University:
East Georgia State College
Course:
PHYS 2211 | Physics I: Classical Mechanics
Academic year:

2022
Views:

97

Pages:

2
Author:

5CelestialChronicle

Dynamic Forces Assessment 1. 2. Name: _________________________ Consider the 1 kg cart on the right. There is a 0.5 N towing force pulling the cart to the right. Assume that friction is zero. Describe the motion of the cart while being pulled? How do you know? S2 The cart accelerates to the right because the forces are unbalanced. 0.5 N 1 kg For each question below one or more features of the system has been changed. Compare, DO NOT SOLVE, the acceleration of the cart in the new situation to the acceleration of the cart at the above. Justify each of your answers. S3 2 kg a. The mass of the cart is increase to 2 kg. The acceleration is cut in half because there is twice the mass. 1.0 N b. The towing force is increased to 1 N. The acceleration doubles because the unbalanced force doubles. 1.0 N c. The mass of the cart is increase to 2 kg and the towing force is increased to 1 N. The acceleration stays the same because both the force and mass double. d. The 0.5 N towing force is applied at an angle as shown at the right. The acceleration is less because the force in the direction of the acceleration is less. e. The 0.5 N towing force is applied to two 500 g carts hooked together as shown. The acceleration is the same because 500 g + 500 g = 1 kg 0.5 N 1 kg 2 kg 0.5 N 1 kg 500 g 500 g 0.5 N A 1500 kg Volkswagen is moving along the road at 20 m/s (45 mph) when the driver applies the breaks. The tires lock up and the car slides to a stop in 4 seconds. 3. In the space under the drawing draw and label (with names and values) a force diagram for the car. S1 4. Which way are the net (unbalanced) force and the acceleration on/of the car? Justify your answer. S2 Both the net unbalanced force and the acceleration are to the right. The car is moving to the left and slowing. 5. Road 15000 N friction 7500 N What is the value of the unbalanced force on the car? Show your work. S3 a = 5 m/s/s Fnet = 7500 N Earth 15000 N 6. While slowing down a bug hits the windshield and “splat”. Which was hit harder, the bug by the windshield or the windshield by the bug? Justify your answer. S5 They apply the same force on each other – so says Newton’s 3rd Law. Normal 24 N Dynamic Forces Assessment A 2.5 kg block sits on a ramp as shown at the right. 7. On the diagram draw and label a force diagram for the block as it accelerates down the ramp. S7 8. Find the value of the acceleration of the block down the ramp if there is NO friction between the block and the surface. Show your work. S3 Page 2 m = 2.5 kg 16.26 Earth 25 N into 24 N down 7N Riding in an elevator a physics student, standing on a bathroom scale, suddenly feels heavier than normal. The scale reads 690 N. 9. Next to the drawing draw and label a force diagram for the student when he feels heavier. S1 10. Which way is he accelerating? How do you know? S2 He is accelerating up – because he feels heavier. We feel the force of the elevator up on us and when it is more than the force of the Earth down we feel heavier. Scale 690 N 11. Find the acceleration of the student. Show your work. S3 Fnet = 690 – 600 = 90 N a = Fnet / m = 90 / 60 = 1.5 m/s/s Earth 600 N m = 60 kg

Phys Dynamic Forces Assessment Answers

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