HCI Assignment 05 Work, energy and power

Work, Energy & Power 1 Name: Index No: Date: Class: Assignment 05: Work, energy and power (IP) For all the questions, take gravitational acceleration, g = 10 m s-2. 1. A simple pendulum consists of a small bob of mass 0.050 kg suspended by an inextensible string. The length of the pendulum is 1. 20 m. The bob is pulled aside for a horizontal distance of 0. 50 m to position A as shown in the diagram below and released. Assume air resistance is negligible. (a) Calculate the decrease in the gravitational potential energy of the bob as it moves from position A to position B. [3] h = 1.20 - (1.202 – 0.502) [1] = 0.1091 m Decrease in G.P.E = (0.050 kg)(10 m s-2)(0.1091 m) [1] = 0.055 J (to 2 s.f.) [1] (b) Calculate the speed of the bob when it passes through position B. [2] K.E.A + P.E.A = K.E.B + P.E.B OR Increase in K.E. = Decrease in G.P.E. 0 + 0.05455 J = ½ mv2 + 0 [Assume reference level at B.] ½ (0.050 kg) v2 = 0.05455 J [1] v = 1.5 m s-1 (to 2 s.f.) [1] (c) The diagram below shows the variation of the kinetic energy of the pendulum with time. HWA CHONG INSTITUTION Sec 3 Physics A 0.50 m 1.20 m B 0 energy time h

Work, Energy & Power 2 (i) Indicate, on the kinetic energy-time graph, the first instant when the bob returns to position A after it had been released. Label this point A. [1] (ii) Draw, on the above diagram, a. the corresponding variation of the gravitational potential energy of the pendulum with time and label it G, b. the total energy of the pendulum with time and label it T. [2] Reference: https://serpmedia.org/scigen/e1.3.html (d) Consider the pendulum bob and the earth as the system. E xplain the graph G you had drawn in (c)(ii). [2] As the pendulum bob swings down, energy from the gravitational potential store is transferred to the kinetic store and vice versa as it swings up. [1] At any point in time, the sum of energy in the gravitational potential store and the kinetic store is constant. [1] Resources: 2. Fig 2.1 shows a gymnast jumping on a trampoline. Fig 2.2 shows the variation of her speed with time during her jump. The mass of the gymnast is 48.0 kg. Ignore the effects of air resistance. Sinusoidal nature of pendulum motion: https://www.physicsclassroom.com/class/waves/lesson- 0/pendulum-motion https://www.desmos.com/calculator Animation: https://www.myphysicslab.com/pendulum/pendulum-en.html 0 energy time G T A A B, D C Fig. 2.1 gymnast trampoline s ground s: vertical displacement of gymnast’s CG

Work, Energy & Power 3 (a) The points A, B, C, D and E on Fig. 2.2 indicate the various positions of the gymnast during her motion. (i) At which position(s) did the gymnast achieve the maximum gravitational potential energy? [2] At positions A [1] and E. [1] (ii) At which position(s) did the gymnast cause the trampoline to achieve the maximum elastic potential energy? [1] At position C. [1] (b) From Fig. 2.2, determine (i) the maximum kinetic energy of the gymnast, [2] Maximum K.E. = ½ (48.0 kg)(5.0 m s-1)2 [1] = 600 J (to 2 s.f.) [1] (ii) the maximum vertical change in displacement of the gymnast after she loses contact with the trampoline. Leave your answer to 2 significant figures. [2] Maximum s = area under DE of v-t graph = ½ (1.2 – 0.7)(5.0) [1] = 1.25 m = 1.3 m (to 2 s.f.) [1] (c) (i) The change in gravitational potential energy of the gymnast when she first touches the trampoline to the lowest point of her jump is 192 J. Calculate the average power developed by the trampoline. Leave your answer to 2 significant figures. [2] Average power developed = energy transferred / time taken = (600 J + 192 J) / 0.1 s [1] = 7.9 kW (to 2 s.f.) [1] (ii) Describe the useful energy changes in of the system gymnast between the position when she first touches the trampoline to the lowest point in her jump. Consider the gymnast, the trampoline and the ground as the system. [1] Energy in gravitational potential store and kinetic store of gymnast is transferred to elastic store of trampoline. (This is B to C on the graph.) E C A 5.0 2.5 speed/ m s-1 time/s 0.5 0.6 0.7 1.2 0 B D Fig. 2.2 How would its velocity-time graph look like?

Work, Energy & Power 4 3. The diagram below shows the energy transfers in a small hydroelectric power station of 55% efficiency. Source: Cambridge O Level Physics, Heather Kennett, Tom Duncan The hydroelectric power station i s used for one hour. During the hour, the level of water in the upper lake falls by 0.050 m. The area of the upper lake is 5 .0 x 103 m2 and it is 80.0 m above the turbine. No additional water enters the lake. Calculate (a) the mass of water which leaves the lake in one hour, (density of water = 1.0 x 103 kg m-3) [2] density = mass / volume mass = (1.0 x 103 kg m-3)( 5.0 x 103 m2)(0.050 m) [1] = 250 000 kg (to 2 s.f.) [1] (b) the decrease in the gravitational potential energy of the water which leaves the lake in one hour given that, on average, the water falls 80.0 m, [2] decrease in G.P.E. = (250 000 kg)(10 m s-2)(80.0 m) [1] = 2.0 X 108 J (to 2 s.f.) [1] (c) the useful energy output of the power station in one hour, [2] efficiency = (useful energy output / total energy input) x 100 55 = [useful energy energy / (2.0 X 108 J)] x 100 [1] useful energy output = 1.1 x 108 J (to 2 s.f.) [1] (d) the average electrical power produced by the generator during the hour. [2] average electrical power = 1.1 x 108 J / 3600 s [1] = 31 kW (to 2 s.f.) [1] *Students should draw a diagram to visualize the problem.

Work, Energy & Power 5 4. A large swinging ball is used to drive a horizontal iron spike into a vertical wall. The centre of the ball falls through a vertical height of 1.6 m before striking the spike in the position as shown in the diagram. (a) The mass of the ball is 3.5 kg and the mass of the spike is 0.80 kg. Immediately after striking the spike, the ball and spike move together. (i) Show that the speed of the ball , on striking the spike, is 5.7 m s–1. [1] v = (2 X 10 X 1.6) [1] = 5.7 m s-1 (shown) (ii) Calculate the energy dissipated as a result of the collision , if the ball and the spike move together at 4.6 m s-1, after the inelastic collision. [2] K.E. before collision = ½ [3.5 × 5.6572] (also equals G.P.E. at start = 56 J) K.E. after collision = ½ [(3.5 + 0.80) × 4.62] [1] Energy dissipated = 56.0 – 45.5 = 11 J (to 2 s.f.) [1] (b) As a result of the ball striking the spike, the spike is driven a distance 7.3 × 10–2 m into the wall. Calculate the magnitude of the constant net force F acting on the spike. [3] Method 1: Using concept of work Wnet by F on spike = K.E.final – K.E.initial = 0 J – 45.5 J = - 45.5 J Fnet (s) = - 45.5 J [1] Fnet = - 45.5 J / (7.3 X 10-2 m) [1] = - 620 N (to 2 s.f.) Magnitude of F = 620 N [1] Method 2: Using equations of motion For the spike, u = 4.6 m s-1, v = 0 m s-1; a = [- (u2 / 2(s)] = [- 4.62 / 2(7.3 X 10-2)] [1] = - 144.9 m s-2 Fnet = ma = (3.5 + 0.8) × (-144.9) [1] = - 620 N (to 2 s.f.) Magnitude of F = 620 N [1] (c) The machine