# Homework Help: How to Calculate the Coefficient of Friction

1. Jun 4, 2012

### ewarsaw

1. The problem statement, all variables and given/known data

Calculate the coefficient of friction between the wooden block and the table top, using a pendulum to hit the block.

Mass of pendulum weight = 1 kg
Mass of wooden block (with weight on top) = 2.073 kg
Release height of pendulum = .06 m
Pendulum's distance from starting point when released = .34 m
Distance block moves = .036 m

2. Relevant equations

Ff=μFN
ΔPE=mgΔh ?

3. The attempt at a solution

I have absolutely no idea where to start. My teacher said it involves a lot of energy transfers, so I calculated the change in potential energy to be 58.86 J, but I'm not even sure if that's right. If you could just explain it to me or point me in the right direction, I would really appreciate it. Thanks!

2. Jun 4, 2012

### Villyer

So the situation is a pendulum is being released, it strikes the box, and then the box slides a little bit. You have to use the data to find μ?

I'm just clarifying, no scenario explanation was posted but that's what I infer from everything.

3. Jun 4, 2012

### ewarsaw

Yeah, that's the situation. Sorry for not explaining.

4. Jun 4, 2012

### Villyer

It's fine, I just wanted to make sure.

So you said that the change in potential energy is 58.86J, where is that energy now?

5. Jun 4, 2012

### ewarsaw

It transferred to the block.

6. Jun 4, 2012

### Villyer

I'm more interested in the moment right before the pendulum and the block collide. (Collisions generally tend to lose energy)

7. Jun 4, 2012

### ewarsaw

It turned into kinetic energy?

8. Jun 4, 2012

### Villyer

Right, so how fast is the pendulum going before it hits the block?

9. Jun 4, 2012

### ewarsaw

2.58 m/s ?

10. Jun 4, 2012

### Villyer

I got 1.08. I think your change in PE is wrong.

11. Jun 4, 2012

### ewarsaw

Probably. How did you get that?

12. Jun 4, 2012

### ewarsaw

Ok, I got that and I got .523 m/s for the velocity of the block after it's hit, but I don't know what to do after that.

13. Jun 4, 2012

### Villyer

$\Delta PE = \Delta KE$
$mgh = \frac{1}{2}mv^{2}$
$v = \sqrt{2gh}$

g = 9.8, h = 0.06

14. Jun 4, 2012

### Villyer

How did you get to that?

15. Jun 4, 2012

### ewarsaw

I used conservation of momentum.

m1v1=m2v2
1 kg(1.08 m/s)=2.073 kg(v)
v=.523 m/s

16. Jun 4, 2012

### Villyer

Okay, looks good to me

17. Jun 4, 2012

### ewarsaw

I think I got the answer, but I'm not sure if it's right. I found the force applied to the box, the force of the block on the table, and the acceleration of the block.

FN=FA-μ(mg)
FN=ma

FA-μ(mg)=ma
9.81 N-μ(9.81 m/s2)(2.073 kg)=2.073 kg(3.799 m/s2)
μ=.095

Do you think that's right?

18. Jun 4, 2012

### Villyer

I got a μ of 0.388.

From your work, I'm not sure where the 9.81 comes from (gravity times a mass of 1 kilogram?). And I'm assuming that 3.799 is the calculated acceleration from the block coming to a stop over a known distance.

19. Jun 4, 2012

### ewarsaw

Yeah. How did you get that? I'm so confused.

20. Jun 4, 2012

### Villyer

I used work-energy principals.

You know the initial kinetic energy of the block, and it gets dissipated by friction, so you can write an equivalency statement from that to find μ.