Finding work done by F given coefficient of kinetic friction

1. Oct 25, 2008

RAKINMAZID

1. The problem statement, all variables and given/known data
A 2.55 kg block is pushed 1.54 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 69.9 degrees with the horizontal. The acceleration of gravity is 9.8 m/s^2. If the coefficient of kinetic friction between the block and wall is 0.613, find work done by F.

2. Relevant equations
W = F*x

3. The attempt at a solution
t = theta
u = coefficient of kinetic friction

Fsint - uN = ma + mg

ma = 0 (since it's constant speed)
N = Fcost

F = (mg + uFcost)/(sint)

W = F * x

I think my method is wrong because I end up with a F in the equation which I don't know. Please help me asap.

2. Oct 26, 2008

kreil

just factor the F out and then divide:

$$Fsin \theta- \mu F cos \theta = mg$$

$$F= \frac{mg}{sin \theta-{\mu} cos \theta }$$

$$W=Fx$$

you know m, g, u, x, and the angle, so you can plug in and solve for F, then plug in and solve for W.

3. Oct 26, 2008

RAKINMAZID

I tried to solve the problem before your response, and I got the same equation. I solved for F and got 34.307 N. Then I put that into the second equation to solve for W and got 52.832 J as my final answer. But this is still wrong (online submit)! I don't understand what I'm doing wrong.

4. Oct 26, 2008

borgwal

The angle is given with respect to the *horizontal*

5. Oct 26, 2008

RAKINMAZID

So would the equation for work be:
W= Fcost * x

Because I did that and I got the answer 18.156 J but that is also wrong! Do you understand what I'm doing wrong?

6. Oct 26, 2008

borgwal

W=Fcos t *x is almost correct (Fcost t is the normal force)

Last edited: Oct 26, 2008