SPH3U1 March 29, 2016 Chapter 3 Test – Forces ANSWERS Part 1 – Multiple Choice 1. 2. 3. 4. B B B E Knowledge and Understanding – 12 marks 5. 6. 7. 8. E C B C 9. 10. 11. 12. Part 2 – Diagrams A C B A Communication – 9 marks 1. The graph has a constant positive slope. As mass is decreasing at a constant rate, so acceleration will increase at the same constant rate. π (π/π 2 ) π‘ (π ) 2. The net force on Block A is to the right. The net force acting on the system is not. Block A accelerates at the same rate as the system, however it has a smaller mass. Therefore, the net force acting on Block A must have a smaller magnitude than that acting on the system. 3. The net force acting on this object is 2.0 N [R]. Therefore, the object will accelerate to the right. In other words, the velocity will change, increasing to the right. Part 3 – Problem Solving Thinking & Investigation – 18 marks 1. The maximum force will be equal to the weight of the woman and boxes. πΉπ = ππππ₯ π (950) = ππππ₯ (9.8) 96.9 ππ = ππππ₯ ππππ₯ = ππ€ππππ + ππππ₯ππ (96.9) − (85) = ππππ₯ππ 11.9 ππ = ππππ₯ππ 11.9 = 2.83 4.2 ∴ π‘βπ π€ππππ πππ πππππ¦ 2 πππ₯ππ . SPH3U1 Wednesday April 12, 2011 2. π£π = π£π + πΔπ‘ (18.6) = (0.0) + π(5.1) (18.6) =π (5.1) 3.65 π/π 2 = π πΉππΈπ = ππ πΉππΈπ = (1480)(3.65) πΉππΈπ = 5402 ∴ π‘βπ πππ‘ πππππ ππ πππππ₯ππππ‘πππ¦ 5400 π 3. a. Block B accelerates with the system, so aB = aS πΉππΈππ = ππ ππ (8.0) = (6.6 + 9.5)ππ 8.0 = ππ 6.6 + 9.5 0.497 = ππ ∴ π΅ππππ π΅ πππππππππ‘ππ ππ‘ ππππππ₯ππππ‘πππ¦ 0.50 π/π 2 b. The force of tension is responsible for accelerating Block A, which also accelerates at the same rate as the system. πΉππΈππ΄ = ππ΄ ππ΄ πΉπ = ππ΄ ππ΄ πΉπ = (6.6)(0.497) πΉπ = 3.28 ∴ π‘βπ π‘πππ πππ ππ π‘βπ π π‘ππππ ππ ππππππ₯ππππ‘πππ¦ 3.3 π 4. πΉππΈπ = ππ (75.0) = (340)π (75.0) =π (340) 0.221 π/π 2 = π π£π2 = π£π2 + 2πΔπ SPH3U1 Wednesday April 12, 2011 π£π = √π£π2 + 2πΔπ π£π = √(0.0)2 + 2(0.221)(48) π£π = 4.61 ∴ π‘βπ πππππ π ππππ ππ ππππππ₯ππππ‘πππ¦ 4.6 π/π 5. πΉππΈπ = ππ πΉππΈπ = πΉπππ − πΉπ πΉπππ − πΉπ = ππ πΉπππ = ππ + πΉπ πΉπππ = (5.0)(1.4) + (5.0)(9.8) πΉπππ = 56 ∴ π‘βπ πππππ πππππππ π πππππ ππ 56 π Part 4 – Short Answer Application – 10 marks 1. Answers may vary. 2. “Action at a distance” forces such as the force of gravity, magnetic force, and electric force can all accelerate objects without contact. 3. According to Newton’s First Law, the object must be experiencing uniform motion. It may be moving at a constant velocity. 4. Answers may vary. ο· Weight is the force of gravity. In space, the force of gravity is present, so objects still have weight ο· Humans cannot perceive the force of gravity, so this feeling of weightlessness is the absence of the normal force ο· The normal force is absent (or diminished) because the environment (i.e. space capsule) is accelerating with the person. ο· This relates to Newton’s 3rd Law because the normal force is a “reaction” force that results when the force of gravity pulls an object down onto on a stationary surface.