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Homework Help: Work done by frition on a crate!

  1. Sep 20, 2009 #1
    A 70 kg crate starting at rest slides down a rough ramp 4.0 meters long, inclined at 30 degrees above horizontal. At the bottom, it is moving at 5.0 m/s. what was the work done by friction on the crate???

    A. -500 Joules
    B. 500 Joules
    C. 460 Joules
    D. -460 Joules
    E. 0.0 Joules

    I got this answer wrong and i was wondering why... so here is my work. i got the height to be 2meters by using sin 30

    Mgh - Wf = 1/2MV^2

    70 * 9.81 * 2 - wf = .5 * 70 * 5.0^2
    875 - wf = 1373.4
    wf =1373.4-875
    wf = 500 joules

    so i got positive 500 joules. can any body tell me what the answer truly is and what i did wrong?? thank you alot!!
  2. jcsd
  3. Sep 20, 2009 #2


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    Welcome to PF!

    Hi cbeeson23! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Work done = force "dot" displacement …

    the friction is up, but the displacement is down. :wink:
  4. Sep 20, 2009 #3
    so the answer then would be -500 joules?

    should my equation have been mgh = 1/2mv - work
  5. Sep 20, 2009 #4
    Or are you saying that the work done is actually 500 joules, but since friction is what goes up the ramp the work done by friction is actually - 500 joules?
  6. Sep 20, 2009 #5
    can any1 explain this answer to me a little better
  7. Sep 20, 2009 #6


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    Frictional force acts in the opposite direction of the displacement of the crate.
    When the work is done on a body, its kinetic energy increases.
    When the work is done by a body, its kinetic energy decreases.
    When there is no friction, mgh = 1/2*mv^2
    Here mgh > 1/2*mv^2
    So due to friction kinetic energy of the crate decreases.
    Hence work done by the friction = final energy - initial energy.
  8. Sep 21, 2009 #7


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    Hi cbeeson23! :smile:

    (just got up :zzz: … )

    If you're confused, then don't treat potential energy as part of the energy, treat it as part of the work done (PE is just another name for minus work done in a conservative field, such as gravity).

    The formula to remember is ∆(kinetic energy) = total work done

    which in this case is mgh - Fd :smile:
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