# Homework Help: Work of friction force

1. Dec 16, 2012

### golanor

1. The problem statement, all variables and given/known data
A body of mass m is moving through a bumpy road. the distance between each 2 bumps is 2h and the height of every bump is h. The body starts at velocity V0 and stops on top of the third bump. The friction coefficient is μ. Ignore the acceleration needed to change the direction on every peak, and the shape of the bumps is unknown.
What is the work of the friction force between the second and third peaks?
What is V0 and what is the velocity on every peak?

2. Relevant equations

Work-Energy theorem

3. The attempt at a solution

I tried to find the the work using the work-energy theorem, but i realized I couldn't find the normal force since i don't know the shape of the path, and I have no idea how to do the line integral if i don't know the shape.
In other words, i'm stumped.

#### Attached Files:

• ###### Untitled.jpg
File size:
19.8 KB
Views:
103
Last edited: Dec 16, 2012
2. Dec 16, 2012

### Staff: Mentor

Show what you've tried. What does the work-energy theorem state? (Hint: You won't need to know the details of the shape of the path.)

3. Dec 16, 2012

### golanor

After wrestling with it for a few hours, this is what I have:
1)The normal force depends on the angle of the bump, N=mgcosθ
2)The friction force doesn't depend on the path taken.
3)
(mv0^2)/2=mgh+∫fds

I think i can do something like:
f=(μmgcosθ, 0)
ds = (dx,dy)
limits of integration are y=0, y=h, x=0, x=6h
But that will only give me the entire course.
I also know that the angle is 0 at the top and at the bottom so i can eliminate the cosθ after computing the limits.
I have no idea how to get only the part from 2nd to 3rd.

4. Dec 16, 2012

### Staff: Mentor

Forget about trying to calculate the work done directly. Do it the easy way--take advantage of the given information. Again, in general terms what does the work-energy theorem tell you?

5. Dec 16, 2012

### golanor

Eki=Ep+Ek+Work
Eki = initial Kinetic Energy.

6. Dec 16, 2012

### Staff: Mentor

Good. That's all you need to solve for the work.

7. Dec 16, 2012

### golanor

But then i will have the total work. Does that mean i should first try to answer the second question?

8. Dec 16, 2012

### Staff: Mentor

Right. Which answers the first question.
No.

9. Dec 16, 2012

### golanor

Oh i didn't write the full question...
What is the work of the friction between the second and third peaks?

10. Dec 16, 2012

### Staff: Mentor

Start by finding the total work.

11. Dec 16, 2012

### golanor

W = m((V0^2)/2-g*h)
so, if 40% of the distance id between 2 and 3, does it mean that 40% of the work is there as well?
W(2->3) = m/5(2(V0^2)-4gh)
?

12. Dec 16, 2012

### Staff: Mentor

Good.
That's a reasonable assumption.
Careful how you write it. W(2->3) = 2/5(Wtotal)

13. Dec 16, 2012

### golanor

I was a little excited :)

Thanks for the tips!

14. Dec 16, 2012

### golanor

By the way, how do I find the velocity?

15. Dec 16, 2012

### Staff: Mentor

Figure out the KE at each peak. (You know how to find the work done.)