This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: pampalone (tp8327) HW10 swaminathan (10500) 1 This printout should have 18 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 skier of mass 72 . 7 kg is pulled up a slope by a motordriven 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 ....
View
Full
Document
 Spring '08
 KEN
 Mass

Click to edit the document details