# What is the energy lost to friction?

1. Oct 31, 2005

### Kumar9

The question is: A 15.7 kg block is dragged over a horizontal surface by a 68.7N force acting 17 degrees to the horizontal. The block is displaced 4.65m, and the kinetic friction coefficient is 0.3. What is the energy lost to friction?

My thinking is that it would be equal to the negative work done by the frictional force (E taken out of the system), but the computer does not recognize this as correct. What am I missing?

2. Oct 31, 2005

### Kumar9

I need to have this figured out soon; if someone could help I would greatly appreciate it.

3. Oct 31, 2005

### Andrew Mason

AM

4. Oct 31, 2005

### Kumar9

This is what I have so far:
W=(mgUk)(d)cos180
Where Uk is the coefficient. Plugging in the appropriate values, I get a value of ~-215 J, the E lost due to friction. It seems right, but clearly I'm missing something.

5. Nov 1, 2005

### Andrew Mason

I don't understand why you are using 180 degrees. The angle is 17 degrees from horizontal. The normal force perpendicular to the surface is equal and opposite to the component of gravity in that direction. What is the magnitude of that component of gravity? (Hint it is a little less than the weight).

AM