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Unformatted text preview: chaney (glc568) Work and Energy murthy (21118) 1 This print-out should have 54 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 5) 10.0 points A 18 . 7 kg block is dragged over a rough, hor- izontal surface by a constant force of 185 N acting at an angle of angle 30 . 4 above the horizontal. The block is displaced 67 . 6 m, and the coefficient of kinetic friction is 0 . 22. The acceleration of gravity is 9 . 8 m / s 2 . 18 . 7 kg = 0 . 22 1 8 5 N 3 . 4 Find the work done by the 185 N force. Correct answer: 10786 . 6 J. Explanation: Consider the force diagram F mg n f k Work is vector W = vector F vectors , where vectors is the distance traveled. In this problem vectors = 5 x is only in the x direction. W F = F x s x = F (cos ) s x = (185 N)(cos30 . 4 )(67 . 6 m) = 10786 . 6 J . 002 (part 2 of 5) 10.0 points Find the magnitude of the work done by the force of friction. Correct answer: 1333 . 18 J. Explanation: To find the frictional force, F friction = N , we need to find N from vertical force balance. Note: N is in the same direction as the y component of F and opposite the force of gravity. Thus, F sin + N = mg . so that N = mg- F sin . Thus the friction force is vector F friction =- N x =- ( mg- F sin ) x. and the work done by friction is W = vector F friction vectors =-| F f || s | =- ( mg- F sin ) s x =- (0 . 22)[(18 . 7 kg)(9 . 8 m / s 2 )- (185 N) sin30 . 4 ](67 . 6 m) =- 1333 . 18 J | W | = 1333 . 18 J . 003 (part 3 of 5) 10.0 points What is the sign of the work done by the frictional force? 1. negative correct 2. zero 3. positive Explanation: See Part 2. 004 (part 4 of 5) 10.0 points Find the work done by the normal force. Correct answer: 0 J. Explanation: Since the normal force points in the y di- rection and vectors is in the x direction W N = vector N vectors = N y x = 0 005 (part 5 of 5) 10.0 points What is the net work done on the block? Correct answer: 9453 . 41 J. Explanation: chaney (glc568) Work and Energy murthy (21118) 2 The net work done on the body is the al- gebraic sum of the work done by the external force F and the work done by the frictional force W net = W F + W = 10786 . 6 J + (- 1333 . 18 J) = 9453 . 41 J . 006 10.0 points The amount of work done by two boys who apply 340 N of force in an unsuccessful at- tempt to move a stalled car is 1. 680 N-m. 2. 0. correct 3. 340 N. 4. 680 N. 5. 340 N-m. Explanation: W = F d , so with an unsuccessful attempt to move the car, no work was done. 007 10.0 points A force acting on an object does no work if 1. the object accelerates....
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