Internal Energy and kinetic friction

[hats_06]

Homework Statement

Please help, I've been struggling with this question for so long! We haven't done internal energy so I am 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]

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

 

[hats_06]

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]

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.