GENERAL-PHYSICS-1-Q1-Week-6_v2.pdf

WORK, ENERGY AND POWER for GENERAL PHYSICS 1/ Grade 12 Quarter 1/ Week 6 NegOr_Q1_GenPhysics1_SLKWeek6_v2 NegOr_Q1_GenPhysics1_SLKWeek6_v2 1  FOREWORD This Self Learning-Kit will serve a...

WORK, ENERGY AND POWER for GENERAL PHYSICS 1/ Grade 12 Quarter 1/ Week 6 NegOr_Q1_GenPhysics1_SLKWeek6_v2 NegOr_Q1_GenPhysics1_SLKWeek6_v2 1  FOREWORD This Self Learning-Kit will serve as your guide to explain the concepts of work as dot product as well as the effects of conservative forces; solve problems involving work, energy, and power in contexts and integrate the concepts learned in work, energy and power to real-life situations. This Self Learning Kit will provide a short and learner-friendly content that stirs your curiosity, develop understanding, and support critical thinking. The writer hopes that this Self Learning-Kit can serve its purpose to you as target learners. Mastery of the content is encouraged before proceeding to the next learning competency. NegOr_Q1_GenPhysics1_SLKWeek6_v2 2  OBJECTIVES At the end of this self-learning kit, the students should be able to: K: explain the concepts of work as dot product as well as the effects of conservative forces; infer force using potential energy diagrams; S: solve problems involving work, energy, and power in contexts; and A: integrate the concepts learned in work, energy and power to real-life situations. LEARNING COMPETENCIES Calculate the dot or scalar product of vectors. (STEM_GP12WE-If40) Determine the work done by a force acting on a system. (STEM_GP12WE-If41) Define work as a scalar or dot product of force and displacement. (STEM_GP12WE-If42) Interpret the work done by a force in one- dimension as an area under a Force vs. Position curve. (STEM_GP12WE-If43) Relate the gravitational potential energy of a system or object to the configuration of the system. (STEM_GP12WE-Ig48) Relate the elastic potential energy of a system or object to the configuration of the system. (STEM_GP12WE-Ig49) Explain the properties and the effects of conservative forces. (STEM_GP12WE-Ig50) Use potential energy diagrams to infer force; stable, unstable, and neutral equilibria; and turning points. (STEM_GP12WE-Ig53) Solve problems involving work, energy, and power in contexts such as, but not limited to, bungee jumping, design of rollercoasters, 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-Ihi- 55) NegOr_Q1_GenPhysics1_SLKWeek6_v2 3  I. WHAT HAPPENED PRE-TEST Directions: Identify the words described by the brief description. Use the jumbled words as your clue to the words being described. Write your answers on your notebook. 1. This is the result of the scalar multiplication of two vectors. 2. It is the product of the force magnitude in the direction of the displacement F and the displacement magnitude Δx due to a constant force. 3. It is a regularly interacting or interdependent group of items forming a unified whole. 4. It explains the relation between the sides of a right-angled triangle. It is basically used to find the length of an unknown side and angle of a triangle. 5. It is the capacity for doing work. 6. It is the rate of doing work. It is the amount of energy consumed per unit time. 7. It is the unit of work or energy in the International System of Units (SI). 8. It is the unit of power in the International System of Units (SI) which is equal to one joule of work performed per second (J/s). 9. It is just its weight 𝑚𝑔 near the surface of Earth, acting vertically down. 10. It is energy stored as a result of applying a force to deform an elastic object. PRE-ACTIVITY: A Box Being Pushed Consider a coordinate system such that we have x as the abscissa and y as the ordinate. Imagine that a box is being pushed along the x direction. What happens in the following scenarios? Write your answers in your Science notebook/Answer Sheet. The box is being pushed parallel to the x direction? The box is being pushed at an angle of 45 degrees to the x direction? The box is being pushed at an angle of 60 degrees to the x direction? The box is being pushed at an angle of 90 degrees to the x direction? NegOr_Q1_GenPhysics1_SLKWeek6_v2 4 II. WHAT I NEED TO KNOW DISCUSSION: A. Scalar Product (the Dot Product) Did you know that a vector can be multiplied by another vector but may not be divided by another vector? In Physics and Engineering, two kinds of products of vectors are widely used. One of these is the scalar multiplication of two vectors. The result of taking the product of two vectors is a number (a scalar), as its name indicates. In this lesson, we will use scalar products to define work and energy relations. Scalar multiplication of two vectors yields a scalar product. The scalar product 𝐀 ⃗ of two vectors 𝐀 ⃗⃗ ∙ 𝐁 ⃗ and 𝐁 ⃗ is a number defined by the equation, ⃗ ∙𝐁 𝐀 ⃗ = 𝐴𝐵 𝑐𝑜𝑠 𝝋 where 𝝋 is the angle between the vectors. The scalar product is also called the dot product because of the dot notation that indicates it. The direction of angle 𝝋 does not matter in the definition of the dot product, and 𝝋 can be measured from either of the two vectors to the other because 𝑐𝑜𝑠 𝝋 = 𝑐𝑜𝑠 (−𝝋) = 𝑐𝑜𝑠 (𝟐𝝅 − 𝝋). The dot product is a negative number when 90° < 𝝋 ≤ 𝟏𝟖𝟎° and is a positive number when 0° ≤ 𝝋 < 𝟗𝟎°. Moreover, the dot product of two parallel vectors is 𝐀 ⃗⃗ ∙ 𝐁⃗ = 𝐴𝐵 𝑐𝑜𝑠 0° = 𝑨𝑩. The scalar product of two orthogonal vectors vanishes: 𝐀⃗ ∙𝐁 ⃗⃗ = 𝐴𝐵 𝑐𝑜𝑠 𝟗𝟎° = 0. The scalar product of a vector with itself is the square of its magnitude: ⃗⃗ 𝟐 ≡ 𝐀 𝐀 ⃗⃗ ∙ 𝐀 ⃗⃗ = 𝐴𝐴 𝒄𝒐𝒔 𝟎° = 𝑨𝟐 Figure 1 NegOr_Q1_GenPhysics1_SLKWeek6_v2 5  The scalar product of two vectors. (a) The angle between the two vectors. (b) The orthogonal projection 𝑨∥ of vector ⃗𝑨 onto the direction of vector 𝑩⃗⃗. (c) The orthogonal projection 𝑩∥ of vector 𝑩 ⃗⃗ onto the direction of vector ⃗𝑨 ⃗. Sample Problem 1: The Scalar Product Figure 2 For the vectors shown in Figure 2, find the scalar product of 𝐀 ⃗ ∙ 𝐅. Strategy: The magnitudes of vectors 𝐀 ⃗⃗ and 𝐅 are 𝐴 = 10.0 and 𝐹 = 20.0. Angle 𝜃, between them, is the difference: 𝜃 = 𝜑 − 𝛼 = 110° − 35° = 75°. Substituting these values into equation 𝐀 ⃗ = 𝐴𝐵 𝑐𝑜𝑠 𝝋 gives a scalar product. ⃗ ∙𝐁 Solution: A straightforward calculation gives us ⃗ ∙ ⃗F = 𝐴𝐹 𝑐𝑜𝑠 𝝋 = (𝟏𝟎. 𝟎)(𝟐𝟎. 𝟎) 𝒄𝒐𝒔 𝟕𝟓° = 𝟓𝟏. 𝟕𝟔 𝐀 B. Work Done by a Constant Force We define the work W done by this constant force as the product of the force magnitude in the direction of the displacement F and the displacement magnitude Δx: W = F │Δx│ (constant force in direction of straight-line displacement) Joule (J) is the unit of work or energy in the International System of Units (SI); it is equal to the work done by a force of one newton acting through one meter (N.m). The work done on the body increases if either the force F or the displacement Δx increases. NegOr_Q1_GenPhysics1_SLKWeek6_v2 6  Figure 3 Work = W, weight = w Don’t confuse uppercase W (work) with lowercase w (weight). Though the symbols are Take note similar, work and weight are different quantities. If you attach a string to the box and pull it, as shown in the illustration below, the force now acts at an angle to the displacement. In this situation, the work done on the box by the force is the product between the force in the direction of the displacement and the magnitude of the displacement: W = F │∆x│= Fcosθ │∆x│or W = Fcosθ │∆x│ where: F = magnitude of the constant force │∆x│= magnitude of the displacement of the point of application of the force θ = angle between the directions of the force and displacement vectors Figure 4 Energy, in physics, is the capacity for doing work. All forms of energy are associated with motion. Power is the rate of doing work. It is the amount of energy consumed per unit time. Watt, is the unit of power in the International System of Units (SI) equal to one joule of work performed per second (J/s). NegOr_Q1_GenPhysics1_SLKWeek6_v2 7 Sample Problem 2: A force of 15N is exerted on a box at an angle of θ = 25°. How much work is done by the force on the box as the box moves along the table a distance of 5.0 m? STEP 1. Draw the object first at its initial position and second at its final position. For ease, the object can be represented as a dot or a box. Label the initial and final positions of the object. Figure 5 STEP 2. Identify the given values in the problem. In the sample problem, the values to be identified are the force applied which is represented by Fx we use the subscript x since the direction of the force is horizontal, next is the value of the angle θ between the directions of the force and vectors and lastly, the magnitude of the displacement │∆x│. Therefore: Given: Fx = 15N θ = 25° │∆x│= 5.0 m STEP 3. Now that we have identified the given values in the problem, we can now solve for work. Observe that the problem involves an angle, therefore the formula that we will be using here is W = Fcosθ │∆x│. 3.a. Supply the given values to the formula W = Fcosθ │∆x│. Now we will have: W = (15N) (cos25°) (5.0m) 3.b. Solve for cos25° using your scientific calculator, round off the result to 2 decimal places. After which the result will be: W = (15N) (0.91) (5.0m) NegOr_Q1_GenPhysics1_SLKWeek6_v2 8  3.c. Multiply (15N) (0.91) (5.0m). The result is: W = 68.25 Nm 3.d. Note that (Newton-meter) Nm is equivalent to Joule (J). Therefore, our final answer is: W = 68.25 J What if there are several forces doing work on a system? How are we going to compute the work done? The total work is found by computing the work done by each force and adding each individual work together. Total work is computed by using the formula: WTotal = F1xΔx1 + F2xΔx2 + F3xΔx3 +... A system is a regularly interacting or interdependent group of items forming a unified whole. The system can be described as particle when the system moves thus all the parts undergo equal displacements. The points of application of the forces are identical when several of these forces do work on such a particle. Let the displacement of the point of application of any one of the forces be Δx. Then, WTotal = F1xΔx + F2xΔx + F3xΔx +... Since the displacements of all forces acting on the system are equal, we will only use one value for Δx. The formula will be: WTotal = (F1x + F2x + F3x +...) Δx or WTotal = Fnet xΔx For a particle controlled to move along the x axis, the net force has only an x component. That is, ⃗𝑭𝒏𝒆𝒕 = 𝑭𝒏𝒆𝒕 𝒙𝒊̂. Thus, for a particle, the x component of the net force times the displacement of any part of the object is equal to the total work done on the object. (Tipler and Mosca 2008) NegOr_Q1_GenPhysics1_SLKWeek6_v2 9 Sample Problem 3: A 4000 kg truck is to be loaded onto a ship by a crane that applies an upward force of 50 kN on the truck. This force, which is strong enough to overcome the gravitation force and keep the truck moving upward, is applied over a distance of 4.0 m. Find: (a) the work done on the truck by the crane; (b) the work done on the truck by gravity; and (c) the net work done on the truck. STEP 1. Draw the object first at its initial position and second at its final position. Choose the +y to be the direction of the displacement. Since the displacement direction is upward. STEP 2. Identify the given values in the problem. In the sample problem, we can identify 4 values: mass(m) of the truck, force (Fapp y) applied by the crane to the truck, displacement(∆y) and the gravitational force(gy) since the displacement is in upward direction. Given: m = 4000 kg Fapp y = 50 kN Figure 6 ∆y = 4.0 m gy = -9.8 N/kg Note: As you may have observed, the unit used for acceleration due to gravity is N/kg, it is because m/s2 is equivalent to N/kg. To further explain; 𝑚 𝑵 𝑘𝑔 ∙ 2 𝒎 = 𝑠 = 𝒌𝒈 𝑘𝑔 𝒔𝟐 Furthermore, the sign for acceleration due to gravity is negative (-) because it is directed downward. NegOr_Q1_GenPhysics1_SLKWeek6_v2 10 STEP 3. After identifying the values. We can now start solving. Let’s first solve for (a) the work done on the truck by the crane (Wapp). We will use this equation, W = F │∆x│. We will only use the given values for force applied by the crane (Fapp y), and the displacement (∆y). Let’s also consider the direction of the displacement. Take note that the direction of the displacement is upward, therefore it is in +y axis. So, the final formula here will be, Wapp = Fapp y ∆y SOLUTION for (a) the work done on the truck by the crane: Given: Fapp y = 50 kN ∆y = 4.0 m Formula: Wapp = Fapp y ∆y 3.a. Supply the given values to the formula Wapp = Fapp y ∆y, now we will have: Wapp = (50 kN) (4.0 m) 3.b. Multiply (50 kN) (4.0 m). The result is, Wapp = 200 kNm 3.c. Note that (Newton-meter) Nm is equivalent to Joule (J). Therefore, our final answer is: Wapp = 200 Kj Step 4. Solve for (b) the work done on the truck by gravity (W g). The given values we will use here are the mass (m) of the truck, acceleration due to gravity (gy), and the displacement (∆y). Take note the Force (F) is equal to mass(m) times acceleration due to gravity (g). Therefore, the formula Wapp = Fapp y ∆y will become, Wg = mg y ∆y. NegOr_Q1_GenPhysics1_SLKWeek6_v2 11 SOLUTION for (b) the work done on the truck by gravity (Wg): Given: m = 4000 kg gy = -9.8 N/kg ∆y = 4.0 m Formula: Wg = mg y ∆y 4.a. Supply the given values to the formula Wg = mgy ∆y, we will have; 𝑁 𝑊𝑔 = (4000 kg) ( -9.8 )(4.0𝑚) 𝑘𝑔 4.b. Cancel out the unit kg in 4000 kg and -9.8 N/kg, since they can be divided. The result will be. 𝑁 𝑊𝑔 = (4000 kg) ( -9.8 )(4.0𝑚) 𝑘𝑔 𝑊𝑔 = (4000) ( -9.8 N)(4.0𝑚) 4.c. Multiply (4000) (-9.8 N) (4.0 m), the result is; 𝑊𝑔 = -156,800 Nm 4.d. Note that (Newton-meter) Nm is equivalent to Joule (J). Therefore, 𝑊𝑔 = -156,800 J 4.e. Since the value is too big, we will convert it to kJ. To do that, 1𝑘𝐽 𝑊𝑔 = (-156,800 J)( ) 1000𝐽 -156,800𝑘𝐽 𝑊𝑔 = 1000 Wg = -156.8 kJ Step 5. Solve for (c) the net work done on the truck (Wnet). The net work done on the truck is the sum of the work done by the crane (Wapp y) and the work done by the acceleration due to gravity (Wg) to the truck. NegOr_Q1_GenPhysics1_SLKWeek6_v2 12 There are two ways to solve for the net work done on the truck. A. Using the formula Wnet = Wapp y + Wg B. Using Eq. (4) WTotal = Fnet x∆x SOLUTION for (c) the net work done on the truck (W net): A. Using the formula Wnet = Wapp y + Wg Based on our solution for (a) the work done on the truck by the crane, Wapp y = 200 kJ and for (b) the work done on the truck by gravity (W g), Wg = - 156.8 kJ. Supply this to the formula Wnet = Wapp y + Wg, we have, Wnet = 200 kJ + (-156.8 kJ) Wnet = 43.2 kJ SOLUTION for (c) the net work done on the truck (W net): B. Using Eq. (4) WTotal = Fnet x∆x Let’s solve for the for applied on the truck by the acceleration due to 𝑁 gravity. Therefore, 𝐹𝑔 = (4000 𝑘𝑔) (−9.8 ), 𝐹𝑔 = −39,200 𝑁 or 𝑭𝒈 = −𝟑𝟗. 𝟐 𝒌𝑵. 𝑘𝑔 WTotal = {(50kN + (-39.2kN)} (4.0m) = (10.8kN) (4.0m) = 43.2 kNm Wnet = 43.2 kJ C. Work in Scalar Product Notation ⃗ 𝒗 𝑭∥ ⃗⃗⃗⃗⃗ 𝒅𝓵 𝝋 ⃗𝑭 ⃗𝑭 𝑭⊥ (𝒂) Figure 7 (𝒃) (Tipler and Mosca 2008) NegOr_Q1_GenPhysics1_SLKWeek6_v2 13  Study the particle moving along the arbitrary curve shown on the illustration above. The component 𝐅∥ in Figure (b) is related to the angle 𝝋 between the directions of 𝑭⃗ and ⃗⃗⃗⃗⃗ 𝒅𝓵 by 𝐅∥ = 𝑭 𝐜𝐨𝐬 𝝋, so the work dW by 𝑭⃗ for the displacement ⃗⃗⃗⃗⃗ 𝒅𝓵 is 𝒅𝑾 = 𝐅∥ 𝒅𝓵 = 𝑭 𝐜𝐨𝐬 𝝋 𝒅𝓵 In scalar-product notation, the work 𝒅𝑾 done by force ⃗𝑭 on a particle over an infinitesimal displacement 𝒅𝓵 ⃗⃗⃗⃗⃗ is ⃗⃗⃗⃗⃗ ⃗ ∙ 𝒅𝓵 𝒅𝑾 = 𝐅∥ 𝒅𝓵 = 𝑭 𝐜𝐨𝐬 𝝋 𝒅𝓵 = 𝑭 Where: 𝒅𝓵 = the magnitude of ⃗⃗⃗⃗⃗ 𝒅𝓵 𝐅∥ = is the component of 𝑭 ⃗ in the direction of 𝒅𝓵 ⃗⃗⃗⃗⃗ The work done on a particle as it moves from point 1 to point 2 is 𝟐 𝑾 = ∫ ⃗𝑭 ∙ ⃗⃗⃗⃗⃗ 𝒅𝓵 𝟏 (If the force remains constant, the work can be expressed as 𝑾 = 𝑭 ⃗ ∙ 𝓵 , where ⃗⃗⃗ ⃗⃗⃗ 𝓵 is the displacement.) When several forces 𝑭 ⃗ 𝒊 act on a particle whose displacement is 𝒅𝓵 ⃗⃗⃗⃗⃗ , the total work done on it is 𝒅𝑾𝒕𝒐𝒕𝒂𝒍 = ⃗𝑭 𝟏 ∙ ⃗⃗⃗⃗⃗ 𝒅𝓵 + ⃗𝑭 𝟐 ∙ ⃗⃗⃗⃗⃗ ⃗ 𝟏 + ⃗𝑭 𝟐 + ∙∙∙) ∙ ⃗⃗⃗⃗⃗ 𝒅𝓵 +∙∙∙ = (𝑭 𝒅𝓵 = (∑ ⃗𝑭 𝒊 ) ∙ ⃗⃗⃗⃗⃗ 𝒅𝓵 Sample Problem 4: You push a box up a ramp using a constant horizontal 150-N force 𝑭 ⃗. For each distance 5.00 m along the ramp, the box gains 3.00 m of height. Find the work done by 𝑭 ⃗ for each 5.00 m the box moves along the ramp (a)by directly computing the scalar product from the components of ⃗𝑭 and 𝓵 , where ⃗⃗⃗ ⃗⃗⃗ 𝓵 is the displacement, (b) by multiplying the product of the magnitudes of ⃗𝑭 and ⃗𝓵⃗ by 𝐜𝐨𝐬 𝝋, where 𝝋 is the angle between the direction of ⃗𝑭 and the direction of ⃗⃗⃗ 𝓵 , (c)by finding 𝐅∥ (the component ⃗𝑭 in the direction of 𝓵⃗⃗ ) and multiplying it by 𝓵 (the magnitude of 𝓵 ⃗⃗ ), and (d)by finding 𝓵∥ (the component ⃗𝓵⃗ in the direction of 𝑭 ⃗ )and multiplying it by the magnitude of the force. NegOr_Q1_GenPhysics1_SLKWeek6_v2 14  Draw a sketch of the box in its initial and final positions. Place coordinate axes on the sketch with the x axis horizontal. Express the force and displacement vectors in component form and take the scalar product. Then find the component of the force in the direction of the displacement, and vice versa. y ⃗⃗𝓵⃗ 3.00 ⃗ 𝑭 𝝋 m x 4.00 m is obtained by using Pythagorean 4.00 Theorem (𝑐 2 = ξ𝑎2 + 𝑏2 ), where c=5.00 m and m b=3.00 m. Figure 8 Pythagorean Theorem explains the relation between the sides of a right-angled triangle. It is basically used to find the length of an unknown side and angle of a triangle. Note: There are four (4) solutions to this problem. Study each solution. First Solution: Express ⃗𝑭 (read as “force vector”) and ⃗⃗⃗ 𝓵 (read as “displacement vector”) in component form and take the scalar product. Note, 𝒊̂ (read as “i-hat”) is a unit vector pointing the +x direction and the unit vector 𝒋̂ pointing the +y direction. (http://www.cbphysics.org/ n.d.) SOLUTION: ⃗𝑭 = (𝟏𝟎𝟎𝒊̂ + 𝟎𝒋̂)𝐍 ⃗⃗ = (𝟒. 𝟎𝟎𝒊̂ + 𝟑. 𝟎𝟎𝒋̂)𝐦 𝓵 Supply these values to the formula 𝑾 = ⃗𝑭 ∙ ⃗𝓵⃗ = 𝑭𝒙 ∆𝒙 + 𝑭𝒚 ∆𝒚 (This formula is the formula for total work). 𝑊 = 𝐹 ∙ ⃗⃗ℓ = 𝐹𝑥 ∆𝑥 + 𝐹𝑦 ∆𝑦 = (100N)(4.00m) + 0(3.00m) 𝑾 = 𝟒. 𝟎𝟎 × 𝟏𝟎𝟐 𝑱 NegOr_Q1_GenPhysics1_SLKWeek6_v2 15 Second Solution: Calculate 𝑭𝓵 𝐜𝐨𝐬 𝝋, where 𝝋 is the angle between the directions of the two vectors as shown. Equate this expression with the Part-(a) result and solve for 𝐜𝐨𝐬 𝝋. Then solve for the work: SOLUTION: ⃗⃗ = 𝑭𝓵 𝐜𝐨𝐬 𝝋 ⃗ ∙𝓵 𝑭 and ⃗⃗ = 𝑭𝒙 ∆𝒙 + 𝑭𝒚 ∆𝒚 ⃗ ∙𝓵 𝑭 So, 𝑭𝓵 𝐜𝐨𝐬 𝝋 = 𝑭𝒙 ∆𝒙 + 𝑭𝒚 ∆𝒚. To solve for 𝐜𝐨𝐬 𝝋, we will rearrange the formula. Therefore 𝑭𝓵 𝐜𝐨𝐬 𝝋 = 𝑭𝒙 ∆𝒙 + 𝑭𝒚 ∆𝒚 (Divide both sides with 𝑭𝓵) 𝑭𝓵 𝐜𝐨𝐬 𝝋 𝑭𝒙 ∆𝒙 + 𝑭𝒚 ∆𝒚 = 𝑭𝓵 𝑭𝓵 𝑭𝒙 ∆𝒙 + 𝑭𝒚 ∆𝒚 𝐜𝐨𝐬 𝝋 = 𝑭𝓵 Supply the needed values to the formula. (𝟏𝟎𝟎 𝑵)(𝟒. 𝟎𝟎 𝒎) + 𝟎 𝐜𝐨𝐬 𝝋 = (𝟏𝟎𝟎 𝑵)(𝟓. 𝟎𝟎 𝒎) 𝐜𝐨𝐬 𝝋 = 𝟎. 𝟖𝟎𝟎 Now that we have the value for 𝐜𝐨 𝐬 𝝋, we can now solve for work using the formula, 𝑾 = 𝑭𝓵 𝐜𝐨𝐬 𝝋. Supplying the given to the formula we have, 𝑾 = (𝟏𝟎𝟎 𝑵)(𝟓. 𝟎𝟎 𝒎)𝟎. 𝟖𝟎𝟎 𝑾 = 𝟒. 𝟎𝟎 × 𝟏𝟎𝟐 𝑱 Third Solution: Find 𝐅∥ and multiply it by 𝓵. To find 𝐅∥, we will use the formula 𝐅∥ = 𝑭 𝐜𝐨𝐬 𝝋. Supply the given values to the formula, we will have, 𝐅∥ = (𝟏𝟎𝟎 𝑵)𝟎. 𝟖𝟎𝟎 𝐅∥ = 𝟖𝟎. 𝟎 𝑵 We will use the result 𝐅∥ = 𝟖𝟎. 𝟎 𝑵 to solve for work by multiplying it by 𝓵 = 𝟓. 𝟎𝟎𝒎 by using the formula 𝑾 = 𝐅∥ 𝓵 NegOr_Q1_GenPhysics1_SLKWeek6_v2 16  𝑾 = (𝟖𝟎. 𝟎 𝑵)(𝟓. 𝟎𝟎 𝒎) 𝑾 = 𝟒. 𝟎𝟎 × 𝟏𝟎𝟐 𝑱 Fourth Solution: Multiply 𝑭 and 𝓵∥ , where 𝓵∥ is the component of 𝓵 ⃗⃗ in the direction of 𝑭 ⃗. We will use the formula 𝓵∥ = 𝓵 𝐜𝐨𝐬 𝝋 for solving for 𝓵∥ and 𝑾 = 𝑭𝓵∥. To solve for the work done. 𝓵∥ = (𝟓. 𝟎𝟎 𝒎)𝟎. 𝟖𝟎𝟎 𝓵∥ = 𝟒. 𝟎𝟎 𝒎 To solve for the work done, 𝑾 = 𝑭𝓵∥ 𝑾 = (𝟏𝟎𝟎 𝑵)(𝟒. 𝟎𝟎 𝒎) The four different calculations give the same result for work which is 𝑾 = 𝟒. 𝟎𝟎 × 𝟏𝟎𝟐 𝑱. D. Work Done by a Varying Force, Straight-Line Motion There are various situations in which a body moves along a curved path and is acted on by a force that varies in magnitude, direction, or both. We need to be able to compute the work done by the force in these more general cases. The work–energy theorem holds true even when varying forces are considered and when the body’s path is not straight. Let’s consider straight-line motion along the x-axis with a force whose x- component 𝑭𝒙 may change as the body moves. (A real-life example is driving a car along a straight road with stop signs, so the driver must alternately step on the gas and apply the brakes.) Supposing a particle moves along the x-axis from point 𝒙𝟏 to 𝒙𝟐 (Figure 9). Figure 10 is a graph of the x-component of force as a function of the particle’s coordinate x. To find the work done by this force, we divide the total displacement into small segments ∆𝒙𝒂, ∆𝒙𝒃 , and so on (Figure 11). We approximate the work done by the force during segment ∆𝒙𝒂 as the average x-component of force 𝑭𝒂𝒙 in that segment multiplied by the x-displacement ∆𝒙𝒂. We do this for each segment and then add the results for all the segments. The work done by the force in the total displacement from 𝒙𝟏 to 𝒙𝟐 is approximately 𝑾 = 𝑭𝒂𝒙 ∆𝒙𝒂 + 𝑭𝒃𝒙 ∆𝒙𝒃 +∙ ∙ ∙ NegOr_Q1_GenPhysics1_SLKWeek6_v2 17 Figure 9. Particle moving from 𝑥1 to 𝑥2 in response to a changing force in the x-direction Figure 10. Figure 11. E. Gravitational potential energy near Earth’s surface The system of interest consists of our planet, Earth, and one or more particles near its surface (or bodies small enough to be considered as particles, compared to Earth). The gravitational force on each particle (or body) is just its weight 𝒎𝒈 near the surface of Earth, acting vertically down. According to Newton’s third law, each particle exerts a force on Earth of equal magnitude but in the opposite direction. Newton’s second law tells us that the magnitude of the acceleration produced by each of these forces on Earth is mg divided by Earth’s mass. Since the ratio of the mass of any ordinary object to the mass of Earth is vanishingly small, the motion of Earth can be completely neglected. Therefore, we consider this system to be a group of single-particle systems, subject to the uniform gravitational force of Earth. The work done on a body by Earth’s uniform gravitational force, near its surface, depended on the mass of the body, the acceleration due to gravity, and the difference in height the body traversed. This work is the negative of the difference in the gravitational potential energy, so that difference is NegOr_Q1_GenPhysics1_SLKWeek6_v2 18  ∆𝑼𝒈𝒓𝒂𝒗 = −𝑾𝒈𝒓𝒂𝒗,𝑨𝑩 = 𝒎𝒈(𝒚𝐁 − 𝒚𝐀 ) You can see from this that the gravitational potential energy function, near Earth’s surface, is 𝑼 (𝒚) = 𝒎𝒈𝒚 + 𝒄𝒐𝒏𝒔𝒕 F. Elastic potential energy Elastic potential energy is energy stored as a result of applying a force to deform an elastic object. In Work, we saw that the work done by a perfectly elastic spring, in one dimension, depends only on the spring constant and the squares of the displacements from the unstretched position. This work involves only the properties of a Hooke’s law interaction and not the properties of real springs and whatever objects are attached to them. Therefore, we can define the difference of elastic potential energy for a spring force as the negative of the work done by the spring force in this equation before we consider systems that embody this type of force. Thus, 1 ∆𝑼 = −𝑾𝑨𝑩 = 𝒌(𝒙𝟐𝑩 − 𝒙𝟐𝑨) 2 where the object travels from point A to point B. The potential energy function corresponding to this difference is 𝟏 𝟐 𝑼(𝒙) = 𝒌𝒙 + 𝒄𝒐𝒏𝒔𝒕 𝟐 If the spring force is the only force acting, it is simplest to take the zero of potential energy at 𝑥 = 0, when the spring is at its unstretched length. Then, the constant in the previous equation is zero. (Other choices may be more convenient if other forces are acting.) G. Conservative Forces A conservative force is one for which work done by or against it depends only on the starting and ending points of a motion and not on the path taken. The work done by a conservative force always has four properties: 1. It can be expressed as the difference between the initial and final values of a potential-energy function. 2. It is reversible. 3. It is independent of the path of the body and depends only on the starting and ending points. NegOr_Q1_GenPhysics1_SLKWeek6_v2 19 4. When the starting and ending points are the same, the total work is zero. H. Potential Energy Diagrams and Stability You can get a good deal of useful information about the dynamical behavior of a mechanical system just by interpreting a graph of its potential energy as a function of position, called a potential energy diagram. This is most easily accomplished for a one-dimensional system, whose potential energy can be plotted in one two-dimensional graph—for example, U(x) versus x—on a piece of paper or a computer program. For systems whose motion is in more than one dimension, the motion needs to be studied in three-dimensional space. We will simplify our procedure for one- dimensional motion only. First, let’s look at an object, freely falling vertically, near the surface of Earth, in the absence of air resistance. The mechanical energy of the object is conserved, 𝐸 = 𝐾 + 𝑈, and the potential energy, with respect to zero at ground level, is 𝑼 (𝒚) = 𝒎𝒈𝒚, which is a straight line through the origin with slope 𝒎𝒈. Figure 12. The potential energy graph for an object in vertical free fall, with various quantities indicated. The line at energy E represents the constant mechanical energy of the object, whereas the kinetic and potential energies, 𝑲𝑨 and 𝑼𝑨, are indicated at a particular height 𝒚𝑨. You can see how the total energy is divided between kinetic and potential energy as the object’s height changes. Since kinetic energy can never be negative, there is a maximum potential energy and a maximum height, which an object with the given total energy cannot exceed: 𝐾 = 𝐸−𝑈 ≥0 𝑈≤𝐸 NegOr_Q1_GenPhysics1_SLKWeek6_v2 20 Quartic and Quadratic Potential Energy Diagram The potential energy for a particle undergoing one-dimensional motion along the x-axis is 𝑼(𝒙) = 𝟐(𝒙𝟒 − 𝒙𝟐 ), where 𝑼 is in joules and 𝒙 is in meters. The particle is not subject to any non-conservative forces and its mechanical energy is constant at 𝑬 = −𝟎. 𝟐𝟓 𝑱. (a) Is the motion of the particle confined to any regions on the x-axis, and if so, what are they? (b) Are there any equilibrium points, and if so, where are they and are they stable or unstable? Strategy: First, we need to graph the potential energy as a function of x. The function is zero at the origin, becomes negative as x increases in the positive or negative directions (𝒙𝟐 is larger than 𝒙𝟒 for 𝒙 < 𝟏), and then becomes positive at sufficiently large |𝑥 |. Your graph should look like a double potential well, with the zeros determined by solving the equation 𝑼 (𝒙) = 𝟎, and the extremes determined by examining the first and second derivatives of 𝑼 (𝒙), as shown in Figure 13. Figure 13. The potential energy graph for a one-dimensional, quartic and quadratic potential energy, with various quantities indicated. You can find the values of (a) the allowed regions along the x-axis, for the given value of the mechanical energy, from the condition that the kinetic energy can’t be negative, and (b) the equilibrium points and their stability from the properties of the force (stable for a relative minimum and unstable for a relative maximum of potential energy). NegOr_Q1_GenPhysics1_SLKWeek6_v2 21  You can just eyeball the graph to reach qualitative answers to the questions in this example. That, after all, is the value of potential energy diagrams. You can see that there are two allowed regions for the motion (𝐸 > 𝑈) and three equilibrium points (slope 𝒅𝑼/𝒅𝒙 = 𝟎), of which the central one is unstable (𝒅𝟐 𝑼/𝒅𝒙𝟐 < 𝟎), and the other two are stable 𝒅𝟐 𝑼/𝒅𝒙𝟐 > 𝟎. Solution: a. To find the allowed regions for x, we use the condition 𝟏 𝑲 = 𝑬 − 𝑼 = − − 𝟐(𝒙 𝟒 − 𝒙 𝟐 ) ≥ 𝟎 𝟒 1 2 If we complete the square in 𝒙𝟐 , this condition simplifies to 2 (𝑥 2 − 2) ≤ 1 , which we can solve to obtain 4 1 1 1 1 − √ ≤ 𝑥2 ≤ + √ 2 8 2 8 This represents two allowed regions, 𝑥𝑝 ≤ 𝑥 ≤ 𝑥𝑅 and −𝑥𝑅 ≤ 𝑥 ≤ −𝑥𝑝 , where 𝑥𝑝 = 0.38 and 𝑥𝑅 = 0.92 (in meters). b. To find the equilibrium points, we solve the equation 𝑑𝑈⁄𝑑𝑥 = 8𝑥 3 − 4𝑥 = 0 and find 𝑥 = 0 and 𝑥 = ±𝑥𝑄 , where 𝑥𝑄 = 1⁄ξ2 = 0.707 (meters). The second derivative 𝑑 2 𝑈⁄𝑑𝑥 2 = 24𝑥 2 − 4 is negative at 𝑥 = 0, so that position is a relative maximum and the equilibrium there is unstable. The second derivative is positive at 𝑥 = ±𝑥𝑄 , so these positions are relative minima and represent stab equilibria. Significance: The particle in this example can oscillate in the allowed region about either of the two stable equilibrium points we found, but it does not have enough energy to escape from whichever potential well it happens to initially be in. The conservation of mechanical energy and the relations between kinetic energy and speed, and potential energy and force, enable you to deduce much information about the qualitative behavior of the motion of a particle, as well as some quantitative information, from a graph of its potential energy. NegOr_Q1_GenPhysics1_SLKWeek6_v2 22  PERFORMANCE TASK Problem Solving: Directions: Write your solution on your notebook. Remember to follow the steps in solving problem. A constant force 𝑭 ⃗ = (𝟒. 𝟎𝟎𝒊̂ + 𝟓. 𝟎𝟎̂)𝑵 𝒋 is applied to a particle that undergoes a displacement 𝓵 ⃗⃗ = (𝟑. 𝟎𝟎𝒊̂ − 𝟔. 𝟎𝟎̂)𝒎. 𝒋 (a) the work done by the force, and (b) the component of the force in the direction of the displacement. Written Work: Directions: Answer what is asked. Write your answer in your notebook. Compare and contrast gravitational potential energy and elastic potential energy. III. WHAT I HAVE LEARNED EVALUATION/POST-TEST A. Directions: Write true if the statement is correct and false if not. 1. Elastic potential energy is energy released as a result of applying a force to deform an elastic object. 2. Conservative force is reversible. 3. Conservative force is dependent of the path of the body and depends only on the starting and ending points. 4. According to Newton’s third law, each particle exerts a force on Earth of equal magnitude but in the opposite direction. 5. The work done on the body increases if either the force F or the displacement Δx increases. B. Directions: Solve the given problem. Show your solution. You joined a “dogsled” race during your winter break. To start, you pulled the sled (total mass of 85 kg) with a force of 195N at 45° above the horizontal. Find the work done on the sled after it moves at a distance of 7m. NegOr_Q1_GenPhysics1_SLKWeek6_v2 23  REFERENCES Britannica, The Editors of Encyclopaedia. n.d. https://www.britannica.com/. https://www.britannica.com/science/work-physics. Ling, Samuel J., Jeff Sanny Loyola, and William Moebs. 2016. University Physics. Vol. 1. OpenStax. https://openstax.org/details/books/university-physics- volume-1 in your citation. n.d. Merriam-Webster.com Dictionary, s.v. Accessed September 6, 2020. https://www.merriam- n.d. http://www.cbphysics.org/. http://www.cbphysics.org/downloadsI/UnitVectors.pdf. n.d. https://a1384-240719.cluster8.canvas-user- content.com/courses/1384~1159/files/1384~240719/course%20files/apb 11o/resources/guides/G07-2.work_as_area.htm. n.d. https://byjus.com/maths/pythagoras-theorem/. n.d. https://courses.lumenlearning.com/boundless-physics/chapter/power/. Paul Peter Urone, Roger Hinrichs. 2012. June 21. https://openstax.org/books/college-physics/pages/1-introduction-to- science-and-the-realm-of-physics-physical-quantities-and-units. Tipler, Paul A., and Gene Mosca. 2008. PHYSICS FOR SCIENCTISTS AND ENGINEERS, with Modern Physics. 6th. New York: Susan Finnemore Brennan. Young, Hugh D., and Roger A. Freedman. 2012. Sears and Zemansky's university physics : with modern physics. -- 13th ed. 13th. Edited by Nancy Whilton. Jim Smith. NegOr_Q1_GenPhysics1_SLKWeek6_v2 24  DEPARTMENT OF EDUCATION Division of Negros Oriental SENEN PRISCILLO P. PAULIN, CESO V Schools Division Superintendent JOELYZA M. ARCILLA EdD OIC - Assistant Schools Division Superintendent MARCELO K. PALISPIS EdD JD OIC - Assistant Schools Division Superintendent NILITA L. RAGAY EdD OIC - Assistant Schools Division Superintendent/CID Chief ROSELA R. ABIERA Education Program Supervisor – (LRMDS) ARNOLD R. JUNGCO PSDS-Division Science Coordinator MARICEL S. RASID Librarian II (LRMDS) ELMAR L. CABRERA PDO II (LRMDS) GENEVA FAYE L. MENDOZA Writer ROSEWIN P. ROCERO Illustrator/Lay-out Artist _________________________________ ALPHA QA TEAM LIEZEL A. AGOR EUFRATES G. ANSOK JR. JOAN Y. BUBULI MA. OFELIA I. BUSCATO DEXTER D. PAIRA LIELIN A. DE LA ZERNA BETA QA TEAM ZENAIDA A. ACADEMIA RANJEL D. ESTIMAR ALLAN Z. ALBERTO MARIA SALOME B. GOMEZ EUFRATES G. ANSOK JR. JUSTIN PAUL ARSENIO C. KINAMOT DORIN FAYE D. CADAYDAY LESTER C. PABALINAS MERCY G. DAGOY ARJIE T. PALUMPA ROWENA R. DINOKOT DISCLAIMER The information, activities and assessments used in this material are designed to provide accessible learning modality to the teachers and learners of the Division of Negros Oriental. The contents of this module are carefully researched, chosen, and evaluated to comply with the set learning competencies. The writers and evaluator were clearly instructed to give credits to information and illustrations used to substantiate this material. All content is subject to copyright and may not be reproduced in any form without expressed written consent from the division. NegOr_Q1_GenPhysics1_SLKWeek6_v2 25  ANSWER KEY 1. 𝟗𝟔𝟗. 𝟏𝟓 𝑱 SYNOPSIS AND ABOUT THE AUTHOR B. This Self Learning Kit is designed to 5. True aid students to independently learn the 4. True important concepts about Work, Energy 3. False and Power. 2. True 1. False The discussion and tasks related to A. the topic are arranged systematically and Evaluation: explained in detail so that the students are (Answers may vary) guided. The students are expected to WRITTEN WORKS master this lesson and value its application (b) 𝑭∥ = −𝟐. 𝟔𝟖𝑵 to the physical world. 11. (a) 𝑾 = −𝟏𝟖 𝑱 PERFORMANCE TASK: 10. Elastic potential energy 9. gravitational force 8. Watt 7. Joule 6. Power 5. Energy 4. Pythagorean Theorem 3. System 2. Work 1. Scalar product PRE-TEST: Geneva Faye L. Mendoza completed her BSE – Physical Science at NORSU-Bayawan Campus and is currently continuing her Master’s Degree at NORSU- Main Campus. She taught Science 7 to 10 at Eligio T. Monte de Ramos High School, Santa Catalina District 1. Now, she teaches Science 8 and 10 at Casiano Z. Napigkit National High School, Santa Catalina District 1. NegOr_Q1_GenPhysics1_SLKWeek6_v2 26