Calculating Work of Friction Force on a Bumpy Road

[golanor]

Homework Statement

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?

Homework Equations

Work-Energy theorem

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.

 

[Doc Al]

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.)

 

[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.

 

[Doc Al]

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?

 

[golanor]

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

 

[Doc Al]

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

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

 

[golanor]

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

 

[Doc Al]

But then i will have the total work.

Right. Which answers the first question.

Does that mean i should first try to answer the second question?

No.

 

[golanor]

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

 

[Doc Al]

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

Start by finding the total work.

 

[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)
?

 

[Doc Al]

W = m((V0^2)/2-g*h)

Good.

so, if 40% of the distance id between 2 and 3, does it mean that 40% of the work is there as well?

That's a reasonable assumption.

W(2->3) = m/5(2(V0^2)-4gh)
?

Careful how you write it. W(2->3) = 2/5(W_total)

 

[golanor]

Good.


Careful how you write it. W(2->3) = 2/5(W_total)

I was a little excited :)

Thanks for the tips!

 

[golanor]

By the way, how do I find the velocity?

 

[Doc Al]

By the way, how do I find the velocity?

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