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Jan 9, 2024

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Assessment Questions Task 1: Motion problems Do two (of three) problems. Ignore any frictional forces not specified in the problem. Please show all steps of your solutions. 1. Ancilla (mass 57 kg) goes skydiving. At one point in her descent, the force of air resistance on Ancilla and her parachute is 670 N [up]. (a) What is the force of gravity on Ancilla? m = 57 kg g = 9.8 N/kg F g = mg F g = (57) (9.8) F g = 558.6 N [down] Therefore, the force of gravity on Ancilla is 558.6 N [down]. (b) What is the net force on Ancilla? F down = 558.6 F up = 670 F net = F up - F down F net = 670 - 558.6 F net = 111.4 N [up] Therefore, the net force on Ancilla is 111.4 N [up]. (c) What is Ancilla’s acceleration? F net = 111.4 N m = 57 kg a = F net / m a = 111.4 N / 57 kg a = 1.95 m/s 2 [up] Therefore, Ancilla’s acceleration is 1.95 m/s 2 [up]. 2. Ancilla (mass 57 kg) goes skating with Wayne Gretzky (mass 83 kg). At one point when both are standing motionless on the ice, she gives Wayne a push of 200 N [S]. (a) What is Wayne’s acceleration? (b) What force does Wayne exert on Ancilla? (c) What is Ancilla’s acceleration? TVO ILC SPH4C Learning Activity 3.5 Assessment Questions Copyright © 2021 The Ontario Educational Communications Authority. All rights reserved. 1

3. The following graph shows the relationship between velocity and time for Ancilla’s car (mass 1120 kg). (a) What is the car’s acceleration? a =  v /  t a = 7.5 m/s / 15 s a = 0.5 m/s 2 Therefore, the car accelerates at 0.5 m/s 2 [W] (b) What is the car’s displacement? d = l x w d = 15 m/s [W] x 30s d = 450 m/s 2 Therefore, the cars displacement is 450 m/s 2 [W] (c) What is the net force on the car? F net = ma F net = (1120 kg) (0.5 m/s 2 ) F net = 560 N Therefore, the net force on the car is 560 N [W] Velocity vs. Time For Ancilla’s Car 15 12 9 6 3 0 0 5 10 15 20 25 30 Time (s) Velocity (m/s) [W] TVO ILC SPH4C Learning Activity 3.5 Assessment Questions Copyright © 2021 The Ontario Educational Communications Authority. All rights reserved. 2

Task 2: Force problems Do one (of two) problems. Please show all steps of your solution. 4. Ancilla applies a force of 200 N [E] horizontally to pull a 38 kg sled across the snow at a constant velocity for 200 m. (a) Draw an FBD of the sled F = mg F = 38 kg x 9.8 F = 372.4 N F k F = 200 N [E] F g = 372.4 N (b) What is the magnitude of the force of friction on the sled? m = 38 kg a = 0 m/s 2. f = 200N f k =? F net = ma = f - fk 0 = f - f k f = f k Therefore, the force of friction is 200N [E] (c) What is the magnitude of the force of gravity on the sled? m = 38 kg g = 9.8 m/s 2 F g = mg F g = (38 kg) (9.8m/s 2 ) F g = 372.4 N Therefore, the magnitude of the force of gravity on the sled is 372.4 N (d) What is the magnitude of the normal force on the sled? Normal force = force of gravity F n = F g 372.4 N = 372.4 N

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Therefore, the magnitude of normal force on the sled is 372.4 N (e) What is the coefficient of kinetic friction between the sled and the snow? F k = 200 N F n = 372.4 μk = F k / F n μk = 200 N / 372.4 N μk = 0.537 Therefore, the coefficient of kinetic friction is 0.537 (f) How much work did Ancilla do on the sled? F = 200N d = 200m W = F  d W = (200N) (200m) W = 40,000 J Therefore, Ancilla did about 40,000 J of work on the sled 5. Dimas Pyrros (four-time Olympic medalist) raises a 215 kg barbell in two stages. In the first stage he raises it to his shoulders, a height of 1.60 m, by exerting a maximum force of 3000 N. In the second stage he raises the barbell above his head at a constant speed to a maximum height of 2.10 m. He then holds it above his head for 2.0 s before dropping it to the mat. (a) What is the magnitude of the force of gravity on the barbell? (b) How much work did Pyrros do on the barbell to raise it to a height of 1.60 m? (c) What is the magnitude of the force which the weightlifter must apply to raise the barbell the last 0.60 m? (d) How much work did the weightlifter do on the barbell to raise it the last 0.60 m? (e) Draw an FBD of the barbell while it is held aloft for 2.0 s (f) How much work did the weightlifter do on the barbell during those 2.0 s?

Task 3: Power and efficiency problems Do two (of three) problems. Please show all steps of your solutions. 6. A 60 W light bulb is left on for 8.0 h. (a) How much electrical energy will it use? P = 60W  t (in seconds) = 8.0h = 22800s  E =? P =  E /  T  E = (60w) (22800s)  E = 1368000 J Therefore, the lightbulb uses 1.4 x 10 6 J or 1.4 MJ of energy (b) If the light bulb is only 15% efficient, how much electrical energy will be converted into radiant energy (light)? P = 60W t = 8 hours = 22800s efficiency = 15% P out = P x efficiency P out = 60 W x 15% P out = 60 W x 0.15 P out = 9.0 W  E = P x t  E = 9.0 x 22800  E = 205,200 J Therefore, 205,200 J of electrical energy will be converted into radiant energy (c) What happens to the energy that is not converted electrical energy? Energy can’t be created or destroyed so it is always conserved. Energy that is not converted into electrical energy is converted into other types of energy. Such as kinetic energy or potential energy. Energy that is not converted into electrical energy is converted into heat energy. 7. A baseball bat has 120 J of kinetic energy as it hits a baseball ( m = 0.145 kg). (a) If the baseball bat transfers 70 J of its kinetic energy to the baseball, what is the efficiency of the energy transformation? (b) Find the speed of the baseball immediately after it has been hit.

8. A skier (mass = 65 kg) is at rest at the top of a 75 m hill. She pushes off and when she reaches the bottom of the hill she has a speed of 25 m/s. (a) What type of energy transformation has taken place? When the skier is at the top of the hill, she has potential energy since she is at rest. And then as she goes down and reaches the bottom, she gains kinetic energy since she’s in motion as she pushes off. (b) Find the efficiency of the energy transformation. m = 65 kg g = 9.8 m/s 2 h = 75m vi = 0/ms 2 vf = 25m/s 2 Potential Energy E p = mgh E p = (65) (9.8) (75) E p = 47,775 J Kinetic Energy E k = ½ mv 2 E k = ½ (65 kg) (25m/s) 2 E k = ½ (65) (625 m/s 2 ) E k = 20,312.5 J Efficiency E = Kinetic Energy / Potential x 100 E = 20312.5 / 47775 x 100 E = (0.42517) (100) E = 42.517 % Therefore, the efficiency of the energy transformation is 42.5% (c) What happened to the energy that was lost? There is roughly 27462.5 J of energy lost. Potential energy (47775 J) - Kinetic energy (20312.5 J). The lost energy was converted into heat energy. This is due to the air friction and hill friction. The energy is not lost, it is just transformed into another form of energy. Task 4: Energy problems Do one (of two) problem. Please show all steps of your solution.

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9. Skateboarder Toby (mass 103 kg) starts his descent from the top of a 6.0 m half-pipe which has negligible friction. (a) What is Toby’s gravitational potential energy at his maximum height? m = 103 kg g = 9.8m/s 2 h = 6m F g = mgh F g = (103) (9.8m/s 2 ) (6m) F g = 6056.4 J Therefore, the gravitational potential energy at his max height is 6056.4 J (b) What is Toby’s gravitational potential energy at a height of 3.5 m? m = 103 kg g = 9.8 m/s 2 h = 3.5m F g = mgh F g (103) (9.8m/s 2 ) (3.5m) F g = 3532.9 J Therefore, the gravitational potential energy at height of 3.5m is 3532.9 J (c) As Toby descends the pipe, what energy transformation occurs? As Toby descends the pipe, gravitational potential energy is transformed into kinetic energy. (d) What is Toby’s speed at the height of 3.5 m? KE = decrease in potential energy ½ mv 2 = 6056.4 - 3532.9 ½ (103) V 2 = 2523.5 v 2 = 2523.5 / 51.5 v 2 = 49 v = 7.0 m/s Therefore, at the height of 3.5m, Toby’s speed is 7.0 m/s 10. Please read Problem #5 again. (a) What is the gravitational potential energy of the barbell at the intermediate height of 1.60 m? (b) What is the gravitational potential energy of the barbell at its maximum height?

(c) As Pyrros drops the barbell, what energy transformation occurs? (d) As the barbell passes his shoulders, at the intermediate height of 1.60 m, what is its speed, assuming it starts from rest on the ground?

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