HW 10

HW 10 - pampalone(tp8327 – HW10 – swaminathan –(10500...

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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 , θ = 37 . 2 ◦ , and d = 63 . 5 m . Δ K = 0 since the speed is constant. The skier rises a vertical distance of Δ y = d sin θ . The work done against gravity is W nc = Δ U = mg Δ y = mg d sin θ = (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 Δ v = Δ d Δ t , so the power output is P = W nc Δ t = W nc Δ v Δ 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 , μ = 0 . 576 , θ = 62 . 4 ◦ , and Δ 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 θ- N = 0 , so that N = F cos θ Since the block moves with constant velocity, summationdisplay F y = F sin θ- mg- f k = 0 F sin θ- mg- μ F cos θ = 0 F (sin θ- μ cos θ ) = mg F = mg sin θ- μ cos θ = (3 . 01 kg) (9 . 8 m / s 2 ) sin 62 . 4 ◦- . 576 cos 62 . 4 ◦ = 47 . 6277 N Thus W F = ( F sin θ ) (Δ 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 ◦ )Δ 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 ◦ )Δ y = 0 J ....
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HW 10 - pampalone(tp8327 – HW10 – swaminathan –(10500...

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