physics3

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Lambton College *

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SPH4C

Subject

Physics

Date

Jan 9, 2024

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docx

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8

Uploaded by ProfessorDiscoveryWolverine30

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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