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Chapter (1) Introduction to basic concepts of physics 1  Content of this chapter Types of motion (translational motion and periodic motion). Displacement, Velocity, Acceleration, momentum, Forces Newton's laws of motion...

Chapter (1) Introduction to basic concepts of physics 1  Content of this chapter Types of motion (translational motion and periodic motion). Displacement, Velocity, Acceleration, momentum, Forces Newton's laws of motion Work and energy. 2  1.1 What is Physics Physics is the study of the natural phenomena and expressed these phenomena as a mathematical equations which is called Laws of nature. The laws of physics can be used to predict the results of future experiments. Physics is based on experimental observations and quantitative measurements. Physicist is a scientist who studies physics, through observation of natural phenomena and try to find Physics is the science of patterns and principles that relate these phenomena. measurements  1.1.2 SI Units System & Standards The measurement of any quantity is made relative to a particular standard or unit. All measured physical quantities have units. Units are VITAL in physics!! The SI system of units: SI = “ Systéme International ” (French) More commonly called the “MKS system” (meter-kilogram-second) or more simply, “the metric system” 4  1.1.3 SI or MKS System Defined in terms of standards (a standard  one unit of a physical quantity) for length, mass, time, …. Length : The standard unit of length is Meter (m) (kilometer = km = 1000 m) Time : The standard unit of time is the second (s). Mass : The standard unit of mass is the Kilogram (kg) (kilogram = kg = 1000 g) 5  1.1.4 Base Quantities versus Derived Quantities Physical quantities can be divided into: Base quantities It is the seven quantities in the SI system and must be define in terms of standard 6  1.7 SI Derived Quantities and Units All physical quantities are defined in terms of the base quantities Both the quantity and its unit are derived from a combination of base units, using a defining equation. Example: Derived units for speed, acceleration and force: Distance (m)  velocity (m/s) Speed (m/s)  Acceleration (m/s )  2 Time (s) Time (s) Force (Newton, N)  Mass (kg)  Acceleration (m/s 2 ) 1N  1Kg.m/s 2 Density 𝑘𝑔Τ𝑚3 = A ratio of mass (kg) to volume (𝒎𝟑 ) 7  1.1.5 Unit Prefixes  Prefixes correspond to powers of 10.  Each prefix has a specific name.  Each prefix has a specific abbreviation.  The prefixes can be used with any basic units.  They are multipliers of the basic unit. 8  1.1.6 Other Systems of Units A second metric system is the Centimeter-Gram-Second, abbreviated is (CGS) system. In which the centimeter, gram, and second are the standard units of length, mass, and time. Centimeter = 0.01 meter, Gram = 0.001 kilogram The British engineering system takes as its standards the foot for length, the pound for mass, and the second for time. British (foot-pound-second) system. Still used in some countries like USA. 9  1.1.7 Converting units  When units are not consistent, you may need to convert to appropriate ones.  Units can be treated like algebraic quantities that can cancel each other out.  Chain-link conversion 1day=24 h 1h=60 min 1min=60 s Conversion factor:1min/60s =1, 60s/1min =1 10  Converting feet to meters: 1 m = 3.281 ft (this is a conversion factor) Or: 1 = 1 m / 3.281 ft 316 ft × (1 m / 3.281 ft) = 96.3 m Note that the units cancel properly – this is the key to using the conversion factor correctly. 11 Example: Example (1) Express the speed 50 m/s in (a) km/h and (b) miles / hour (mph). Solution: 12  1.1.8 Vectors and Scalars A scalar quantity can be described by a single number. A vector quantity has both a magnitude and a direction in space. In this book, a vector quantity is represented in boldface italic type with an arrow over it: A. The magnitude of A is written as A or A. 13  Example For Vectors &Scalar Quantities Speed 10 m/s Velocity 10 m/s N Time Distance Displacement 2 sec Scalar 6m Force 2 N.E Vector 6mE Temperature 9 0C Acceleration 4 m/s2 S 14  1. 2 Concept of Motion A body is said to be in rest if its position does not vary with respect to a given referral point as time passes. A body is said to be in motion if there is a continues change in its position with respect to a given referral as time passes. To describe the position (location) of an object we need to specify a reference point called the origin. The motion of a body is observed by attaching a frame of reference to an observer and measuring the change in position of the body relative to that frame. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, speed, and time.  1. 2.1 Types of motion There are many types of motion such as: Periodic or Oscillatory Translatory motion  1. 2.2 Translatory Motion is the motion by which a body shifts from one point in space to another (e.g., the motion of a bullet fired from a gun). motion in which all points of a moving body move uniformly in the same line or direction. Examples: Dropping of an object from height, rays of light, a stone falling, pulling a drawer are some examples of translatory motion.  1. 2.3 Periodic motion Periodic motion is performed as that motion repeated in equal intervals of time. Examples: Rocking chair. Bouncing ball. Vibrating tuning fork. Swing in motion. Earth in its orbit around the Sun. Water wave. Oscillation of simple pendulum  1. 2.4 Concepts of motion Frequency, period, displacement, amplitude, and Phase angle Frequency ( f ): The number of complete cycles (or oscillations, or vibrations) the object completes per unit time (per second), and is given by A frequency of one cycle per second is called one Hertz (Hz), which is the unit of frequency. Period ( T ): The time that it takes for an oscillator to execute one complete cycle of its motion. If it starts at t = 0 at x = A , then it gets back to x = A after one full period at t = T. Displacement ( x ) is the distance of mass from its equilibrium position at any instance of time, measured in meters Amplitude (A ): The maximum distance (maximum value of displacement x.) that an object moves from its equilibrium position.  1.3 Distance and Displacement Distance is the total path length covered by an object from the initial position to the final position. Displacement is the shortest distance between the initial and the final position. Or it is the change in position relative to a reference frame. it expressed as 𝚫𝒙 = 𝒙𝒇 − 𝒙𝟎  Example (2) You drive 20 km east, then turn around and drive 15 km west. What is your displacement? Solution 𝚫𝒙 = 𝒙 − 𝒙𝟎 𝚫𝒙 = 𝟓 𝒌𝒎 − 𝟎 𝒌𝒎 𝟓 𝒌𝒎 5 km east of your starting point What is your distance traveled? 𝟐𝟎 𝒌𝒎 + 𝟏𝟓 𝒌𝒎 𝟑𝟓 𝒌𝒎  1.4 Velocity The slope of a position vs time is the velocity, so velocity is rate of change of position 𝒙−𝒙𝟎 ഥ= 𝒗 and 𝒙 = 𝒗𝒕 + 𝒙𝟎 𝒕−𝒕𝟎 As in the figure if the graph is not a straight line, then use the slope of a tangent line drawn to that point. 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 Velocity is a vector (has direction) 𝒗 = 𝒕𝒊𝒎𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 Speed is a scalar (no direction) 𝒗 = 𝒕𝒊𝒎𝒆 Units of both are m/s  Example (3) The figure at right shows the height of a ball thrown straight up vs time. Find the velocity of the ball at 2 seconds. Soulution 𝒙−𝒙𝟎 𝒗= 𝒕−𝒕𝟎 𝟒.𝟐 𝒄𝒎−𝟏.𝟕 𝒄𝒎 𝒗= 𝟑 𝒔−𝟏 𝒔 𝟐.𝟓 𝒄𝒎 𝒄𝒎 𝒗= = 𝟏. 𝟑 𝟐𝒔 𝒔  Example (4) The spine-tailed swift is the fastest bird in powered flight. On one flight a particular bird flies 306 m east, then turns around and flies 406.5 m back west. This flight takes 15 s. What is the bird’s average velocity? 𝜟𝒙 𝟑𝟎𝟔 𝒎−𝟒𝟎𝟔.𝟓 𝒎 𝒗= = = −𝟔. 𝟕 𝒎/𝒔 𝜟𝒕 𝟏𝟓 𝒔 6.7 m/s west Average speed? 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒗= 𝒕𝒊𝒎𝒆 𝟑𝟎𝟔 𝒎+𝟒𝟎𝟔.𝟓 𝒎 𝒎 𝒗= = 𝟒𝟕. 𝟓 𝟏𝟓 𝒔 𝒔 Which of these would we use to say how fast the bird is? Average speed  1.5 Acceleration Acceleration is defined as the rate of change of velocity 𝜟𝒗 𝒗𝒇 − 𝒗𝟎 𝒂= = 𝜟𝒕 𝒕𝒇 − 𝒕𝟎 𝒗 = 𝒂𝒕 + 𝒗𝟎 The unit of acceleration is 𝒎/𝒔𝟐 If the acceleration is same direction as motion, then the object is increasing speed. If the acceleration is opposite direction as motion, then the object is decreasing speed.  Example (5) A dropped object near the earth will accelerate downward at 9.8 m/s2. (Use -9.8 m/s2.) If the initial velocity is 1 m/s downward, what will be it’s velocity at the end of 3 s? Is it speeding up or slowing down? solution 𝒗𝒇 −𝒗𝟎 𝒂= 𝒕𝒇 −𝒕𝟎 𝒎 𝒎 𝒗𝒇 − −𝟏 𝒔 −𝟗. 𝟖 = 𝒔𝟐 𝟑𝒔 𝒎 𝒎 −𝟐𝟗. 𝟒 = 𝒗𝒇 + 𝟏 𝒔 𝒔 𝒎 −𝟑𝟎. 𝟒 = 𝒗𝒇 𝒔 30.4 m/s downward (minus signed)  1.6 Motion with Constant Acceleration Assume 𝒕𝟎 = 𝟎, so 𝚫𝒕 = 𝒕 and acceleration is constant 𝒙−𝒙𝟎 𝒗= 𝒕 𝒗𝟎 +𝒗 𝒙 = 𝒗𝒕 + 𝒙𝟎 and 𝒗= 𝟐 𝟏 𝒙= 𝒗𝟎 + 𝒗 𝒕 + 𝒙𝟎 𝟐 𝟏 𝒙= 𝒂𝒕𝟐 +𝒗𝟎 𝒕 + 𝒙𝟎 𝟐 𝒗−𝒗𝟎 𝒂= 𝒕 𝒗 − 𝒗𝟎 = 𝒂𝒕 𝒗 = 𝒂𝒕 + 𝒗𝟎  Example (6) A plane starting from rest accelerates to 𝟒𝟎 𝒎/𝒔 in 𝟏𝟎 𝒔. How far did the plane travel during this time? Solution 𝒗 = 𝟒𝟎 𝒎/𝒔, 𝒕 = 𝟏𝟎 𝒔, 𝒗𝟎 = 𝟎, 𝒙𝟎 = 𝟎, 𝒙 = ? 𝒎 𝒗𝟎 +𝒗 𝟎+𝟒𝟎 𝒔 𝒗= 𝒗= = 𝟐𝟎 𝒎/𝒔 𝟐 𝟐 𝒙 = 𝒗𝒕 + 𝒙𝟎 𝒎 𝒙 = 𝟐𝟎 𝟏𝟎 𝒔 + 𝟎 𝒔 𝒙 = 𝟐𝟎𝟎 𝒎  Example (7) To avoid an accident, a car decelerates at 𝟎. 𝟓𝟎 𝒎/𝒔𝟐 for 𝟑. 𝟎 𝒔 and covers 𝟏𝟓 𝒎 of road. What was the car’s initial velocity? Solution 𝒂 = −𝟎. 𝟓 𝒎/𝒔𝟐 , 𝒕 = 𝟑 𝒔, 𝒙 = 𝟏𝟓 𝒎, 𝒙𝟎 = 𝟎, 𝒗𝟎 = ? 𝟏 𝒙 = 𝒂𝒕𝟐 +𝒗𝟎 𝒕 + 𝒙𝟎 𝟐 𝟏 𝒎 𝟐 𝟏𝟓 𝒎 = −𝟎. 𝟓 𝟐 𝟑 𝒔 + 𝒗𝟎 𝟑 𝒔 + 𝟎 𝟐 𝒔 𝟏𝟓 𝒎 = −𝟐. 𝟐𝟓 𝒎 + 𝒗𝟎 𝟑 𝒔 𝟏𝟕. 𝟐𝟓 𝒎 = 𝒗𝟎 𝟑 𝒔 𝒗𝟎 = 𝟓. 𝟕𝟓 𝒎/𝒔  Example (8) A cheetah is walking at 1.0 m/s when it sees a zebra 25 m away. What acceleration would be required to reach 20.0 m/s in that distance? Solution 𝒎 𝒎 𝒗 = 𝟐𝟎. 𝟎 , 𝒗𝟎 = 𝟏. 𝟎 ,𝒙 = 𝟐𝟓 𝒎, 𝒙𝟎 = 𝟎, 𝒂 = ? 𝒔 𝒔 𝒗𝟐 = 𝒗𝟐𝟎 + 𝟐𝒂 𝒙 − 𝒙𝟎 𝒎 𝟐 𝒎 𝟐 𝟐𝟎 = 𝟏. 𝟎 + 𝟐𝒂 𝟐𝟓 𝒎 − 𝟎 𝒔 𝒔 𝒎𝟐 𝒎𝟐 𝒎𝟐 𝟒𝟎𝟎 𝟐 = 𝟏 𝟐 + 𝟓𝟎 𝒎 𝒂 » 𝟑𝟗𝟗 𝟐 = 𝟓𝟎 𝒎 𝒂 » 𝒂 = 𝟕. 𝟗𝟖 𝒎/𝒔𝟐 𝒔 𝒔 𝒔  Example (9) The left ventricle of the heart accelerates blood from rest to a velocity of +26 cm/s. (a) If the displacement of the blood during the acceleration is +2.0 cm, determine its acceleration (in cm/s2). (b) (b) How much time does blood take to reach its final velocity? Solution 𝒄𝒎 𝒄𝒎 𝒗𝟎 = 𝟎 , 𝒗 = 𝟐𝟔 , 𝜟𝒙 = 𝟐 𝒄𝒎, 𝒂 =? 𝒔 𝒔 𝒗𝟐 = 𝒗𝟐𝟎 + 𝟐𝒂 𝒙 − 𝒙𝟎 𝒄𝒎 𝒂 = 𝟏𝟔𝟗 𝟐 𝒔 𝒄𝒎 𝒄𝒎 𝒗𝟎 = 𝟎 , 𝒗 = 𝟐𝟔 , 𝜟𝒙 = 𝟐 𝒄𝒎, 𝒕 =? 𝒔 𝒔 𝒗 +𝒗 𝒙 = 𝒗𝒕 + 𝒙𝟎 ; 𝒗 = 𝟎 » 𝒕 = 𝟎. 𝟏𝟓 𝒔 𝟐  1.7 Momentum Momentum is a vector quantity: it has both magnitude and direction. Since momentum has a direction, it can be used to predict the resulting direction and speed of motion of objects after they collide. Momentum has direction: the exact same direction as the velocity. But many examples here only use speed (velocity without direction) to keep it simple. 𝒑 = 𝒎𝒗 𝒑 = 𝑭. 𝒅𝒕. The unit for momentum is: kg m/s (kilogram meter per second), or N s (Newton second) They are the same! 1 kg m/s = 1 N s, We will use both here. Momentum is defined as the mass (m) times the velocity (v). If an object is steady so its velocity is zero resulting in zero momentum. 32  Examples (10) Example (10.1 ) What is the momentum of a 1500 kg car going at highway speed of 28 m/s (about 100 km/h or 60 mph)? Solution p=mv p = 1500 kg × 28 m/s » p = 42,000 kg m/s Example (10.2 ) Determine the momentum of a 110-kilogramme football player moving at 8 metres per se Solution p=mv » p = 110 kg × 8 m/s » p = 880 kg m/s. 33  1.8 Force Force can be defined as a push or a pull. (Technically, force is something that can accelerate objects.), and can be measured by unit called Newton (N), which is the force that causes an object with a mass of 1 kg to accelerate at 1 m/s. Force can also be measured by a spring scale as an example. 𝑭 = 𝒎𝒂 𝒅 𝑭= 𝒑. 𝒅𝒕 Where (m) is the mass of the object, a is the acceleration and p is the momentum. Examples: When you throw a baseball, you apply a force to the ball. Using force and other concepts one can study Newton’s laws of motion. 34  1.9 Newton’s Laws of Motion 1.9.1 (Newton’s First Laws of Motion) the law of Inertia) Newton's First Law of Motion, is also known as the law of Inertia. This law is stated as: "Every object continues in its state of rest or of uniform speed in a straight line unless it is compelled to change that state by a net force acting on the object." or “A body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force”. As an example of this law a book sitting on a table has First note what is not stated or a net force equal to zero because it is in a state of rest. implied by Newton’s first law. This doesn’t say that every moving Inertia can be known as the property of objects to let object has a force acting on it. the objects remain in constant motion or rest. One This doesn’t say that a stationary object has no forces acting on it. can consider mass as a measure of inertia.  1.9.2 (Newton’s Second Law of Motion) Newton's Second Law talks about an object that has net force, or it also gives an explanation for what causes an object to undergo an acceleration. It is the net force applied to the object. It states that "The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The direction o f the acceleration is in the direction of the applied net force". 𝑭𝒏𝒆𝒕 = 𝒎𝒂 Where 𝐹𝑛𝑒𝑡 is the net force which is the vector sum of all the forces. When the net force acting on an object is not zero, the object will accelerate at the direction of the exerted force.  Example (11) A car of mass 1000 kg accelerates from rest to 30 m/s in 4 seconds. (a) compute its acceleration (assumed constant) (b) compute the force exerted on the car by the road during this time. Solution (a) a = Δv/Δt = (30 m/s – 0) / 4 s = 7.5 m/s2 (b)F = ma = 1000 kg  7.5 m/s2 = 7500 kg m/s2 = 7500 N Example (11.1) An object moving at 40 m/s with no forces acting on it will (circle answer) (a) fall with an acceleration of 9.8 m/s2 (b) fall with an acceleration of -9.8 m/s2 (c) continue at this same velocity as long as there are no forces (d) slow down, reaching 20 m/s after 9.8 seconds Solution (c ) continue at this same velocity as long as there are no forces 37  Example (12) A football player named Ali is blocking a player on the other team named Bob. Al applies a 1500 N force on Bob. If Bob's mass is 100 kg, what is his acceleration? 1- What is the size of the force on Ali? 2- If Al's mass is 75 kg, what is his acceleration? Solution 𝑭 = 𝒎𝒂 𝟏𝟓𝟎𝟎 𝑵 = 𝟏𝟎𝟎 𝒌𝒈 𝒂 → 𝒂 = 𝟏𝟓 𝒎/𝒔𝟐 1500 N (Newton’s 3rd Law) 𝑭 = 𝒎𝒂 𝟏𝟓𝟎𝟎 𝑵 = 𝟕𝟓 𝒌𝒈 𝒂 → 𝒂 = 𝟐𝟎 𝒎/𝒔𝟐 38  Example (13) A 0.046 kg golf ball hit by a driver can accelerate from rest to 67 m/s in 1 ms while the driver is in contact with the ball. How much average force does the golf ball experience? Solution 𝒗 = 𝒂𝒕 + 𝒗𝟎 𝒎 𝟔𝟕 = 𝒂 𝟏 × 𝟏𝟎−𝟑 𝒔 + 𝟎 𝒔 𝒎 𝟔𝟕𝟎𝟎𝟎 𝟐 = 𝒂 𝒔 𝑭 = 𝒎𝒂 𝒎 𝑭 = 𝟎. 𝟎𝟒𝟔 𝒌𝒈 𝟔𝟕𝟎𝟎𝟎 𝟐 𝒔 𝑭 = 𝟑𝟎𝟖𝟐 𝑵 39  1.9.3 (Newton’s third Law of Motion) When you kick the wall in your room, you will probably end up hurting your foot. Newton's Third Law of Motion can explain why. It states that " Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first". So every force has an equal and opposite reaction force. You push down on your chair, so the chair pushed back up on you F12= - F21 In other words, when you kick the wall, the wall kicks you The minus singe is for the back with equal force. As a result you will get hurt. These opposite direction forces are called action-reaction forces. Therefore, the net force on the wall is not zero and the net force on your foot is not zero neither.  Example (14) A person of mass 85 kg is standing in a lift which is accelerating downwards at 0.45 m s-2. Draw a diagram to show the forces acting on the person and calculate the force the person exerts on the floor of the lift. Solution The weight of the person is 85 g, where g is the acceleration due to gravity. The resultant force is W - R and using Newton’s second law gives: F = ma W - R = 85 × 0.45 R = W - 85 × 0.45 = 85 × 9.81 - 85 × 0.45 = 795.6 N = 800 N (2 d.p.) Therefore, using Newton’s third law the force the person exerts on the floor of the lift is equal to the force of the floor acting on the person, i.e. R, which equals 800 N. 41  Example (15) A person of mass 94 kg is standing in a lift which is accelerating upwards at 0.54 m s-2. Draw a diagram to show the forces acting on the person and calculate the force, F, the person exerts on the floor of the lift. Solution 42  Example (16) A person of mass 65 kg is standing in a lift of mass 400 kg, which is accelerating upwards at 0.6 m s-2. Draw a diagram to show the forces acting on the lift and calculate their respective values. Solution 43  Example (17) A tug boat, of mass 8000 kg, is towing a smaller boat, of mass 2000 kg. Given that the tug boat has a horizontal driving force of D N and that they are accelerating at 1.2 m s-2, draw a diagram to show the forces acting on both boats and calculate the driving force. Solution 44  1.10 Work Work is depending on the force and the distance the force moves the object, the force is being in the direction of the distance. Work can be measured by using the following formula: 𝑊 = 𝐹റ ⋅ 𝑑റ Question 𝑊 = 𝐹𝑑 cos 𝜃 𝑘𝑔⋅𝑚2 The SI unit of work is (𝑁 ⋅ 𝑚 = 𝐽 ) which called (Joule), (𝐽 = 𝑠2 ). One must note that work is a Scalar quantity. Which of the following is NOT work? 1. Pushing a Stalled Car 2. Pulling a Wagon 3. Climbing stairs 4. Falling Down 5. Carrying a Heavy Backpack Down the Hall  Example (18) Marcy pulls a backpack on wheels down the 100-m hall. The 60-N force is applied at an angle of 30° above the horizontal. How much work is done by Marcy? Solution 𝑾 = 𝑭𝒅 𝐜𝐨𝐬 𝜽 𝑾 = 𝟔𝟎 𝑵 𝟏𝟎𝟎 𝒎 𝐜𝐨𝐬 𝟑𝟎° = 𝟓𝟏𝟗𝟔 𝑱 W = 5200 J 46  Example (19) Drew is carrying books (200 N) down the 100-m hall. How much work is Drew doing on the books? Solution 𝑾 = 𝑭𝒅 𝐜𝐨𝐬 𝜽 𝑾 = 𝟐𝟎𝟎 𝑵 𝟏𝟎𝟎 𝒎 𝐜𝐨𝐬 𝟗𝟎° = 𝟎 𝑱 W=0J The force is vertical and the displacement is horizontal 47  1.11 Kinetic Energy The energy of motion of a particle is called kinetic energy: Like work, the kinetic energy of a particle is a scalar quantity; it depends on only the particle’s mass and speed, not its direction of motion. Kinetic energy can never be negative, and it is zero only when the particle is at rest. The SI unit of kinetic energy is the joule. 48  Example (20) An object with 10 Kg mass moving with 4 m/s. Calculate its kinetic energy. Solution:  1.12 Energy and work Do work means  W = Fd F = ma So work by a net force gives an object some acceleration and acceleration means the velocity changes. 𝑭 = 𝒎𝒂 𝑭𝒅 = 𝒎𝒂𝒅 𝒗𝒇𝟐 = 𝒗𝟎𝟐 + 𝟐𝒂𝒅  solve for ad 𝟏 𝟐 𝒂𝒅 = 𝒗𝒇 − 𝒗𝟐𝟎 𝟐 𝟏 𝟐 𝑭𝒅 = 𝒎 𝒗𝒇 − 𝒗𝟐𝟎 𝟐 𝟏 𝟐 𝟏 𝑾 = 𝒎𝒗𝒇 − 𝒎𝒗𝟐𝟎 𝟐 𝟐 50  1.13 Energy work theorem Energy is the ability to do work Kinetic Energy is defined as that energy due to motion If something in motion hits an object, it will move it some distance. 𝟏 𝐊𝐄 = 𝐦𝒗𝟐 𝟐 KE is the kinetic energy which is scalar, its unit is the joule (J). Work of Net external force = change in kinetic energy 𝟏 𝟏 𝐖= 𝐦𝒗𝟐𝒇 − 𝐦𝒗𝟐𝟎 𝟐 𝟐 𝐖 = 𝐊𝑬𝒇 − 𝐊𝑬𝟎 51  Example (21) A 0.075-kg arrow is fired horizontally. The bowstring exerts a force on the arrow over a distance of 0.90 m. The arrow leaves the bow at 40 m/s. What average force does the bow apply to arrow? Solution 𝒎 𝒎 𝒎 = 𝟎. 𝟎𝟕𝟓 𝒌𝒈, 𝒔 = 𝟎. 𝟗𝟎 𝒎, 𝒗𝟎 = 𝟎 , 𝒗𝒇 = 𝟒𝟎 𝒔 𝒔 𝑾 = 𝑲𝑬𝒇 − 𝑲𝑬𝟎 𝟏 𝒎 𝟐 𝟏 𝒎 𝟐 𝑭 𝟎. 𝟗𝟎 𝒎 = 𝟎. 𝟎𝟕𝟓 𝒌𝒈 𝟒𝟎 − 𝟎. 𝟎𝟕𝟓 𝒌𝒈 𝟎 𝟐 𝒔 𝟐 𝒔 𝑭 𝟎. 𝟗𝟎 𝒎 = 𝟔𝟎 𝑱 𝑭 = 𝟔𝟔. 𝟕 𝑵 52  Example (22) How much net work is required to accelerate a 1600 kg car from 10 m/s to 50 m/s. Solution:  1.14 Conservation of Energy Energy can neither be created nor destroyed. It can only change from form to form. Mechanical Energy, PE and KE Conservation of Mechanical Energy PE1 + KE1 = PE2 + KE2  Example (23) A 1500-kg car is driven off a 50-m cliff during a movie stunt. If it was going 20 m/s as it went off the cliff, how fast is it going as it hits the ground? Solution 𝑷𝑬𝟎 + 𝑲𝑬𝟎 = 𝑷𝑬𝒇 + 𝑲𝑬𝒇 𝟏 𝟏 𝒎𝒈𝒉𝟎 + 𝒎𝒗𝟎 = 𝒎𝒈𝒉𝒇 + 𝒎𝒗𝟐𝒇 𝟐 𝟐 𝟐 𝟐 𝒎 𝟏 𝒎 𝟏 𝟏𝟓𝟎𝟎 𝒌𝒈 𝟗. 𝟖 𝟐 𝟓𝟎 𝒎 + 𝟏𝟓𝟎𝟎 𝒌𝒈 𝟐𝟎 = 𝟎 + 𝟏𝟓𝟎𝟎 𝒌𝒈 𝒗𝟐𝒇 𝒔 𝟐 𝒔 𝟐 𝒗𝒇 = 𝟑𝟕. 𝟏 𝒎/𝒔 55 1.14.1 Energy consistently changes forms  1.14.1 Energy consistently changes forms Am I above the ground? NO, h = 0, U = 0 J Am I moving? Yes, v = 8 m/s, m = 60 kg K  1 mv 2 1 (60)(8) 2 2 2 K  1920 J Position m v U K ME (= U+K) 1 60 kg 8 m/s 0J 1920 J 1920 J  1.14.1 Energy consistently changes forms Energy Before = Energy After KO =U+K 1920= (60)(9.8)(1) + (.5)(60)v2 1920= 588 + 30v2 1332 = 30v2 44.4 = v2 v = 6.66 m/s Position m v U K ME 1 60 kg 8 m/s 0J 1920 J 1920 J 2 60 kg 6.66 m/s 588 J 1332 J 1920 J  1.14.1 Energy consistently changes forms Am I moving at the top? No, v = 0 m/s EB = EA Using position 1 Ko = U 1920 = mgh 1920 =(60)(9.8)h h = 3.27 m Position m v U K ME 1 60 kg 8 m/s 0J 1920 J 1920 J 2 60 kg 6.66 m/s 588 J 1332 J 1920 J 3 60 kg 0 m/s 1920 J 0J 1920 J  1.15 POWER * A quantity that measures the rate at which work is done or energy is transformed * Power = work / time interval P = W/Δt  W = Fd P = Fd/Δt  v = d/Δt * Power = Force x speed P = Fv SI Unit for Power: Watt (W)  Defined as 1 joule per second (J/s) Horsepower = Another unit of power 1 hp = 746 watts  Example (24) A 193 kg curtain needs to be raised 7.5 m, in as close to 5 s as possible. The power ratings for three motors are listed as 1 kW, 3.5 kW, and 5.5 kW. What motor is best for the job? Solution m = 193 kg Δt = 5s d =7.5m P=? P = W/Δt  = Fd/Δt  = mgd/Δt = (193kg)(9.8m/s2)(7.5m)/5s = 280 W  2.8 kW ** Best motor to use = 3.5 kW motor. The 1 kW motor will not lift the curtain fast enough, and the 5.5 kW motor will lift the curtain too fast  Example (25) A rock climber wears a 7.5 kg backpack while scaling a cliff. After 30.0 min, the climber is 8.2 m above the starting point. b.How much power does the climber expend in this effort? a. How much work does the climber do on the backpack? Solution W = Fd = mgd = 7.5kg(9.8m/s2)(8.2m) P = W = 603 J = 0.34 watt t 1800 s  Example (26) What power is required to lift a v = 4 m/s 900-kg elevator at a constant speed of 4 m/s? Solution P = F v = mg v P = (900 kg)(9.8 m/s2)(4 m/s) P = 35.3 kW  Example (27) What work is done by a 4-hp mower in one hour? The conversion factor is needed: 1 hp = 746 watt. Solution 𝟒×𝟕𝟒𝟔 𝒘𝒂𝒕𝒕 4hp= =2984 watt 𝟏 𝒉𝒑 𝑾𝒐𝒓𝒌 P= » work = P ×t 𝒕 work = 2984 ×60×60 Work = 10.7 MJ  Example (28) What power is consumed in lifting a 70-kg man 1.6 m in 0.50 s? Solution Fh mgh P  t t (70 kg)(9.8 m/s 2 )(1.6 m) P 0.50 s Good luck 66