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Unformatted text preview: chaney (glc568) – Work and Energy – murthy – (21118) 1 This printout should have 54 questions. Multiplechoice 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 μ = 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 θ mg n f k Work is vector W = vector F · vectors , where vectors is the distance traveled. In this problem vectors = 5 ˆ x is only in the ˆ x direction. ⇒ W F = F x s x = F (cos θ ) 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 = μ 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 θ + N = mg . so that N = mg F sin θ . Thus the friction force is vector F friction = μ N ˆ x = μ ( mg F sin θ )ˆ x. and the work done by friction is W μ = vector F friction vectors = F f  s  = μ ( mg F sin θ ) s x = (0 . 22)[(18 . 7 kg)(9 . 8 m / s 2 ) (185 N) sin30 . 4 ◦ ](67 . 6 m) = 1333 . 18 J  W μ  = 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 ˆ y di rection and vectors is in the ˆ x direction W N = vector N · vectors = N ˆ y · ˆ 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 μ = 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 Nm. 2. 0. correct 3. 340 N. 4. 680 N. 5. 340 Nm. 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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 Spring '08
 MURTHY
 Energy, Force, Friction, Work, Sin, Cos, Correct Answer, Chaney

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