Calculating Energy Lost due to Friction

[shrutij]

Homework Statement

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?

Homework Equations

W=F (Δr) cos∅
Friction (kinetic)= μk* m*g

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?

 

[Delphi51]

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

 

[shrutij]

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

 

[Delphi51]

Yes, right on.

 

[shrutij]

Got it. Thanks so much!

 

[Delphi51]

Most welcome! Good luck on the next one.