hw6 - patel (kp9583) HW6 Mackie (20211) 1 This print-out...

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Unformatted text preview: patel (kp9583) HW6 Mackie (20211) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A 1 kg object attached to a horizontal string moves with constant speed in a circle on a frictionless horizontal surface. The kinetic energy of the object is 77 J and the tension in the string is 465 N. Find the radius of the circle. Correct answer: 0 . 331183 m. Explanation: Let : m = 1 kg , K = 77 J , and T = 465 N . T N m g r The kinetic energy is K = 1 2 m v 2 . The centripital force is supplied by the string: T = F c = m v 2 r r = m v 2 T = 2 K T = 2 (77 J) 465 N = . 331183 m . keywords: 002 (part 1 of 5) 10.0 points A 16 . 3 kg block is dragged over a rough, hor- izontal surface by a constant force of 81 N acting at an angle of angle 25 . 8 above the horizontal. The block is displaced 6 . 23 m, and the coefficient of kinetic friction is 0 . 117. 16 . 3 kg = 0 . 117 8 1 N 2 5 . 8 Find the work done by the 81 N force. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 454 . 328 J. Explanation: Consider the force diagram F m g 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 = (81 N) cos25 . 8 (6 . 23 m) = 454 . 328 J . 003 (part 2 of 5) 10.0 points Find the magnitude of the work done by the force of friction. Correct answer: 90 . 7393 J. Explanation: To find the frictional force, F friction = N , we need to find N from vertical force balance. N is in the same direction as the y component of F and opposite the force of gravity, so F sin + N = m g and N = m g F sin . Thus the friction force is vector F friction = N x = ( m g F sin x patel (kp9583) HW6 Mackie (20211) 2 and the work done by friction is W = vector F friction vectors = | F f | | s | = ( m g F sin ) s x = (0 . 117)[(16 . 3 kg)(9 . 8 m / s 2 ) (81 N) sin25 . 8 ](6 . 23 m) = 90 . 7393 J | W | = 90 . 7393 J . 004 (part 3 of 5) 10.0 points What is the sign of the work done by the frictional force? 1. zero 2. positive 3. negative correct Explanation: 005 (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 . 006 (part 5 of 5) 10.0 points What is the net work done on the block?...
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hw6 - patel (kp9583) HW6 Mackie (20211) 1 This print-out...

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