# Finding the net work done on an object

1. Oct 11, 2009

### Shyamalan

1. The problem statement, all variables and given/known data
A 325 N force acts on an object with a mass of 42.5 kg at an angle of 27° , pulling the object along a flat surface. The coefficient of friction is 0.21 and the object's displacement is 320m.
Calculate the net work done on the object.

2. Relevant equations
W=Fdcos(θ) Fg=mg Ffr=μFN

3. The attempt at a solution
Ok, so I know FN in this case is equal to Fg, and the work done by the two is 0 J, because they perpendicular to the displacement. The work done by the pulling force is 325*320*cos(27) which is 92665 J. The friction force is 0.21*42.5*9.8 which is 87 N. The work done by the friction force is 87*320*cos(180) which is -27840 J. The net work is the sum of these two values, 92665+(-27840) which is 64825 J.

Now, the answer given is 7.46*10^4 J, which is one digit off of my answer. Am I doing something wrong or is there a typo in the answer I was given?

2. Oct 11, 2009

### tiny-tim

Welcome to PF!

Hi Shyamalan! Welcome to PF!
Noooo

3. Oct 11, 2009

### Shyamalan

Pfffft... Whoopsy, I got it now. XP