# Coefficient of Kinetic Friction

1. Oct 12, 2015

### Beanie

1. The problem statement, all variables and given/known data
A horizontal force of 150 N is used to push a 50.0 kg packing crate a distance of 5.00 m on a rough horizontal surface.
The acceleration of gravity is 9.81 m/s2 .

If the crate moves with constant velocity, calculate:
a) the work done by the force. Answer in units of J.
b) the coefficient of kinetic friction.

2. Relevant equations
W=Fdcostheta
Ff=muFn

3. The attempt at a solution
I got part a right. However I am struggling for part B. Here is my work for both parts...

a) W=fdcostheta
W=(150)(5)cos(0)
W=750J

b) Ff=muFn

Fn=Fg
Fn= 50*9.81=490.5N
Ff=0 because there is constant velocity. Therefore, mu=0.

This answer is wrong. Can anyone tell me where I went wrong. Thank you.

2. Oct 12, 2015

### stockzahn

If the velocity is constant and the force to keep it constant is 150 N, what's the force the crate is affected by the roughness of the surface?

3. Oct 12, 2015

### Beanie

The normal force/force of gravity?

4. Oct 12, 2015

### stockzahn

The force of gravity has a vertical direction. You are looking for a force in horizontal direction. So in horizontal direction: No change of velocity (so no acceleration as you already indicated in your first post). But you push with 150 N, what does that mean for other horizontal forces and what kind of force affects an object that is pushed over a rough surface?

5. Oct 12, 2015

### Beanie

Right, okay. So, in the horizontal direction, the force of friction and the force of the push (150N) are both affecting the object. They are also in opposite directions. I drew a FBD for it (attached file), however I am still stuck on how you calculate the Force of Friction.

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6. Oct 12, 2015

### stockzahn

The forces in vertical directions are correct, but if the forces in horizontal direction would be of the kind you drew, the object would accelerate as the pushing force is larger, than the friction force... If you find the right relation between pushing und friction force you promptly will have the answer to the original question.

7. Oct 12, 2015

### Beanie

Thank you for the help! This makes a lot of sense now!