# Work done by a constant force problem

1. Sep 26, 2009

### mooney82

1. The problem statement, all variables and given/known data
A 1.8kg block is moved at constant speed over a surface for which uk = 0.25. The displacement is 2 m. It is pushed with a force at 45 degrees below the horizontal.
Find the work done by: (a) the force F; (b) friction; (c) gravity

2. Relevant equations
W=Fs cos (theta)

s=displacement

3. The attempt at a solution

First I found the force F to be 6.24 N.

Then using W=Fs cos (theta) I put in:

W=6.24 N * 2 m * cos 135
W= -8.82

The answer in the back of the book says 11.8, I'm wondering where I'm making my mistake.

2. Sep 26, 2009

### Staff: Mentor

How did you solve for the force?

Why did you use an angle of 135? (That's why your answer is negative.) But first figure out the force properly.

3. Sep 26, 2009

### mooney82

Well, I summed the x component forces:
Fcos45 - friction = ma

acceleration = 0
friction = uk(N)
N equals mg

so

Fcos45 - uk(mg)=0

F = uk(mg)/cos45

F = 0.25*1.8*9.81/cos45

4. Sep 27, 2009

### Staff: Mentor

Here's the problem. In this case, N ≠ mg. The fact that the applied force has a vertical component changes the normal force.

To find the normal force, analyze the vertical components of the forces acting on the block.

5. Sep 27, 2009

### mooney82

Would you then solve by substitution? I'm having a hard time because F seems to cancel out.

6. Sep 27, 2009

### Staff: Mentor

Yes. You'll have two equations: one for vertical forces; one for horizontal. You can eliminate N by substituting one into the other.
Show what you're doing.

7. Sep 27, 2009

### mooney82

I got it to work out. You tha man Al!