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Work-energy princple

  1. Jul 29, 2009 #1
    1. A crate of mass 10kg is accelerated up a rough incline with an initial speed of 1.5m/s. The pulling force is 100N parallel to the incline, which makes an angle of 20deg with the horizontal. The coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 5.0m. Use the work-energy princple to calculate the speed of the crate after 5m.

    Am I right or wrong?



    2. KE1 + PE1 + Ffr*d = KE2 + PE2
    Coefficient of Friction = Ffr/Fn



    I got 36.83N for the force of friction.

    From my working out I know I have the work done by friction but if i put it on the other side I will have a negative number in my final solution.
    (therefore my equation below doesn't actually make sense to me, if I am understand correctly? I am saying with that equation that the inital energy includes friction energy!)

    3.

    KE1 + PE1 + Ffr*d = KE2 + PE2
    =1/2 X 10 X 2.25 + 0 + 36.83 X 5 = 1/2 X 10 X v2 + 10 X 9.8 X 5sin20
    =11.25 + 184.15 = 5v2 + 167.59
    v2 =(195.4 – 167.59) / 5
    v = sqrt(5.562)
    = 2.538
    v= 2.5 m s-1
     
  2. jcsd
  3. Jul 29, 2009 #2

    rl.bhat

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    First of all find the net force acting on the crate.
    100 N up
    mg*sin20...down
    μ*mg*cos20....down. Find net force.
    work down on the crate = Net force *displacement.....(1)
    Change in KE = 0.5*m*v2^2 - 0.5*m*v1^2...(2)
    Equate 1 and 2 and solve for V2.
     
  4. Jul 29, 2009 #3

    kuruman

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    Gold Member

    Your expression has the wrong sign for the "work done by friction". This work is always negative because the displacement and the force are always at 180 degrees with respect to each other which makes the cosine of the angle between them - 1.

    Another way to look at it: KE2 + PE2 must be less than KE1 + PE1 because some Joules are lost to friction. If the three terms on the left are all positive, then KE2 + PE2 is a greater number than KE1 + PE1.
     
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