# hw6 - patel(kp9583 – HW6 – Mackie –(20211 1 This...

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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 ω 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 μ = 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 θ m g 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 = (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 = μ 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 θ + N = m g and N = m g − F sin θ . Thus the friction force is vector F friction = − μ N ˆ x = − μ ( m g − F sin θ ˆ x patel (kp9583) – HW6 – Mackie – (20211) 2 and the work done by friction is W μ = vector F friction vectors = −| F f | | s | = − μ ( m g − F sin θ ) s x = − (0 . 117)[(16 . 3 kg)(9 . 8 m / s 2 ) − (81 N) sin25 . 8 ◦ ](6 . 23 m) = − 90 . 7393 J | W μ | = 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 ˆ y di- rection and vectors is in the ˆ x direction W N = vector N · vectors = N ˆ y · ˆ 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...

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