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2007f_f

# 2007f_f - Physics 2101 Final Exam Form A Name SOLUTIONS...

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Physics 2101, Final Exam, Form A December 11, 2007 Name: SOLUTIONS Section: (Circle one) 1 (Rupnik, MWF 7:40am) 2 (Rupnik, MWF 9:40am) 3 (Gonz´ alez, MWF 2:40pm) 4 (Pearson, TTh 9:10am) 5 (Pearson, TTh 12:10pm) Please turn OFF your cell phone and MP3 player! Feel free to detach, use and keep the formula sheet. No other reference material is allowed during the exam. You may use scientific or graphing calculators. For the questions, no work needs to be shown (there is no partial credit). For the problems, please write as much as you can explaining to us your reasoning. Lonely right answers will not receive full credit, lonely wrong answers will receive no credit at all. Please carry units through your calculations when needed, lack of units will result in a loss of points. GOOD LUCK! 1

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Question 1 (6 pts) A man pulls a crate of weight equal to 100 N from the bottom to the top of a frictionless 30 slope which is 5 m high, making the crate move with a constant speed . (a) (2pts) What is the work done by gravity on the crate? The work done by gravity is equal to minus the change in gravitational potential energy, W g = - Δ U g = - Mgh = 100N × 5m = - 500J. It is negative, since the gravitational force points down, and the crate is moving upwards along the incline. -500 J -250 J 0 250 J 500 J (b) (1pt) What is the work done by the man on the crate? Since the crate is moving with constant velocity, the work done by external forces must be zero (work-energy theorem). The normal force does not do any work since it is perpendicular to the crate’s motion. The work done by the man’s force must then be equal to the work done by gravity, calculated above: 0 = W g + W m = - 500J + W m W m = +500J. -500 J -250 J 0 250 J 500 J (c) (1 pt) What is the work done by the normal force on the crate? The normal force does not do any work since it is perpendicular to the crate’s motion. -500 J -250 J 0 250 J 500 J (d) (2 pts) If there were friction between the crate and the sloped surface, and the man kept pulling up the crate with constant speed, the magnitude of the work done by gravity on the crate would be... larger than the same as smaller than ... the work calculated in (a). Gravity’s work is always equal to minus the change in gravitational potential energy, friction or not. If there were friction, the man’s work would have to be larger, to compensate for both the work done by gravity and friction forces, since those are both negative. 2
Question 2 (5 pts) A steel ball of mass m is fastened to a massless cord of length L , forming a simple pendulum, fixed at the far end. The ball is released from rest when the cord is horizontal. (a) (2 pts) What is the speed of the ball as the ball moves through the lowest point at the bottom? p 2 g/L 2gL p gL/ 2 p g/ (2 L ) Using conservation of energy between initial position at rest and final position at the bottom oof the swing, we have: Δ KE + Δ U = 1 2 mv 2 - mgL = 0 v 2 = 2 gL (b) (2 pts) What is the tension in the cord as the ball moves through the lowest point at the bottom?

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