# Work Done Question

1. Mar 20, 2010

### Paymemoney

1. The problem statement, all variables and given/known data
A 40.0kg box initially at rest is pushed 5.00m along a rough, horizontal floor with a constant applied horizontal force of 130N. If the coefficient of friction between box and floor is 0.300, find the work done by the normal force?

2. Relevant equations
W=F*d

3. The attempt at a solution
The answer is 0N, however i got 1260(which i know does not seem logically correct).
So can someone explain to me why does work done by the normal force equal to 0N.

2. Mar 20, 2010

### thebigstar25

because the work = F.d, dont forget that F and d (the force and displacment) are both vectors , F.d is a dot product, then you have work=F.d=Fdcostheta , where theta is the angle between the force(in your case is the normal force) and the displacement .. and since theta is 90 degree then w=Fdcos90=0J (the unit of work is J not N)...

3. Mar 20, 2010

### Paymemoney

ok thanks i understand, however another question i would like to ask, Then how would you explain how i got the work done by friction to be -588. Because to find the Frictional force i times the (coefficient of friction by the Normal Force ). However if you say that the Normal Force is mgcostheta then i don't understand why the answer would be -588, because W=-392cos90 * 5 then it would be 0 Joules?

4. Mar 20, 2010

### thebigstar25

who said that the normal force is mgcostheta???? ...
If u carefully draw the freebody diagram u can see that u have 4 forces acting on the box ..
1- the force that horizontally applied which is 130 N ..
2- the force of friction which is opposing the force you applied and = coeffient of friction * normal force..
3-the weight of the box which is downward and = m*g ..
4-the normal force which is upward..

Sice the resulant motion is in the horizontal axes, then according to Newtons 2nd law .. The net force in the vertical axes = zero .. Which implies that : the normal force - the weight of the box = zero ..
Then, N = mg (and not mgcostheta) ...
So, the friction force = coeffient of friction* mg ..

Hopefully this was clear enough :) .. If u still have question ask again ....

5. Mar 20, 2010

### Paymemoney

yep thanks i understand now :)