# Homework Help: Coefficient of Friction and Energy

1. Feb 20, 2014

### Blakeasd

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

A 30kg sled is sliding on a frictionless sheet of ice at a velocity of 4m/s. The sled encounters a rough patch of ice and begins to slow down. After traveling on the rough patch of ice for 3m, the sled's velocity is 2m/s. Determine the coefficient of friction between the rough ice and sled.

2. Relevant equations

KE = (1/2)mv2

FF = uk * FN

W = f * d

v2 = vi2 + 2a(x - xi)

3. The attempt at a solution

I begin by finding the Kinetic energy when the sled is on the frictionless sheet of ice:

KE = (1/2)mv2

KE = (1/2) * (30) * 42

KE = 240J

Then I find the Kinetic Energy when the sled is on the rough patch:

KE = (1/2)mv2

KE = (1/2) * (30) * 22

KE = 60J

I now note the amount of energy released due to friction:

240J - 60J = 180J

I continue to evaluate the force acting on the sled when on the rough patch:

W = fd
60J = f * 3m
f = 20N

Then I solve for acceleration in the rough patch:

v2 = vi2 + 2a(x - xi)

4 = 0 + 2*a*3
a= (2/3)

Now I sum the forces:

20N - (uk * (30kg * 9.8)) = 30kg * (2/3)
uk = 0

This is obviously not correct as it states friction is present. I don't know where I'm making a mistake. Presumably my logic is incorrect, not the math itself. I think finding the kinetic energy for the frictionless surface was useless with the approach I'm taking.

Thanks.

Last edited: Feb 20, 2014
2. Feb 20, 2014

### tiny-tim

Hi Blakeasd!
No, W = 180J.

(and what is the relevance of acceleration? )

3. Feb 20, 2014

### Blakeasd

When I sum up the forces, I set them equal to ma. I need acceleration so I can solve for the coefficient of friction.

Can you please explain how you found work to be 180J .

Thanks!

4. Feb 20, 2014

### tiny-tim

Hi Blakeasd!

The work energy theorem says that the change in mechanical energy equals the work done

That change is 180J

(and you can get µ from W = Fd and F = µmg)

5. Feb 20, 2014

### Blakeasd

Ok, I see why it is 180J, thanks! Here is my revamped work:

W = fd
180J = f * 3m
f = 60N

60N - (u * (30*9.8)) = (2/3) * 30
u = -.14

This is not a choice, though:

a.) 0.07
b.) 0.12
c.) 0.20
d.) 0.27
e.) 0.60

When I use W = Fd and F = umg , I get .204 (an answer choice).

So why doesn't this work when I sum forces? If I were to assume constant velocity, then summing forces would work, but how could I tell it is constant velocity? Why is it appropriate to assume this?

6. Feb 20, 2014

### tiny-tim

No, that should be
16 = 4 + 2*a*3
a= 2

corrected for a, that's 60N - (µ * (30*9.8)) = 2 * 30

which is F - µmg = ma

that's wrong because you've used the same force twice

there is only one force (= ma), and it's either F or µmg

7. Feb 20, 2014

### Blakeasd

I don't understand how I'm using the same force twice. I take the force pushing the object forward and subtract the force of friction pushing in the other direction. When summing forces I thought it is necessary to set their sum equal to ma. Am I missing something fundamental?

Thanks!

8. Feb 20, 2014

### tiny-tim

there is no force pushing the sled!!

if it wasn't for the rough patch, the sled would go on for ever at 4m/s

the only force is the friction!

9. Feb 20, 2014

### Blakeasd

Ah... so if I were to sum the forces:

0 - (u*30*9.8) = 60

u = .204

It works!

If you don't mind me asking another question, can you explain you reasoning for: v2 = vi2 + 2a(x - xi)

16 = 4 + 2*a*3

Why don't you use 2m/s for the final velocity?

Thanks!

10. Feb 20, 2014

### tiny-tim

ohhh, that should have been

4 = 16 + (-2)*a*3
a= -2 …​

the acceleration from the friction is actually -2