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Unformatted text preview: Hood, Charles – Homework 9 – Due: Nov 9 2006, 11:00 pm – Inst: Jan Opyrchal 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A 19 . 1 kg block is dragged over a rough, hor izontal surface by a constant force of 89 . 2 N acting at an angle of 30 . 8 ◦ above the horizon tal. The block is displaced 92 . 7 m, and the coefficient of kinetic friction is 0 . 137. The acceleration of gravity is 9 . 8 m / s 2 . 19 . 1 kg μ = 0 . 137 8 9 . 2 N 3 . 8 ◦ Find the work done by the force of friction. Correct answer: 1797 . 11 J. Explanation: Consider the force diagram F θ mg n f k To find the frictional force, F friction = μ N , we need to find N from vertical force balance. Note that 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 ~ F friction = μ N ˆ ı = μ ( mg F sin θ )ˆ ı. The work done by friction is then W μ = ~ F friction · ~s = f μ  s  = μ ( mg f μ sin θ ) s x = . 137[(19 . 1 kg)(9 . 8 m / s 2 ) (89 . 2 N) sin30 . 8 ◦ ](92 . 7 m) = 1797 . 11 J . keywords: 002 (part 1 of 1) 10 points A 2 . 18 kg block is pushed 1 . 69 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 69 . 7 ◦ with the horizontal....
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 Fall '08
 KEN
 Force, Mass, Work, kg, Hood, KF

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