This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 = 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 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....
View
Full
Document
This note was uploaded on 10/22/2008 for the course PHY 260 taught by Professor Murthy during the Spring '08 term at Kentucky.
 Spring '08
 MURTHY
 Energy, Work

Click to edit the document details