hw sol 5 - Solutions to Homework #5 1) To work this...

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1) To work this problem, consider the fact it will require just as much energy to stop the car as the car already has in kinetic energy. If F is the friction force, d is the stopping distance, and v is the initial velocity, then: Required stopping energy: Car’s kinetic energy: d F E = 2 2 1 mv E = So, to conserve energy, 2 2 1 mv d F = An easy way to do these comparison problems is to take the equations of motion for each case and then divide one entire equation by the other: 2 2 2 1 2 1 2 1 2 1 mv mv d F d F = now everything cancels out except 2 2 2 1 2 1 v v d d = since the friction force and the masses are the same. Then we know d 1 , v 1 , and v 2 , so: 135 2 1 2 2 1 2 = = v v d d meters. 2) To conserve energy, the big piston must do the same amount of work that is being fed into the little piston. So, Little piston’s work: Big piston’s work: l l d F W = b b d F W = Setting them equal: b b l l d F d F = Then we know F l
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This note was uploaded on 02/13/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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hw sol 5 - Solutions to Homework #5 1) To work this...

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