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Unformatted text preview: garcia (jg46766) – Problem Set 5 – lacosse – (S2107) 1 This printout should have 36 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 830 N. As the elevator moves up, the scale reading increases to 944 N, then decreases back to 830 N. The acceleration of gravity is 9 . 8 m / s 2 . Find the acceleration of the elevator. Correct answer: 1 . 34602 m / s 2 . Explanation: When the elevator is at rest, your true weight is being read, so m = W g At any time, the reading on the scale is equiv alent to the normal force N pushing up on the student. N a mg While the elevator is moving up, the accel eration is directed upward, and by Newton’s Second Law, F net = ma up = Σ F up − Σ F down W g a up = N 1 −W a up = g ( N 1 −W ) W = (9 . 8 m / s 2 )(944 N − 830 N) 830 N 002 (part 2 of 2) 10.0 points As the elevator approaches 74th floor the scale reading drops as low as 792 N. What is the acceleration of the elevator? Correct answer: − . 448675 m / s 2 . Explanation: N a mg While the elevator is slowing down, the ac celeration is directed downward, and by New ton’s Second Law, F net = ma down = Σ F up − Σ F down W g a down = N 2 −W a down = g ( N 2 −W ) W = (9 . 8 m / s 2 )(792 N − 830 N) 830 N = − . 448675 m / s 2 003 10.0 points Consider a man standing on a scale which is placed in an elevator. When the elevator is stationary, the scale reading is S s . Scale Find S up , the scale reading when the el evator is moving upward with acceleration vectora = 1 5 g ˆ , in terms of S s . 1. S up = 7 5 S s garcia (jg46766) – Problem Set 5 – lacosse – (S2107) 2 2. S up = 2 3 S s 3. S up = 5 4 S s 4. S up = 4 5 S s 5. S up = 0 m/s 2 6. S up = 3 5 S s 7. S up = 6 5 S s correct 8. S up = 5 6 S s 9. S up = 3 2 S s 10. S up = 7 6 S s Explanation: Basic Concepts: Newton’s 2nd law summationdisplay vector F = mvectora. (1) Solution We consider the forces acting on the man . Taking up (ˆ ) as positive, we know that mg acts on the man in the downward ( − ˆ ) di rection. The only other force acting on the man is the normal force vector S s from the scale. By the law of action and reaction, the force on the scale exerted by the man ( i.e. , the scale reading) is equal in magnitude but opposite in direction to the vector S s vector. Initially, the el evator is moving upward with constant speed (no acceleration) so S s − mg = 0 , or S s = mg as we would expect. Call the scale reading for this part S up (in units of Newtons). Consider the free body diagram for each the case where the elevator is accelerating down (left) and up (right). The man is represented as a sphere and the scale reading is represented as S ....
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This note was uploaded on 03/28/2011 for the course PHYSICS 101 taught by Professor Carter during the Spring '11 term at University of Texas.
 Spring '11
 carter
 Physics

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