HW 10 - pampalone (tp8327) HW10 swaminathan (10500) 1 This...

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Unformatted text preview: pampalone (tp8327) HW10 swaminathan (10500) 1 This print-out should have 18 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 skier of mass 72 . 7 kg is pulled up a slope by a motor-driven cable. The acceleration of gravity is 9 . 8 m / s 2 . How much work is required to pull him 63 . 5 m up a 37 . 2 slope (assumed to be fric- tionless) at a constant speed of 2 . 48 m / s? Correct answer: 27 . 3528 kJ. Explanation: Given : m = 72 . 7 kg , = 37 . 2 , and d = 63 . 5 m . K = 0 since the speed is constant. The skier rises a vertical distance of y = d sin . The work done against gravity is W nc = U = mg y = mg d sin = (72 . 7 kg) ( 9 . 8 m / s 2 ) (63 . 5 m) (sin37 . 2 ) parenleftbigg 1 kJ 1000 J parenrightbigg = 27 . 3528 kJ . 002 (part 2 of 2) 10.0 points What power must a motor have to perform this task? Correct answer: 1 . 43199 hp. Explanation: The velocity is v = d t , so the power output is P = W nc t = W nc v d = (27 . 3528 kJ) (2 . 48 m / s) 63 . 5 m parenleftbigg 1 hp . 746 kW parenrightbigg = 1 . 43199 hp . 003 (part 1 of 4) 10.0 points A 3 . 01 kg block is pushed 1 . 47 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 62 . 4 with the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . 3 . 01 kg F 6 2 . 4 If the coefficient of kinetic friction between the block and wall is 0 . 576, find the work done by F . Correct answer: 62 . 0455 J. Explanation: Given : m = 3 . 01 kg , = 0 . 576 , = 62 . 4 , and y = 1 . 47 m . F 6 2 . 4 v mg f k N pampalone (tp8327) HW10 swaminathan (10500) 2 The block is in equilibrium horizontally, so summationdisplay F x = F cos - N = 0 , so that N = F cos Since the block moves with constant velocity, summationdisplay F y = F sin - mg- f k = 0 F sin - mg- F cos = 0 F (sin - cos ) = mg F = mg sin - cos = (3 . 01 kg) (9 . 8 m / s 2 ) sin 62 . 4 - . 576 cos 62 . 4 = 47 . 6277 N Thus W F = ( F sin ) ( y ) = (47 . 6277 N) (sin62 . 4 )(1 . 47 m) = 62 . 0455 J . since 1 J = 1 kg m 2 / s 2 . 004 (part 2 of 4) 10.0 points Find the work done by the force of gravity. Correct answer:- 43 . 3621 J. Explanation: W g = mg (cos 180 ) y = (3 . 01 kg) (9 . 8 m / s 2 ) (cos180 ) (1 . 47 m) =- 43 . 3621 J . 005 (part 3 of 4) 10.0 points Find the work done by the normal force be- tween the block and the wall. Correct answer: 0 J. Explanation: W N = ( N cos 90 ) y = 0 J ....
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HW 10 - pampalone (tp8327) HW10 swaminathan (10500) 1 This...

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