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Unformatted text preview: lee (cyl327) – HW05 – gentle – (56245) 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A team of dogs drags a 83 . 4 kg sled 1 . 09 km over a horizontal surface at a constant speed. The coefficient of friction between the sled and the snow is 0 . 103. The acceleration of gravity is 9 . 8 m / s 2 . Find the work done by the dogs. Correct answer: 91 . 7605 kJ. Explanation: Let : m = 83 . 4 kg , μ = 0 . 103 , and d = 1 . 09 km . The dogs must do work to overcome fric tion. The force due to friction is F friction = μ N = μmg. The work done by the dogs traveling a dis tance 1 . 09 km is W = μmg d = (0 . 103) (83 . 4 kg) (9 . 8 m / s 2 ) (1 . 09 km) = 91 . 7605 kJ . 002 (part 2 of 2) 10.0 points Find the energy lost due to friction. Correct answer: 91 . 7605 kJ. Explanation: Since all of the work done by the dogs was done to overcome friction, the energy lost to friction is equal to the work done by the dogs. 003 (part 1 of 3) 10.0 points As shown in the figure, a block of mass 2 . 4 kg is pushed up against the vertical wall by a force of 68 N acting at 50 ◦ to the ceiling. The coefficient of kinetic friction between the block and the wall is 0 . 54. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 4 kg F 5 ◦ μ k =0 . 54 Find the work done by this force in moving the block upward by a distance 1 . 4 m. Correct answer: 72 . 9275 J. Explanation: Given : m = 2 . 4 kg , μ = 0 . 54 , θ = 50 ◦ , bardbl vector F bardbl = 68 N , and s = 1 . 4 m . The angle between vector F and the displacement is 90 ◦ − θ , so W = vector F · vectors = F s cos(90 ◦ − θ ) = F s sin θ = (68 N) (1 . 4 m) sin50 ◦ = 72 . 9275 J . 004 (part 2 of 3) 10.0 points For a force of F = 68 N, find the magnitude of the frictional force. Correct answer: 23 . 6032 N. Explanation: In the horizontal direction, N − F cos θ = 0 N = F cos θ , lee (cyl327) – HW05 – gentle – (56245) 2 so f = μ N = μF cos θ = (0 . 54) (68 N) cos 50 ◦ = 23 . 6032 N . 005 (part 3 of 3) 10.0 points Find the force F needed to keep the block moving up with a constant velocity. Correct answer: 56 . 1418 N. Explanation: m F θ v mg f μ Using the diagram above, the equation of motion in the vertical direction is F sin θ = mg + μF cos θ , where f μ = μF cos θ . Solving for F , F = mg sin θ − μ cos θ = (2 . 4 kg) (9 . 8 m / s 2 ) sin 50 ◦ − (0 . 54) cos50 ◦ = 56 . 1418 N . 006 (part 1 of 2) 10.0 points A 101 g bullet is fired from a rifle having a barrel 0 . 784 m long. Assuming the origin is placed where the bullet begins to move, the force exerted on the bullet by the expanding gas is F = a + bx − cx 2 , where a = 18100 N, b = 11700 N / m, c = 37600 N / m 2 , with x in meters....
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This note was uploaded on 02/28/2012 for the course PHY 316 taught by Professor Markert during the Fall '10 term at University of Texas.
 Fall '10
 MARKERT
 mechanics, Force, Friction, Work

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