# Energy Lost due to Friction

1. Feb 28, 2012

### shrutij

1. The problem statement, all variables and given/known data
A 15.6 kg block is dragged over a rough, horizontal surface by a 68.9 N force acting at 19.7 degrees above the horizontal. The block is displaced 4.59 m, and the coefficient of kinetic friction is 0.286. Find the work done by the 68.9 N force.

How much energy is lost due to friction?

2. Relevant equations
W=F (Δr) cos∅
Friction (kinetic)= μk* m*g

3. The attempt at a solution
I found the work done by the 68.9 N force using the equation above to get 297.74 J.
The energy lost due to friction should be simply the Force of kinetic friction multiplied by the displacement, right?
I did 0.286*15.6 kg* 9.81 to get 43.77 N as the force of kinetic friction.
Energy lost= Force friction * displacement i.e. 43.77 N * 4.59 m = 200.90 J, which is wrong.

What am I missing?

2. Feb 28, 2012

### Delphi51

The normal force is reduced by the "lift" of the vertical component of the 68.9 N pulling force.

3. Feb 28, 2012

### shrutij

So, normal force should be mg-68.9sin19.7 ?

4. Feb 28, 2012

### Delphi51

Yes, right on.

5. Feb 28, 2012

### shrutij

Got it. Thanks so much!

6. Feb 28, 2012

### Delphi51

Most welcome! Good luck on the next one.