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Work done by a man

  1. Feb 26, 2017 #1
    1. The problem statement, all variables and given/known data
    A man is standing on a cart of mass double the mass of man. mass of the man is m. initially the cart is at rest on a smooth ground. now, the man jumps w/ velocity v horizontally towards right wrt the cart. find the work done by the man during the process of jumping.

    2. Relevant equations
    W = Change in kinetic energy
    Initial momentum = Final momentum

    3. The attempt at a solution
    I calculated the change in kinetic energy and equated it to the work done, but i'm not able to get the right answer, which is mv^2/3.
     
  2. jcsd
  3. Feb 26, 2017 #2

    TSny

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    Hello and welcome to PF!
    Your general approach to the problem sounds right. But we can't identify where you are making a mistake unless you show your work in detail.
     
  4. Feb 26, 2017 #3
    This is what I did.

    Conserving momentum: mv=2m(x)
    therefore x = velocity of cart = v/2

    Apply work energy theorem:
    W = mv^2 + (2m)(v^2/4)/2 = 3mv^2/4
     
  5. Feb 26, 2017 #4

    TSny

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    You did not use the fact that v is the velocity of the man wrt the cart, not the velocity of the man wrt the ground. You have three velocities to work with:
    v = velocity of man wrt cart
    vc = velocity of the cart wrt the ground
    vm = velocity of the man wrt the ground.

    How are these related? Which of these velocities should be used in the conservation of momentum equation?
     
  6. Feb 26, 2017 #5
    but the system is initially at rest, no? so wouldn't velocity will be imparted to the cart only after the man has jumped?
     
  7. Feb 26, 2017 #6

    TSny

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    Yes
    Yes.

    Once the man has jumped, he has a certain velocity wrt the ground (vm) and the cart has a certain velocity wrt the ground (vc). Also after the man has jumped, the man has a certain velocity wrt the cart (v).
     
  8. Feb 26, 2017 #7
    I got the right answer! Thanks!
    And thank you for the welcome. :-)
     
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