# Work and energy with normal force

1. Nov 10, 2008

### Maiia

1. The problem statement, all variables and given/known data
Starting from rest, a 5.85 kg block slides 2.07m down a rough 26.4degree incline. The coefficient of kinetic friction between the block and the incline is 0.427. The acceleration of gravity is 9.8m/s^2. Find the work done by the normal force. Answer in units of Joules.

Here's my FBD.

I know the formula for work is Fcos(theta)d. But I don't know how to find the angle for the normal force...
If someone could help out, I would really appreciate it.

2. Nov 10, 2008

### mgb_phys

The normal force acts at right angles to the slope.
You 26.4 is in the wrong place - a slope angle of 26.4 means the angle between the slope and horizontal.

3. Nov 10, 2008

### Maiia

yes, but if you draw some parallel lines, the 26.4 degrees of the slope angle is also equivalent to that angle up there. oh but i see what you said about the normal being perpendicular...wouldnt that mean work is zero then?

4. Nov 10, 2008

### LowlyPion

Yes you are right that forces in the direction of motion are the ones that do work, but in the case of the Normal force there is a resulting force along the direction of motion as determined by the Normal force * the Coefficient of Friction.