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Thermal energy of a system

  1. Oct 23, 2008 #1
    1. The problem statement, all variables and given/known data

    An 8.5 kg crate is pulled 5.5 m up a 30degree incline by a rope angled 16degree above the incline. The tension in the rope is 140 N and the crate's coefficient of kinetic friction on the incline is 0.26. What is the increase in thermal energy of the crate and the incline?

    2. Relevant equations

    F = mg
    W = Fdcos(theta)

    3. The attempt at a solution

    I figured, thermal energy could only come from friction so I calculated the work done by friction.

    N = mgcos(30)

    Ff = uN
    = umgcos(30)

    W = umgcos(30)d
    = (0.26)(8.5)(9.8)cos(30)(5.5)
    = 103J

    but it's wrong!
  2. jcsd
  3. Oct 23, 2008 #2


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    Homework Helper

    Welcome to PF.

    What about the component of the pulling force that was 16 degrees above the slope that lightened the normal force of the crate?
  4. Oct 25, 2008 #3
    I'm not 100% if i'm approaching the question properly but I got the right answer. o_O

    basically I have normal force, and then the lifting force which is the y-component of the force of tension. so...

    W = F*u*d
    =(n - Ty)(u)(d)
    =(mgcosθ1 - Tsinθ2)(u)(d) *θ1 = 30, θ2 = 16
    =(8.5*9.8*cos30 - 140*sin16)(0.26)(5.5)
    =47.9 ~ 48J

    It's the right answer, but normal force (n) and the y-component of Tension (Ty) is going in the same direction, so I should be adding them instead, but then I won't get the right answer. Can someone tell me what I'm doing wrong?
  5. Oct 25, 2008 #4


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    When you calculate Work you calculate it along the direction of it's motion. F * D. The Normal force per se is doing no work on the crate as it is not in the direction of its motion.. What is doing work in the direction of its motion is the frictional resistance that is retarding its motion. Now the retarding effect of friction depends on the normal force as it is related by the coefficient of friction. And the normal force is made up of the m*g component into the incline less (in this case) the Tension component of the rope that is pulling in the upward direction. (If someone was pushing down at an angle on the crate then the normal component of that would be added to the normal m*g component.)
  6. Oct 25, 2008 #5
    I think I got the idea. The lifting force of the y-component of Tension counteracts the force of Gravity and reduces it's magnitude, and because the Normal force depends on the Fg, it will also be reduced. But, I'm still not sure how I could show this on paper.
  7. Oct 25, 2008 #6


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    The Force you are concerned with for work is the Force vector down the plane that has the magnitude = u*(m*g*Cosθ - T*sin16) and of course your m*g*Sinθ force too. You can draw in the other normal forces for reference, but you should draw and label Friction. Your math should demonstrate that you understand what it is made up of.
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