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Cart with Friction

  1. Oct 8, 2008 #1
    A 10 kg cart is pulled by a force of 12 newtons for a distance of 7 meters, then by a force of 4 Newtons for 2 meters, then it is pushed (slowed down) by a force of 3 Newton for 1 meters. All this occurs on a horizontal surface.

    A)Initially assuming the surface to be frictionless, what net work is done on the cart as it moved the 10 meters?
    Work total?
    i know work total = (.5mvfinall^2)-(.5mvinitial^2) I don't know what to plug in for velocity..im stuck here...cause i got to go to work...help please.

    B)If the cart starts from rest, what is its speed after it has moved the 10 m?
    Vf=?

    Now assume a coefficient of friction µ between cart and surface during the entire distance.
    c) For what value of μ does the cart just come to a halt after 10 m?

    µ =??
    D) For the value of µ computed in part (c), what is the speed of the cart after it has gone 4 m?
    v=??
     
  2. jcsd
  3. Oct 8, 2008 #2

    Doc Al

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    Staff: Mentor

    What's the definition of work?
     
  4. Oct 8, 2008 #3
    Work is force*distance
    then i did 19*10=190 it doesn't work
     
  5. Oct 9, 2008 #4

    Doc Al

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    Staff: Mentor

    Good.
    No. Do each segment separately and add them up to find the total work done. For example, the first segment has F = +12 and distance = +7, so the work done for that segment would be: (12)*(7) = +84 Joules. Keep going.
     
  6. Oct 9, 2008 #5
    i tried it before. 84+8+3=95 it still didn't work..:(
     
  7. Oct 9, 2008 #6

    Doc Al

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    Staff: Mentor

    The third segment is incorrect. What's different about it compared to the first two?
     
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