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Work by gravity

  1. Nov 23, 2015 #1
    1. The problem statement, all variables and given/known data
    In the figure, a 700-kg crate is on a rough surface inclined at 30°. A constant external force P = 5600 N is applied horizontally to the crate. As the force pushes the crate a distance of 3.00 m up the incline, the speed changes from 1.40 m/s to 2.50 m/s. How much work does gravity do on the crate during this process?
    pLyECts.jpg

    2. Relevant equations
    [itex]
    W=KE_f - KE_i\\
    W=Fd\cos{\theta}
    [/itex]

    3. The attempt at a solution
    [tex]
    W=KE_f - KE_i
    W_{total}=\frac{1}{2}mv^2_f - \frac{1}{2}mv^2_i\\
    W_{total}=\frac{1}{2}m(v^2_f-v^2_i)\\
    W_{total}=\frac{1}{2}(700)(2.5^2-1.4^2)\\
    W_{total}=1501.5 \ \texttt{J}\\ \ \\
    W=Pd\cos{\theta}\\
    W=5600(3)\cos{30}\\
    W=14549.2 \ \texttt{J}\\ \ \\
    W_{total}=W_g+W\\
    W_g=W_{total}-W\\
    W_g=1501.5-14549.2\\
    W_g=-13047.7 \ \texttt{J}
    [/tex]
    The correct answer is [itex]-10300 \ \texttt{J}[/itex]. I am not sure how I would incorporate friction (if at all).
     
  2. jcsd
  3. Nov 23, 2015 #2

    haruspex

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    You are making it much too complicated.
    How do you usually calculate work done against gravity?
     
  4. Nov 23, 2015 #3

    gneill

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    Staff: Mentor

    They're looking for the work done by gravity. What's the force of gravity on the crate? Sketch in the gravitational force vector on your diagram. What distance does the crate move along the direction of the gravitational force vector?


    Edit: Oops! haruspex got there first!
     
  5. Nov 23, 2015 #4
    [tex]
    W=-mgh\\
    W=-mg(dcos\theta)\\
    W=-(700)(9.8)(3\sin{30})\\
    W=-10290 \ \texttt{J or } -10300 \ \texttt{J}
    [/tex]
    Wow. I guess I was thrown off because they mentioned P. Thanks!

    Out of curiosity, how would I find the work that friction does?
     
  6. Nov 23, 2015 #5

    gneill

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    Staff: Mentor

    Compare the actual results given (change in KE of the crate) with what it would have been if there were no friction. That means sorting out all the works done by the known forces.
     
  7. Nov 23, 2015 #6
    So would it be like this?
    [tex]
    W_f=mgd\sin{\theta}+\frac{1}{2}m(v^2_f-v^2_i)-Pd\cos{\theta}
    [/tex]
     
  8. Nov 23, 2015 #7

    gneill

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    Staff: Mentor

    That looks promising. You might want to ponder on the signs of the terms. Consider, for example, that the work done by friction should be negative. And the force P is adding energy to the crate while gravity is stealing it.
     
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