HW05-solutions

# HW05-solutions - Le (tl8426) – HW05 – gentle –...

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Unformatted text preview: Le (tl8426) – HW05 – gentle – (56245) 1 This print-out should have 27 questions. Multiple-choice 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 114 kg sled 1 . 92 km over a horizontal surface at a constant speed. The coefficient of friction between the sled and the snow is 0 . 124. The acceleration of gravity is 9 . 8 m / s 2 . Find the work done by the dogs. Correct answer: 265 . 983 kJ. Explanation: Let : m = 114 kg , μ = 0 . 124 , and d = 1 . 92 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 . 92 km is W = μmg d = (0 . 124) (114 kg) (9 . 8 m / s 2 ) (1 . 92 km) = 265 . 983 kJ . 002 (part 2 of 2) 10.0 points Find the energy lost due to friction. Correct answer: 265 . 983 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 3 . 1 kg is pushed up against the vertical wall by a force of 78 N acting at 47 ◦ to the ceiling. The coefficient of kinetic friction between the block and the wall is 0 . 6. The acceleration of gravity is 9 . 8 m / s 2 . 3 . 1 kg F 4 7 ◦ μ k =0 . 6 Find the work done by this force in moving the block upward by a distance 1 . 4 m. Correct answer: 79 . 8639 J. Explanation: Given : m = 3 . 1 kg , μ = 0 . 6 , θ = 47 ◦ , bardbl vector F bardbl = 78 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 = (78 N) (1 . 4 m) sin47 ◦ = 79 . 8639 J . 004 (part 2 of 3) 10.0 points For a force of F = 78 N, find the magnitude of the frictional force. Correct answer: 31 . 9175 N. Explanation: In the horizontal direction, N − F cos θ = 0 N = F cos θ , Le (tl8426) – HW05 – gentle – (56245) 2 so f = μ N = μF cos θ = (0 . 6) (78 N) cos47 ◦ = 31 . 9175 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: 94 . 3024 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 = (3 . 1 kg) (9 . 8 m / s 2 ) sin 47 ◦ − (0 . 6) cos47 ◦ = 94 . 3024 N . 006 (part 1 of 2) 10.0 points A 126 g bullet is fired from a rifle having a barrel 0 . 571 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 = 8130 N, b = 6510 N / m, c = 11100 N / m 2 , with x in meters....
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## This note was uploaded on 12/05/2011 for the course PHY 301 taught by Professor Swinney during the Fall '07 term at University of Texas at Austin.

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HW05-solutions - Le (tl8426) – HW05 – gentle –...

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