sol7 - HW Solutions # 7 - 8.01 MIT - Prof. Kowalski...

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HW Solutions # 7 - 8.01 MIT - Prof. Kowalski Potential Energy Curves, Momentum and Center of Mass. 1 ) 7.61 We use conservation of energy K i + U i + W other = E i + W other = E f = K f + U f (1) 1 2 mv 2 1 + mgy 1 + W other = 1 2 mv 2 2 + mgy 2 (2) I will measure the gravitation potential energy with respect to the ground. a ) First obtain the kinetic energy - denote it as K h -asifhestepped o± the platform: When he stepped of a distance h d, W other =0 because of negligible air resistance the energy conservation implies: 0+ mgh = K h +0 (3) For the case he slides down the pole W other after moving down the distance d is just f ave d cos 180 o (force opposes the motion Wis < 0): W other = f ave d (4) The next step is to calculate Kinetic energy when he slides down the pole and as it’s said make it equal to K h : mgd + W other = K d (5) Combining equation (4) with equations (2) and (3) and K d = K h : mgd f ave d = mgh (6) Therefore: f ave = mg (1 h d ) 1
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When h = d, f = 0 ; as expected: when there is no friction sliding down the pole has the same eFect as stepping oF the platform. And when h=0 we expect f ave = mg which is consistent with what we derived. b ) Plugging the numbers give in the problem into the boxed equa- tion: f ave =75 . 80 × (1 1 2 . 5 ) = 441 N . (7) c ) Let’s denote v at position y as v ( y ) Again setting up the equation (1) -energy conservation- : v 1 =0 y 1 = dW other = f ave ( d y )( 8 ) v 2 = v ( y ) y 2 = y (9) 0+ mgd mgf ave ( d y )= 1 2 mv ( y ) 2 + mgy (10) 1 2 mv ( y ) 2 =( mg f ave )( d y ) (11) Using the boxed equation for f ave we get: 1 2 mv ( y ) 2 = mg ( h d )( d y mgh (1 y d ) (12) ±rom which v ( y ± 2 gh (1 y d ) When y ,v = 2 , which is the original condition. When y = d, v = 0: the ²rman is at the top of the pole.
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This note was uploaded on 04/22/2011 for the course PHYS 1441 taught by Professor White during the Spring '08 term at UT Arlington.

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sol7 - HW Solutions # 7 - 8.01 MIT - Prof. Kowalski...

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