Internal Energy and kinetic friction

hats_06 · Apr 11, 2009

1. The problem statement, all variables and given/known data

Please help, Ive been struggling with this question for so long! We havent done internal energy so im not sure even where to really begin?

A 10.0 kg block is dragged over a rough, horizontal surface by a 76.0 N force acting at 20.0° above the horizontal. The block is displaced 3.50 m, and the coefficient of kinetic friction is 0.300.

What is increase in internal energy of the block-surface system due to friction?

Doc Al · Apr 11, 2009

The change in internal energy equals the work done against friction.

hats_06 · Apr 11, 2009

so if Ff= Uk.N...Then since Uk = 0.3 and N = 76.sin 20
So Ff = 7.8, Hence W = 7.8 x 3.5 = 27.3? I tried that answer but it didnt work :S

Doc Al · Apr 11, 2009

hats_06 said: ↑

so if Ff= Uk.N...Then since Uk = 0.3 and N = 76.sin 20

First you must correctly solve for the normal force. 76sin20 is the vertical component of the applied force, not the normal force.

To find the normal force, analyze all the vertical forces that act on the block. Since the block doesn't accelerate vertically, vertical forces must sum to zero. Hint: There are three forces acting on the block that have vertical components; the normal force is one of them.

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Internal Energy and kinetic friction

hats_06

Doc Al

Staff: Mentor

hats_06

Doc Al

Staff: Mentor

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