# Net Work check

## Homework Statement

A 15.8 kg block is dragged over a rough, horizontal surface by a constant force of 90.9 N
acting at an angle of angle 28.6 above the horizontal. The block is displaced 22.8 m,
and the coefficient of kinetic friction is 0.18.

What is the net work done on the block?

W=F d cos(theta)

## The Attempt at a Solution

Work done by the 90.9N Force:
W=F d cos(theta)
=90.9*22.8*cos(28.6)
WF=1819.637 J

Work done by friction:
W=F d cos(theta)
W= friction * d * cos(theta)
W=u Fn * d * cos(theta)
W=u(mg-y component)*d * cos(180)
W=.18[(15.8*9.8)-90.9sin(28.6)]*22.8 * -1
Wf=-456.886 J

Wnet=WF-Wf * d
Wnet= 1819.637-(-456.886) * 22.8
Wnet=51907.724 J

G01
Homework Helper
Gold Member
You have a sign error in your second to last line. You are also multiplying the net work by d. I don't know why you are doing this. Your second to last line should read:

$$W_{net}=1819.637-456.886$$

The negative sign comes from the cos(180) term. You do not need to add another one in this line.

As I said before, you also shouldn't multiply by 22.8 in your second to last line. I don't know why your doing this.

Does this make sense?

1362.771 J

I'm not sure why I multiplied it by d again. I guess I forgot to look at the unit and thought they were forces. Thanks.