# Homework Help: A Physics problem

1. Feb 9, 2010

### Misr

Hello Peace be upon you

http://img8.imageshack.us/img8/9198/workdone.jpg [Broken]

The first part of this problem is OK but I'm asking about the second part
We know that the work done is F *d*cosine the angle between the force direction and
the displacement direction .So why the book calculated the work done in this way?

I know that the work done in this case is the potential energy which is (m*g*d) and this does works!! But my problem is why the book calculated the work done in this way?

Another way :
Its given that the angle between force direction and the displacement direction is 60 then the result is the same.
So Are all of these ways right?
I hope you can understand my problem
Thanks so much

Last edited by a moderator: May 4, 2017
2. Feb 9, 2010

### PhanthomJay

Yes, it's all the same, either it's work = force times displacement times cos 60, where the displacement, d, is the distance along the incline. Since d =15/cos 60, then work = (15mg/cos60)(cos60), which is just Work = 15mg. Thus, the work done against gravity can be calcualted from W=mg(vertical distance) = 15mg. I agree that the theta = 0 in the book solution is somewhat misleading. Note also that the work done by gravity is (- delta U_g).

3. Feb 10, 2010

### Misr

Yeah this was exactly what i'm asking about

Thanks so much
this really helped