NJC 2023 Work, Energy, Power Problem Set Answers

National JC Work, Energy and Power 2023 - 1 - Problemset Solutions Exercises E1(a) Work done by Denise = Fx cos  = (30) (5.0) (cos 60o) = 75 J E1(b) Hilda does not do work on the book because the book’s displacement is zero. E2. (a) Can work done be positive or negative? State the condition for it. Yes, work done can be positive or negative. When force and displacement are in same direction, workdone is positive. When force and displacement are in opposite direction, workdone is negative. (b) Can work done be zero? State the condition(s) for it. Yes. Conditions: - force is zero - Displacement is zero - Force and displacement is perpendicular to each other. E3 The elastic potential energy stored in the bow just before the release of an arrow is 95 J. When the arrow of mass 170 g is fired, 90% of the elastic potential energy is transferred to the arrow. Show that the speed of the arrow as it leaves the bow is 32 ms-1. 90Amount of energy transferred to the arrow 95 85.5 J100=  = 2 -1 1 85.52 32 m s mv v = = E4. An object at rest is pulled by a constant horizontal force, F on a smooth floor to a displacement of x. (i) Sketch a graph of force applied on object against the displacement. Label F and x in the graph. (ii) What does the area under force against displacement graph in (i) represent? Hint: work done. (i) 0 force displacement x F

National JC Work, Energy and Power 2023 - 2 - (ii) Area under force against displacement graph represent the work done by force on object. Work done = F(x) E5. An object at rest is pulled by a varying horizontal force, Fx on a smooth floor to a displacement of x. (i) Can we still use this formula to calculate the work done by Fx, i.e workdone = Fx(x)? No. The formula work done = Fx(x) is only for a constant force. (ii) How can the work done by Fx be calculated? To calculate the work done by a varying force acting on an object that E6(a) Gain of KE of the sack = Work done on the sack = 2.0 0.35 0.70 J= E6(b) 2 -11 0.7 0.37 ms2 mv v=  = E7(a) By Principle of COE, Loss of GPE = Gain in KE m(9.81)(30) – 0 = ½ m (2.80)2 - 0 v = 24.4 m s-1 (3 s.f.) OR By Principle of COE, GPE at A + KE at A = GPE at B + KE at B m(9.81)(30) + ½ m (2.80)2 = 0 + ½ m v2 v = 24.4 m s-1 (3 s.f.) E7(b) By Principle of COE, Loss of GPE = Gain in KE m(9.81)(30) – m(9.81)(25) = ½ m (2.80)2 - 0 v = 10.3 m s-1 (3 s.f.) Work done by a varying force is equal to the area under the force –displacement graph.

National JC Work, Energy and Power 2023 - 3 - OR By Principle of COE, GPE at A + KE at A = GPE at C + KE at C m(9.81)(30) + ½ m (2.80)2 = m(9.81)(25) + ½ m v2 v = 10.3 m s-1 (3 s.f.) E8(a) ( )( ) 2 2 Work done in compressing spring Elastic P E stored in spring 1 2 1 500 0.102 2.5 J kx = = = = E8(b) Assume all the elastic PE stored in the spring is converted to kinetic energy. ( ) 2 -1 1 2.52 2 2.5 1.6 ms2.0 mv v =  = = E9 Power of engine P = Fe(v) where Fe is driving force from engine. At maximum speed, acceleration = 0  D = Fe 2 3 P =F v P =D v, D α v α v e P 72 k 3 12 36 k 3 v one Vone = 9.5 m s-1 (Ans D) Problems Work P1. 2011/P1/Q11, TYS Topic 5: MCQ 12 Gain of kinetic energy = work done by F = Fs (Ans C)

National JC Work, Energy and Power 2023 - 4 - P2. A force F that is parallel to the x-axis acts on a block of ice. The magnitude of the force varies with the x-coordinate of the block as shown. Calculate the work done on the block of ice by the force F when the block moves from x = 0 m to x = 7.0 m. ( )( ) ( )( ) Work done on block Area under graph 112 1 3 1 222 3 J = = + − = Work done: Force-extension graph P3. N14/P1/Q17, TYS: Topic 5: MCQ 5 Area under force-extension graph is the work done P is work done on fibre by stretching force 2 1 0 -1 1 2 3 4 5 6 7 F / N x / m

National JC Work, Energy and Power 2023 - 5 - Q is work done by fibre Hence net work done on fibre = P – Q (Ans: C) Conservation of energy P4. In the process of crossing an obstacle course, a 65 kg student running at 5.0 m s−1 grabs a hanging rope of length 2.0 m, and swings out over a pit of water. He releases the rope when his speed is 2.0 m s−1. What is the angle  when he releases the rope? Using the Principle of Conservation of Energy, Loss of KE = Gain in GPE ( )( ) ( )( ) ( )( ) 221165 5.0 65 2.0 65 9.81 022 h− = − 1.07 mh= OR Using the Principle of Conservation of Energy, student’s initial KE = student’s final KE + student’s increase in GPE ( )( ) ( )( ) ( )( ) 221165 5.0 65 2.0 65 9.8122 1.07 m h h  = + = Now, 2.0cos 2.0 62 h  −=  =  2.0 m  h 2.0 - h 2.0

National JC Work, Energy and Power 2023 - 6 - P5. The top end of a spring is attached to a fixed point and a mass of 4.2 kg is attached to its lower end. The mass is released and after bouncing up and down several times it comes to rest at a distance 0.29 m below its starting point. Which row gives the gain in the gravitational potential energy of the mass Ep and the gain in the elastic potential energy of the spring Es? Ep / J Es / J A –12 +12 B –12 +6 C +12 +12 D +12 +6 The GPE obviously decreased as the final position is lower. Hence either choice A or B. We know that there must be stored energy in the spring, so it’s a matter of how much. Since the spring came a rest after some time due to resistive forces, there must be some loss of energy to the surrounding (and also heat in the spring). The gain in EPE must hence be less than the loss in GPE. Hence, choice B. (Common mistake is choosing A, thinking that mechanical energy is conserved. It is suitable to bring in an initial discussion of SHM) However if we want to verify, we can do so too. At equilibrium position, Spring force = weight = mg EPE = ½ Fx = ½ (mg)(x) = ½ loss in GPE Ans: B P6. N13/P1/Q10, TYS: Topic 5: MCQ 8 Ans: A B: KE must be 0 at bottom

National JC Work, Energy and Power 2023 - 7 - C: Total energy =120 J but total energy at middle = 150 J D: KE cant be 0 at middle. P7. An 80.0 kg sky diver jumps out of a balloon at an altitude of 1000 m and opens the parachute at an altitude of 200 m. The total retarding force on the diver is constant at 50.0 N with the parachute closed and constant at 3600 N with the parachute open. (a) What is the speed of the diver when he lands on the ground? (b) At what height should the parachute be opened so that the final speed of the sky diver when he hits the ground is 5.00 m s-1? (a) Gravitational Potential energy is converted into work done by total retarding force and kinetic energy Loss in GPE = workdone by retarding force + gain in kinetic 80.0 x 9.81 x 1000 = 50.0 x 800 + 3600 x 200 + ½ x 80.0 x v2 ➔v = 24.9 m s-1 (b) Gravitational Potential energy is converted into work done by total retarding force and kinetic energy 80.0 x 9.81 x 1000 = 50.0 x s1 + 3600 x s2 + ½ x 80.0 x 5.002, where s1 + s2 = 1000 Solving for s2 gives that the parachute should be opened at a height of 207 m. Notice what a difference a mere 7 m makes! 8. 2009 H1 P2 Q7

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National JC Work, Energy and Power 2023 - 9 - 2.89 s

National JC Work, Energy and Power 2023 - 10 - 32 m 5.