curved path. If the car is traveling at a constant 40 m/s (about 89 mi/hr or 144 km/h) on level ground, what is the radius R of the tightest unbanked curve it can negotiate? This is the minimum radius because arad is the maximum centripetal acceleration. 2. Centripetal acceleration on a carnival ride. Passengers on a carnival ride move at constant speed in a horizontal circle of radius 5.0 m, making a complete circle in 4.0 s. What is their acceleration? Relative Velocity The velocity seen by a particular observer is called the velocity relative to that observer, or simply relative velocity. Each observer, equipped in principle with a meter stick and a stopwatch, forms what we call a frame of reference. LET’S TRY! Flying in a crosswind: An airplane’s compass indicates that it is headed due north, and its airspeed indicator shows that it is moving through the air at 240 km/h. If there is a 100-km h wind from west to east, what is the velocity of the airplane relative to the earth? E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG33  The plane is pointed north, but the wind blows east, giving the resultant velocity vp/e relative to the earth. WORK ON THIS! Compute for the following unknowns, supplement you answer/s with necessary solutions. Problem Set (Kinematics in 2D or 3D) 1. A squirrel has x- and y-coordinates (1.1m, 3.4m) at time t1= 0 and coordinates (5.3 m, -0.5 m) at time t2=3.0s. For this time interval, find (a) the components of the average velocity, and (b) the magnitude and direction of the average velocity. 2. A jet plane is flying at a constant altitude. At time t1=0 it has components of velocity vx = 90 m/s, vy = 110 m/s. At time t2 = 30.0 s the components are vx = - 170 m/s, vy = 40 m/s. (a) Sketch the velocity vectors at and How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration. 3. A daring 510-N swimmer dives off a cliff with a running horizontal leap, as shown in Fig. E3.10. What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 m wide and 9.00 m below the top of the cliff? 4. You throw a ball from your window 8.0 m above the ground. When the ball leaves your hand, it is moving at at an angle of below the horizontal. How far horizontally from your window will the ball hit the ground? Ignore air resistance. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG34  5. A physics book slides off a horizontal tabletop with a speed of it strikes the floor in 0.350 s. Ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book’s velocity, and the magnitude and direction of its velocity, just before the book reaches the floor. 6. Two crickets, Chirpy and Milada, jump from the top of a vertical cliff. Chirpy just drops and reaches the ground in 3.50 s, while Milada jumps horizontally with an initial speed of How far from the base of the cliff will Milada hit the ground? 7. On level ground a shell is fired with an initial velocity of at 60.0° above the horizontal and feels no appreciable air resistance. (a) Find the horizontal and vertical components of the shell’s initial velocity. (b) How long does it take the shell to reach its highest point? (c) Find its maximum height above the ground. (d) How far from its firing point does the shell land? (e) At its highest point, find the horizontal and vertical components of its acceleration and velocity. 8. A “moving sidewalk” in an airport terminal building moves at and is 35.0 m long. If a woman steps on at one end and walks at relative to the moving sidewalk, how much time does she require to reach the opposite end if she walks (a) in the same direction the sidewalk is moving? (b) In the opposite direction? 9. Our balance is maintained, at least in part, by the endolymph fluid in the inner ear. Spinning displaces this fluid, causing dizziness. Suppose a dancer (or skater) is spinning at a very fast 3.0 revolutions per second about a vertical axis through the center of his head. Although the distance varies from person to person, the inner ear is approximately 7.0 cm from the axis of spin. What is the radial acceleration (in and in g’s) of the endolymph fluid? 10. A Ferris wheel with radius 14.0 m is turning about a horizontal axis through its center (Fig. E3.29). The linear speed of a passenger on the rim is constant and equal to What are the magnitude and direction of the passenger’s acceleration as she passes through (a) the lowest point in her circular motion? (b) The highest point in her circular motion? (c) How much time does it take the Ferris wheel to make one revolution? References: Science TG pages 1-26; MELC General Physics 1 page 537, PIVOT BOW R4QUBE pages 134-135; K to 12 Science Curriculum Guide University Physics with Modern Physics, Author: Young and Freedom, 3rd edition pp.69-95 Physics (Principle with Application), Author: Douglas Giancoli pp. 50-74 Conceptual Physics 12th edition, Global edition, Author: Paul G. Hewitt pp.55-77 Student Workbook for College Physics: A Strategic Approach Volume 1. Pearson New International Edition. Knight et.al. pp 23-32 https://phet.colorado.edu/sims/html/projectile-motion/latest/projectile-motion_en.html http://www.mrmont.com/games/carequationchallenge.html E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG35  https://ophysics.com/k7.html https://openstax.org/books/university-physics-volume-1/pages/4-3-projectile-motion https://d3bxy9euw4e147.cloudfront.net/oscms- prodcms/media/documents/APCollegePhysics-OP.pdf E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG36 E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG37  May The Forces Be With You: NEWTON’S LAW OF MOTION AND ITS APLLICATION Lesson Competencies: 4. Define inertial frames of reference (STEM_GP12N-Id28) 5. Identify action -reaction pairs (STEM_GP12N -Id – 31) 6. Draw free -body diagrams (STEM_GP12N -Id – 32) 7. Apply Newton’s 1st law to obtain quantitative and qualitative conclusions about the contact and noncontact forces acting on a body in equilibrium (STEM_GP12N -Ie – 33) 8. Differentiate the properties of static friction and kinetic friction (STEM_GP12N -Ie – 34) 9. Apply Newton’s 2nd law and kinematics to obtain quantitative and qualitative conclusions about the velocity and acceleration of one or more bodies, and the contact and noncontact forces acting on one or more bodies (STEM_GP12N -Ie – 36) 10. Solve problems using Newton’s Laws of motion in contexts such as, but not limited to, ropes and pulleys, the design of mobile sculptures, transport of loads on conveyor belts, force needed to move stalled vehicles, determination of safe driving speeds on banked curved roads (STEM_GP12N -Ie – 38) What is Force? Force in simple word is a strength use in physical action. On the other hand, force in science is simply a push or a pull to an object. You can apply this force with or without touching each other and can cause objects at rest to accelerate. Force (F ⃗ ) is a vector which can be considered as a push or a pull. The unit of force is Newton (N). 1N=1kg*m/s2 Figure 3: Properties of Forces Table 1. Typical Force Magnitudes Source Force (N) Sun’s gravitational force on the earth 3.5 x 1022 Thrust of a space shuttle during launch 3.1 x 107 Weight of a large blue whale 1.9 x 106 Maximum pulling force of a locomotive 8.9 x 105 Weight of a 250-lb linebacker 1.1 x 103 Weight of a medium apple 1 Weight of smallest insect eggs 2 x 10-6 Electric attraction between the proton and the electron in a hydrogen 8.2 x 10-8 atom Weight of a very small bacterium 1 x 10-18 Weight of a hydrogen atom 1.6 x 10-26 Weight of an electron 8.9 x 10-30 Gravitational attraction between the proton and the electron in a 3.6 x 10-47 hydrogen atom E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG38  A. Contact Force is when the force involves direct contact between the two bodies a. Normal Force (FN or n) is the force each body exerts on the other perpendicular to their contact surface. Figure 4: Normal Force Figure 5: Example of Contact Force b. Frictional Force (Ff or f) is the force exerted on an object by a surface acts parallel to the surface, in the direction that opposes sliding. b.1. Kinetic Friction (𝒇𝒌 ) is the kind of friction that acts when a body slides over a surface: 𝑓𝑘 = 𝜇𝑘 𝑛 , 𝜇𝑘 is the coefficient of kinetic friction b.2. Static Friction (𝒇𝒔 ) is the kind of friction that acts Figure 6: Friction Force when there is no relative motion on the body: 𝑓𝑠 = 𝜇𝑠 𝑛 , 𝜇𝑠 is the coefficient of static friction Rolling friction is similar to kinetic friction where 𝜇𝑟 is called coefficient of rolling friction or tractive resistance. Typical values of 𝜇𝑟 are 0.002 to 0.003 for steel wheels on steel rails and 0.01 to 0.02 for rubber tires on concrete. The force of fluid resistance depends on the speed of an object through a fluid. 𝑚𝑔 𝑓 = 𝑘𝑣 (fluid resistance at low speed) ; 𝑣𝑡 = (terminal speed) 𝑘 where k is a proportionality constant that depends on the shape and size of the body and the properties of the fluid. 𝑚𝑔 𝑓 = 𝐷𝑣 2 (fluid resistance at high speed, air drag) ; 𝑣𝑡 = √ 𝐷 (terminal speed) Table 2. Approximate Coefficient of Friction Coefficient of static friction, Coefficient of kinetic friction, Materials 𝝁𝒔 𝝁𝒌 Steel on steel 0.74 0.57 Aluminum on steel 0.61 0.47 Copper on steel 0.53 0.36 Brass on steel 0.51 0.44 Zinc on cast iron 0.85 0.21 Copper on cast iron 1.05 0.29 Glass on glass 0.94 0.40 Copper on glass 0.68 0.53 Teflon on Teflon 0.04 0.04 Teflon on steel 0.04 0.04 Rubber on concrete 1.0 0.8 (dry) Rubber on concrete 0.30 0.25 (wet) c. Tension Force (T) is the force transmitted through string, rope, and cable or wire when pulled by forces acting from opposite ends. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG39  Figure 7: Tension Force B. Noncontact Force (Action-at-a distance Force) is the force that act even when the bodies are separated by empty space Figure 9: Non-Contact Forces a. Gravitational Force (Fg), also known as weight (w), is the force due to gravity which is directed downwards. b. Electric force is the force due to electric field. c. Magnetic force is the attraction between objects and Figure 8: Weight magnet. Principle of Superposition of Forces states that any number of forces applied at a point on a body have the same effect as a single force equal to the vector sum of the forces. ⃗⃗⃗ = ⃗⃗⃗⃗⃗ 𝑹 𝑭𝟏 + ⃗⃗⃗⃗⃗ 𝑭𝟐 + ⃗⃗⃗⃗⃗ ⃗⃗ ; 𝑭𝟑 + ⋯ = ∑ 𝑭 𝑹𝒙 = ∑ 𝑭𝒙 ; 𝑹𝒚 = ∑ 𝑭𝒚 Any force can be replaced by its component vectors, acting at the same point. Aristotle believed that the natural state of a body was at rest, and a force was believed necessary for the object to continue its motion. Furthermore, he believed that the greater the force on the body, the greater its speed. Galileo concludes that, in an idealized world with no friction, if no force is applied to a moving object, it will continue to move with constant speed in a straight line. Figure 10: (a) The forces acting on the student are due to the chair, the table, the floor, and Earth’s gravitational attraction. (b) In solving a problem involving the student, we may want to consider only the forces acting along the line running through his torso. A free-body diagram for this situation is shown. Equation 3: Net External Force Equation 4: Component Forces E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG40  LET’S TRY! Calculating Net Force Suppose that the forces acting on the hockey puck in Figure 4-2 are A = 9.0 N at 0°, and B = 14.0 N at 45°. What is the acceleration of the puck, given that its mass is 0.1 kg? 1. Convert force A into vector component notation. Use the equation Ax = A cos θ to find the x coordinate of the force: 9.0 cos 0° = 9.0. 2. Use the equation Ay = A sin θ to find the y coordinate of the force: 9.0 sin 0°, or 0.0. That makes the vector A (9.0, 0.0) in coordinate form. 3. Convert the vector B into components. Use the equation Bx = B cos θ to find the x coordinate of the acceleration: 14.0 cos 45° = 9.9. 4. Use the equation By = B sin θ to find the y coordinate of the second force: 14.0 sin 45°, or 9.9. That makes the force B (9.9, 9.9) in coordinate form. 5. Perform the vector addition to find the net force: (9.0, 0.0) + (9.9, 9.9) = (18.9, 9.9). 6. Convert the vector (18.9, 9.9) into magnitude/angle form. Use the equation θ = tan –1(y/x) to find the angle of the net force: tan–1(.52) = 27°. 7. Apply the equation 𝑣 = √(𝑥 2 + 𝑦 2 ) + to find the magnitude of the net force, giving you 21.3 N. 8. Convert 21.3 N into acceleration: a = F/m = 21.3/0.1 = 213 m/sec 2. The correct answer is magnitude 213 m/sec2, angle 27°. WORK ON THIS! Compute for the following unknowns, supplement you answer/s with necessary solutions. 1. Using the particle model, represent the force a person exerts on a table when (a) pulling it to the right across a level floor with a force of magnitude F, (b) pulling it to the left across a level floor with force 2F, and (c) pushing it to the right across a level floor with force F. a. Table pulled right with force F b. Table pulled left with force 2F c. Table pushed right with force F 2. Two or more forces are shown on the objects below. Draw and label the net force. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG41 3. Suppose that you have two forces as shown in Figure 4-1: A = 5.0 N at 40°, and B = 7.0 N at 125°. What is the net force, ΣF? | 4. Add two forces: A is 8.0 N at 53°, and B is 9.0 N at 19°. 5. Add two forces: A is 16.0 N at 39°, and B is 5.0 N at 125°. 6. Assume that the two forces acting on a 0.10 kg hockey puck are as follows: A is 16.0 N at 53°, and B is 21.0 N at 19°. What is the acceleration of the hockey puck? 7. Two forces act on a 1000 kg car. A is 220 N at 64°, and B is 90 N at 80°. Neglecting friction, what is the car’s acceleration? Drawing Free-Body Diagrams The first step in describing and analyzing most phenomena in physics involves the careful drawing of a free-body diagram. Remember that a free-body diagram must only include the external forces acting on the body of interest. Once we have drawn an accurate free-body diagram, we can apply Newton’s first law if the body is in equilibrium E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG42 (balanced forces; that is, Fnet = 0 ) or Newton’s second law if the body is accelerating (unbalanced force; that is, Fnet ≠ 0 ). Problem-Solving Strategy: Constructing Free-Body Diagrams Observe the following rules when constructing a free- body diagram: 1. Draw the object under consideration; it does not have to be artistic. At first, you may want to draw a circle around the object of interest to be sure you focus on labeling the forces acting on the object. If you are treating the object as a particle (no size or shape and no rotation), represent the object as a point. We often place this point at the origin of an xy- coordinate system. 2. Include all forces that act on the object, representing these forces as vectors. Consider the types of forces described in Common Forces—normal force, friction, tension, and spring force—as well as weight and applied force. Do not include the net force on the object. With the exception of gravity, all of the forces we have discussed require direct contact with the object. However, forces that the object exerts on its environment must not be included. We never include both forces of an action-reaction pair. 3. Convert the free-body diagram into a more detailed diagram showing the x- and y-components of a given force (this is often helpful when solving a problem using Newton’s first or second law). In this case, place a squiggly line through the original vector to show that it is no longer in play—it has been replaced by its x- and y-components. 4. If there are two or more objects, or bodies, in the problem, draw a separate free-body diagram for each object. Note: If there is acceleration, we do not directly include it in the free- body diagram; however, it may help to indicate acceleration outside the free-body diagram. You can label it in a different color to indicate that it is separate from the free-body diagram. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG43  LET’S TRY! Two Blocks on an Inclined Plane Construct the free-body diagram for object A and object B in Figure. (a) The free-body diagram for isolated object A. (b) The free-body diagram for isolated object B. Comparing the two drawings, we see that friction acts in the opposite direction in the two figures. Because object A experiences a force that tends to pull it to the right, friction must act to the left. Because object B experiences a component of its weight that pulls it to the left, down the incline, the friction force must oppose it and act up the ramp. Friction always acts opposite the intended direction of motion. WORK ON THIS! Draw a free-body diagram for the object. Be sure to think carefully about the direction of Fnet. Note: Draw individual force vectors with a black or blue pencil or pen. Draw the net force vector with a red pencil or pen. 1. A heavy crate is being lowered 2. A boy is pushing a box across the floor at straight down at a constant speed by a steadily increasing speed. Let the box a steel cable. be the object for analysis. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG44  3. A bicycle is speeding up down a hill. 4. You have slammed on your car brakes Friction is negligible, but air resistance while going down a hill. The car is skidding is not. to a halt. 5. A block rests on the table, as shown. A light rope is attached to it and runs over a pulley. The other end of the rope is attached to a second block. The two blocks are said to be coupled. Block m2 exerts a force due to its weight, which causes the system (two blocks and a string) to accelerate. Newton's Law of Motion compiled by Sir Isaac Newton in his work Philosophiae Naturalis Principia Mathematica published on July 5,1687 Newton's First Law of Motion (Law of Inertia) → states that if the net force of on an object is zero, an object at rest will remain at rest, and an object in motion remains in motion in a straight line with constant velocity, unless acted upon by an external force. Inertia is the tendency of the body to resist a change in state of motion. Mass is a measure of the inertia of a body. (the greater the mass, the greater the inertia) Weight is the force of gravity on a body: 𝐹𝑔 = 𝑚𝑔. Inertial Reference Frames are reference frames in which the Newton’s First Law holds. Non-inertial Reference Frames are reference frames in which the Newton’s First Law does not hold. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG45  Newton's Second Law of Motion (Law of Acceleration) → that the acceleration of a body is directly proportional to the net force acting on it, and inversely proportional to its mass: ∑ 𝐹⃗ = 𝑚𝑎⃗. ∑ 𝐹𝑥 = 𝑚𝑎𝑥 , ∑ 𝐹𝑦 = 𝑚𝑎𝑦 (Force is an action capable of giving rise to acceleration) Different forces exerted on the same mass produce different accelerations. (a) Two students push a stalled car. All external forces acting on the car are shown. (b) The forces acting on the car are transferred to a coordinate plane (free-body diagram) for simpler analysis. (c) The tow truck can produce greater external force on the same mass, and thus greater acceleration. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG46  Sorting Out Weight and Mass One of the forces you are asked to deal with frequently in physics is the force on an object due to gravity. In a gravitational field, all objects are accelerated due to gravity; on the surface of the Earth, the acceleration due to gravity is 9.8 m/sec2, which is about 32 ft/sec2. Because the equation F = ma holds for all forces where the object being forced is free to accelerate, you can calculate the force an object feels due to gravity. Here are a few things to keep in mind when dealing with force and gravity: ✓ The force on an object is proportional to the object’s mass. For example, twice the mass means twice the force. ✓ Because a = F/m, twice the force still means the same acceleration if you have twice the mass. It’s the acceleration due to gravity that is constant in the Earth’s gravitational field at the surface of the Earth, not the force. ✓ The acceleration due to gravity points downward, toward the center of the Earth. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG47  The acceleration due to gravity at the surface of the Earth carries the symbol g. That means F = ma becomes F = mg; as a vector equation, that’s F = mg. In practical terms, unless you’re dealing with points so far apart that the curvature of the Earth matters, g is just considered downward. LET’S TRY! What Rocket Thrust Accelerates This Sled? Before manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust T, for the four-rocket propulsion system shown in Figure. The sled’s initial acceleration is 49 m/s2, the mass of the system is 2100 kg, and the force of friction opposing the motion is 650 N. Strategy. Although forces are acting both vertically and horizontally, we assume the vertical forces cancel because there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in Figure 5.14. Solution. Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. We have defined the direction of the force and acceleration as acting “to the right,” so we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with Fnet = ma where Fnet is the net force along the horizontal direction. We can see from the figure that the engine thrusts add, whereas friction opposes the thrust. In equation form, the net external force is Fnet = 4T – f Substituting this into Newton’s second law gives us Fnet = ma = 4T − f Using a little algebra, we solve for the total thrust 4T: 4T = ma + f Substituting known values yields 4T = ma + f = (2100 kg) (49 m/s2) + 650 N Therefore, the total thrust is 4T = 1.0 × 105 N and the individual thrusts are T = (1.0 × 105 N)/4 = 2.5 × 104 N E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG48  LET’S TRY! Several Forces on a Particle. A particle of mass m = 4.0 kg is acted upon by four forces of magnitudes. F1 = 10.0 N, F2 = 40.0 N, F3 = 5.0 N, and F4 = 2.0 N, with the directions as shown in the free- body diagram in Figure. What is the acceleration of the particle? We draw a free-body diagram as shown in Figure. Now we apply Newton’s second law. We consider all vectors resolved into x- and y-components: Thus, the net acceleration is which is a vector of magnitude 8.4 m/s2 directed at 276° to the positive x-axis. WORK ON THIS! 1. Forces are shown on three objects. For each: a. Draw and label the net force vector. Do this right on the figure. b. Below the figure, draw and label the object’s acceleration vector. 2. In the figures below, one force is missing. Use the given direction of acceleration to determine the missing force and draw it on the object. Do all work directly on the figure. 3. A constant force applied to an object causes the object to accelerate at 10 m/s2. What will the acceleration of this object be if? E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG49  a. The force is doubled? _______________________________________________ b. The mass is doubled? _______________________________________________ c. The force is doubled, and the mass is doubled? _________________________ d. The force is doubled, and the mass is halved? ____________________________ 4. A worker applies a constant horizontal force with magnitude 20 N to a box with mass 40 kg resting on a level floor with negligible friction. What is the acceleration of the box? Newton's Third Law of Motion (Law of Action-Reaction) → states that whenever one body exerts a force on a second body, the second body always exerts a force on the first body with equal magnitude but opposite in direction: 𝐹12 = −𝐹21. In the statement of Newton’s third law, “action” and “reaction” are the two opposite forces; we sometimes refer to them as an action–reaction pair. This is not meant to imply any cause-and-effect relationship; we can consider either force as the “action” and the other as the “reaction.” We often say simply that the forces are “equal and opposite,” meaning that they have equal magnitudes and opposite directions. The simple act of walking depends crucially on Newton’s third law. To start moving forward, you push backward on the ground with your foot. As a reaction, the ground pushes forward on your foot (and hence on your body as a whole) with a force of the same magnitude. This external force provided by the ground is what accelerates your body forward. LET’S TRY! Applying Newton’s third law: Objects at rest An apple sits at rest on a table, in equilibrium. What forces act on the apple? What is the reaction force to each of the forces acting on the apple? What are the action– reaction pairs? E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG50 Getting Up to Speed: Choosing the Correct System A physics professor pushes a cart of demonstration equipment to a lecture hall. Her mass is 65.0 kg, the cart’s mass is 12.0 kg, and the equipment’s mass is 7.0 kg. Calculate the acceleration produced when the professor exerts a backward force of 150 N on the floor. All forces opposing the motion, such as friction on the cart’s wheels and air resistance, total 24.0 N. Solution: Newton’s second law is given by The net external force on System 1 is deduced from Figure and the preceding discussion to be The mass of System 1 is These values of Fnet and m produce an acceleration of WORK ON THIS! Show the interacting objects, with a small gap separating them. Draw the force vectors of all action/reaction pairs. Label the force vectors, using a notation like and. 1. A bat hits a ball. Draw your picture from 2. A boy pulls a wagon by its handle. Rolling the perspective of someone seeing the friction is not negligible. The objects are end of the bat at the moment it strikes the boy, the wagon, and the ground. the ball. The objects are the bat and the ball. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG51  3. A crate is in the back of a truck as the 4. You find yourself in the middle of a frozen truck accelerates forward. The crate lake with a surface so slippery that you does not slip. The objects are the truck, cannot walk. However, you happen to the crate, and the ground. have several rocks in your pocket. The ice is extremely hard. It cannot be chipped, and the rocks slip on it just as much as your feet do. Can you think of a way to get to shore? Use pictures, forces, and Newton’s laws to explain your reasoning. 5. Boxes A and B are in contact on a horizontal, frictionless surface, as shown in Figure. Box A has mass 20.0 kg and box B has mass 5.0 kg. A horizontal force of 100 N is exerted on box A. What is the magnitude of the force that box A exerts on box B? REFERENCES: Science TG pages 1-26; MELC General Physics 1 page 537, PIVOT BOW R4QUBE pages 134-135; K to 12 Science Curriculum Guide University Physics with Modern Physics, Author: Young and Freedom, 3rd edition pp.35- 59 Physics (Principle with Application), Author: Douglas Giancoli pp. 50-74 Conceptual Physics 12th edition, Global edition, Author: Paul G. Hewitt pp.55-77 Student Workbook for College Physics: A Strategic Approach Volume 1. Pearson New International Edition. Knight et.al. pp 1-8 http://www.mrmont.com/games/carequationchallenge.html https://ophysics.com/k7.html I can calculate the motion of heavenly bodies but not the madness of people. -Sir Isaac Newton E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG52 E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG53  Being Energetic: Work Lesson Competencies: 1. Calculate the dot or scalar product of vectors (STEM_GP12WE -If – 40) 2. Determine the work done by a force acting on a system (STEM_GP12WE -If – 41) 3. Define work as a scalar or dot product of force and displacement (STEM_GP12WE -If – 42) 4. Interpret the work done by a force in one - dimension as an area under a Force vs. Position curve (STEM_GP12WE -If – 43) 5. Relate the gravitational potential energy of a system or object to the configuration of the system (STEM_GP12WE -Ig – 48) 6. Relate the elastic potential energy of a system or object to the configuration of the system (STEM_GP12WE -Ig – 49) 7. Explain the properties and the effects of conservative forces (STEM_GP12WE -Ig – 50) 8. Use potential energy diagrams to infer force; stable, unstable, and neutral equilibria; and turning points (STEM_GP12WE -Ig – 53) 9. Solve problems involving work, energy, and power in contexts such as, but not limited to, bungee jumping, design of roller -coasters, number of people required to build structures such as the Great Pyramids and the rice terraces; power and energy requirements of human activities such as sleeping vs. sitting vs. standing, running vs. walking. (STEM_GP12WE -Ih - i – 55) Working the Physics Way Work is done on an object when energy is transferred to the object. In other words, work is done when a force acts on something that undergoes a displacement from one position to another. Forces can vary as a function of position, and displacements can be along various paths between two points. We first define the increment of work Figure 11: These people are doing work as they push on dW done by a force F → acting through an infinitesimal the stalled car because they exert a force on the car as it moves. displacement d →r as the dot product of these two vectors: The SI unit of work is the joule (abbreviated J, pronounced “jool,” and named in honor of the 19th-century English physicist James Prescott Joule). The unit of work is the unit of force multiplied by the unit of distance. In SI units the unit of force is the newton and the unit of distance is the meter, so 1 joule is equivalent to 1 newton-meter (Nm). In the British system the unit of force is the pound (lb), the unit of distance is the foot (ft), and the unit of work is the foot-pound (ftlb) The following conversions are useful: E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG54  Work Done by a Force The work done by a force is the integral of the force with respect to displacement along the path of the displacement: Figure 12: Work done by a constant force. (a) A person pushes a lawn mower with a constant force. The component of the force parallel to the displacement is the work done, as shown in the equation in the figure. (b) A person holds a briefcase. No work is done because the displacement is zero. (c) The person in (b) walks horizontally while holding the briefcase. No work is done because cos θ is zero. Figure 13: A constant force can do positive, negative, or zero work depending on the angle between and the displacement s. Figure 14: This weightlifter’s hands do negative work on a barbell as the barbell does positive work on his hands. Because the weightlifter’s hands and the barbell have the same displacement, the work that his hands do on the barbell is just the negative of the work that the barbell does on his hands. In E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG55 general, when one body does negative work on a second body, the second body does an equal amount of positive work on the first body. LET’S TRY! A. Calculating the Work You Do to Push a Lawn Mower How much work is done on the lawn mower by the person in Figure 2(a) if he exerts a constant force of 75.0 N at an angle 35° below the horizontal and pushes the mower 25.0 m on level ground? The equation for the work is Substituting the known values gives B. Work done by a constant force (a) Steve exerts a steady force of magnitude 210 N (about 47 lb) on the stalled car in Figure below as he pushes it a distance of 18 m. The car also has a flat tire, so to make the car track straight Steve must push at an angle of 30° to the direction of motion. How much work does Steve do? (b) In a helpful mood, Steve pushes a second stalled car with a steady force F=(160N)i – (40N)j. The displacement of the car is s=(14m)i + (11m)j. How much work does Steve do in this case? Solution: WORK ON THIS! Compute for the following unknowns, supplement you answer/s with necessary solutions. Identify all forces acting on the object. Determine if the work done by each of these forces is positive (+), negative (−), or zero (0). Make a little table beside the figure showing every force and the sign of its work. 1. An elevator moves upward. 2. An elevator moves downward. E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG56 3. An object experiences a force while undergoing the displacement shown. Is the work done positive (1), negative (2), or zero (0)? 4. Moving a Couch. You decide to move your couch to a new position on your horizontal living room floor. The normal force on the couch is 1 kN and the coefficient of friction is 0.6. (a) You first push the couch 3 m parallel to a wall and then 1 m perpendicular to the wall (A to B in Figure). How much work is done by the frictional force? (b) You don’t like the new position, so you move the couch straight back to its original position (B to A in Figure). What was the total work done against friction moving the couch away from its original position and back again? 5. Work done by several forces. A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance of 20 m along level ground (Fig. 6.7a). The total weight of sled and load is 14,700 N. The tractor exerts a constant 5000-N force at an angle of 36.9˚ above the horizontal. A 3500-N friction force opposes the sled’s motion. Find the work done by each force acting on the sled and the total work done by all the forces. Getting Energetic: Kinetic Energy The kinetic energy of a particle is one-half the product of the particle’s mass m and the square of its speed v: We then extend this definition to any system of particles by adding up the kinetic energies of all the constituent particles: E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG57  The work done by the net force on a particle equals the change in the particle’s kinetic energy: Unit for Kinetic Energy: (SI UNIT) (British System) Figure 15: The relationship between the total work done on a body and how the body’s speed changes LET’S TRY! Getting Kinetic Energy from Work. In Figure, a 1000.0 kg safe full of gold bars is sliding down a 3.0 m ramp that meets the horizontal at an angle of 23°. The kinetic coefficient of friction is 0.15. What will the refrigerator’s speed be when it reaches the bottom of the ramp? E-MODULE FOR GENERAL PHYSICS 1 | FIRST QUARTER|PG58