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Mass-string system problem

  1. Oct 1, 2013 #1
    1. The problem statement, all variables and given/known data
    A body of mass 5kg is suspended at the lower end of a mass-less string, which passing through a mass-less pulley and is being pulled down at the other end by a force of 500N.
    The force of the pull reduces at a rate of 30N per meter, through which the body is raised.

    What is the velocity of the body when it reaches a height 10m, starting from rest?
    At what height will the direction of velocity change?


    2. Relevant equations
    a=dv/dt=v.dv/dx
    T-mg=ma
    T=f, where f = F-30y
    so basically, F-mg=ma


    3. The attempt at a solution
    starting from F-mg=ma, i went to deriving:
    (F-30y)-mg=m(dv/dt)=m.v.dv/dy
    so, (F-mg)*y -15(y^2) = (m/2)v^2
    => (500-5*10)*10 - 15*(100) = (5/2)v^2
    => 4500 - 1500 = (5/2)v^2
    => 1200 = v^2

    so v = 10√12 (that's my answer)

    for the velocity to change direction, it must first become zero again - so putting zero as the initial and final values limits of integration, i got:
    (F-mg)*y -15(y^2) = 0
    => [F -mg -15y]*y = 0

    so y = 0 or (F-mg)/15 = 30

    as y=0 is starting point, it cant be accepted as the required answer

    hence, at y=30, the block will change it's velocity


    IS WHAT I DID CORRECT?
     
  2. jcsd
  3. Oct 1, 2013 #2
    Everything looks good.

    Note you could have obtained your results more "physically" by using work/energy explicitly.
     
  4. Oct 1, 2013 #3
    Can you please show me how to do that - just a short, brief set of instructions will do
     
  5. Oct 1, 2013 #4
    The change in the kinetic energy of the body is given by the work of the resultant force. This leads directly to the equation you obtained through integration.
     
  6. Oct 1, 2013 #5

    haruspex

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    But since the force is changing it will still require integration, no?
     
  7. Oct 1, 2013 #6
    zibs.shirsh...This is how you would approach using work energy method

    Let T be the variable tension in the string .Displacement of mass is considered positive upwards.

    ΔT/Δx=-30 N/m

    Integrate with proper limits and you would get T as the function of x .

    Now apply work energy theorem ,∫Tdx = ΔKE + ΔPE .
     
  8. Oct 1, 2013 #7
    so the fact that the question mentions a spring is just a red herring...?
     
  9. Oct 1, 2013 #8
    There is no spring in the question. There is a string.
     
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