# Work done by friction

1. Oct 20, 2012

### DLH112

1. The problem statement, all variables and given/known data
There's a mass (16.6 kg) on a surface with kinetic coeficient of friction 0.181. it's being pulled by a constant force of 161 N at 26 degrees above the horizontal. The block is displaced 39.1 meters. Calculate the work done by friction.

2. Relevant equations
Friction = (mu)Fn, F=ma
W = fd

3. The attempt at a solution
I've tried to do this 2 ways now, both ended up wrong:
First I thought the force doesn't matter since you know mu and the mass and the displacement. so i thought the work done by friction would equal (16.6)(9.8)(0.181)(39.1)
since friction = mu Fn and W = Fd.

My second attempt, which i only tried because the first was wrong, factored out the vertical component of the pulling force when finding the normal and friction.

2. Oct 20, 2012

### cepheid

Staff Emeritus
Yeah, to get the frictional force, you need to know the normal force, and the normal force is going to depend on the other vertical forces that act on the object.

Can you post the work you did for the second method?

3. Oct 20, 2012

### DLH112

sure.
mg = (9.8)(16.6)= -162.68 N
Fy = 161 sin(26.6) = 72.08921314 N
so Fnet down/ -Fn = -90.59078686 N
then W(fric) = (39.1)(0.181)(90.59078686) = -641.1200577 J

4. Oct 20, 2012

### frogjg2003

I understand that you're trying to indicate direction with the negative, you should make sure that it appears in all parts of the equation.

I can't find anything wrong with your answer, try using 9.81 instead of 9.8, and possibly a positive answer instead of a negative answer. It might have been asking (not clearly) for the magnitude of the work, or was expecting you to use a specific value of g.

5. Oct 22, 2012

### DLH112

I must've been tired and mixed up the 16.6 kg and 26 degrees, so i used 26.6 degrees when calculating instead. i feel dumb xD