Kinetic Energy and Work: Ramp With Friction

  • Thread starter CaptFormal
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  • #1
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Homework Statement



When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.15 m/s. The mass stops a distance S2 = 2.5 m along the level part of the slide. The distance S1 = 1.22 m and the angle θ = 32.70°. Calculate the coefficient of kinetic friction for the mass on the surface.
PhysicsGraph.jpg


Homework Equations





The Attempt at a Solution


Not quite sure how to start on this one. Any help will be appreciated. Thanks.

Sincerely,
Captformal
 

Answers and Replies

  • #2
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First, find the energy of the mass in its initial position.

[tex]E = K+U = \frac{1}{2}mv_{0}^{2} + mgS_{1}\sin(\theta)[/tex]

Now note that the block has lost all of its energy once it reaches S_{2}.

[tex]E + W_{other} = 0 \iff W_{other} = -E[/tex]

The work done on the block has to be described in 2 parts: one along the path S_{1} and another along the path S_{2}. If you draw a free body diagram, you'll find that:

[tex]W_{1} = [mg\cos(\theta)-u_{k}mg\sin(\theta) ]S_{1}[/tex]

[tex]W_{2} = -u_{k}mgS_{2}[/tex]

After going through an algebraic mess, you'll find:

[tex] u_{k} = \frac{ E + mg\cos(\theta)S_{1} }{mgS_{2} + mg\sin(\theta)S_{1} } [/tex]

Hopefully I didn't make a mistake in there somewhere.
 
Last edited:
  • #3
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Thanks Vykan12. That helped out a lot.
 

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