Calculating Work of Friction Force on a Bumpy Road

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Homework Help Overview

The problem involves a body of mass m moving over a bumpy road, with specific distances and heights of bumps defined. The task is to calculate the work done by the friction force between the second and third peaks, while also determining the initial velocity V0 and the velocity at each peak. The work-energy theorem is mentioned as a relevant concept.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the work-energy theorem but struggles with the lack of information regarding the normal force and the path shape. Some participants question the necessity of knowing the path details to apply the theorem effectively.
  • Participants discuss the relationship between normal force, friction force, and the work done, with some suggesting that the friction force is independent of the path taken.
  • There is a consideration of how to isolate the work done specifically between the second and third peaks, with some participants proposing to calculate total work first.
  • Questions arise about the implications of dividing work based on distance traveled between bumps.

Discussion Status

The discussion is ongoing, with various approaches being explored. Some participants have provided guidance on using the work-energy theorem and calculating total work, while others are contemplating how to derive specific values for work done between the peaks. There is a mix of interpretations regarding the calculations needed to find velocities at each peak.

Contextual Notes

Participants are working within the constraints of the problem's parameters, including the unknown shape of the bumps and the requirement to ignore certain forces. The original poster has not provided complete information about the question, leading to further clarification requests.

golanor
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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.
 

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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.)
 
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.
 
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?
 
Eki=Ep+Ek+Work
Eki = initial Kinetic Energy.
 
golanor said:
Eki=Ep+Ek+Work
Eki = initial Kinetic Energy.
Good. That's all you need to solve for the work.
 
But then i will have the total work. Does that mean i should first try to answer the second question?
 
golanor said:
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.
 
Oh i didn't write the full question...
What is the work of the friction between the second and third peaks?
 
  • #10
golanor said:
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.
 
  • #11
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
golanor said:
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(Wtotal)
 
  • #13
Doc Al said:
Good.


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

I was a little excited :)

Thanks for the tips!
 
  • #14
By the way, how do I find the velocity?
 
  • #15
golanor said:
By the way, how do I find the velocity?
Figure out the KE at each peak. (You know how to find the work done.)
 

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