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Energy and a Frictional Incline

  1. Jan 26, 2004 #1
    I was hoping I could get some help on this particular exam review question:

    [4] A 3.0kg crate slides down a ramp at a loading dock. The ramp is 1.0m in length and inclined at an angle of 30 degrees. The crate starts from rest at the top of the ramp and experiences a constant frictional force of magnitude 5N, and continues to move for a short distance on the flat floor. (a) Use the energy methods to determine the speed of the crate when it reaches the bottom of the ramp. (b) Determine the acceleration of the crate.

    For part A, I found the initial Potential Energy to be 14.75 (PE=mgh), and since PI-initial is 14.75 and KE-initial is 0, KE-final must be 14.75 as well (since PE-final will be 0). I then used KE=(1/2)mv^2 to find the velocity, which I found to be 3.1 m/s.

    For part B, I used W=F*d to find Work to be 5J (F=5, d=1). I then used W=m*a*d to find the acceleration, which I think I did incorrectly. Using that formula I found the acceleration to be a=1.667 m/s^2.

    I don't think these are correct, but I cannot think of a way to better include the frictional force of 5N. Any suggestions?

    Thanks for your time.
  2. jcsd
  3. Jan 26, 2004 #2

    Doc Al

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    Staff: Mentor

    You forgot to include the energy "lost" due to friction. You calculated the initial energy correctly. The final energy (KE) equals the original energy minus that which is transformed into thermal energy due to friction (this is the "work" done by friction).
    This time you did the opposite: you ignored gravity and only considered friction! You can use Fnet=ma to find the acceleration. Be sure to include both gravity and friction. But, since you've just calculated the final KE in part a, you can also make use of the kinematic equation: V2=2ad.
  4. Jan 26, 2004 #3
    Thanks very much Doc,

    I recalculated part A with friction included, so KE-final changed from 14.715J to 9.715J after I subtracted the 5J lost to friction. This changed my velocity value to 2.55 m/s using the KE=(1/2)mv^2 formula.

    So as you suggested I plugged in that velocity value into V^2=2ad and found a to equal a=3.25 m/s^2. Do you know if those numbers are correct?

    Again, thanks for the boost Doc.
  5. Jan 26, 2004 #4

    Doc Al

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    Staff: Mentor

    Close enough. And you are welcome. :smile:
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