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2 carts, a string, what is the work done?

  1. Aug 7, 2007 #1
    1. The problem statement, all variables and given/known data

    A cart of mass M1 = 6 kg is attached to a block of mass M2 = 3 kg by a string that passes over a frictionless pulley. The system is initially at rest and the table is frictionless. After the block has fallen a distance h = 1 m, What is the work Ws done on the cart by the string?

    2. Relevant equations
    heres how it looks...

    http://www.imagination3.com/LaunchP...7_211715035_342819690_usa&transcript=&_lscid=



    3. The attempt at a solution
    I know work is F * distance moved, and i can surmise that M1 moves the same distance as M2 (1m) cause theyre contrained by the string.
    I know that force on M2 is 3 * 9.8 = 29.4N,
    No friction so force on M1 (pulling horizontally) is also 29.4N
    distance is 1 meter, so 29.4N * 1m = 29.4J
    = wrong!

    Please help!
    ]thanks
     
    Last edited: Aug 7, 2007
  2. jcsd
  3. Aug 7, 2007 #2

    learningphysics

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    Hint: The force acting on M1 is not the weight of M2... Draw the freebody diagram of M1... and also draw the freebody diagram of M2... label all the forces... What are the forces acting on M1... what are the forces acting on M2?
     
    Last edited: Aug 7, 2007
  4. Aug 7, 2007 #3
    Yes i drew fbd for both, so i need to find the tension in string (on M1).

    But I dont understand why force on m1 is not downwhard force (weight) on m2?

    pully and surface are frictionless
    strings are massless
    i dont get it?
     
  5. Aug 7, 2007 #4

    learningphysics

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    You would be right if the acceleration was 0... But that's not the case here. You have two equations with two unknowns... T (tension in the string) and a the acceleration (which is same for both carts).
     
  6. Aug 7, 2007 #5
    why do i need the acceleration i just need the tension in the string dont i?
     
  7. Aug 7, 2007 #6

    Dick

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    As in the last post I looked at you simply aren't writing the equations down. If you do, then what learningphysics is telling you will be much clearer.
     
  8. Aug 8, 2007 #7

    learningphysics

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    That's right, you don't need to calculate acceleration... But your two equations will have the two variables tension and acceleration. You only need to calculate the tension...

    As Dick suggested, you should go ahead and write out the equations... it will make things much clearer... what do you get for your two equations?
     
    Last edited: Aug 8, 2007
  9. Aug 8, 2007 #8
    M1
    Fx = T = ma (no other x force cause no friction)
    Fy = 0 (not moving vertically)
     
  10. Aug 8, 2007 #9

    learningphysics

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    That's right... and what equations do you have for M2?
     
  11. Aug 8, 2007 #10
    Fx = 0 (not moving horiz)
    Fy = T-W (mg) = ma
     
    Last edited: Aug 8, 2007
  12. Aug 8, 2007 #11
    so my 2 unkonws are T and a, right?
     
  13. Aug 8, 2007 #12

    learningphysics

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    Yes. I'd write it like this...

    Fy = M2*a, therefore
    T - M2*g = M2*a

    and your equation for M1 from before is:

    T = M1*a

    So you have these 2 equations (the other 2 equations Fx=0 and Fy=0, don't help to solve the problem):

    T - M2*g = M2*a
    T = M1*a

    And from here you can solve for T.
     
  14. Aug 8, 2007 #13

    learningphysics

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    Exactly. :smile:
     
  15. Aug 8, 2007 #14
    T-mg = ma (x)
    T = ma (y)

    is this ok so far?
     
  16. Aug 8, 2007 #15

    learningphysics

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    Yes, but be careful about the masses... write M1 and M2 instead of just m so that you aren't mixing up the masses...
     
  17. Aug 8, 2007 #16
    OK so..

    T-M1g = T

    ????

    i combined equations but the Ts will just cancel
     
  18. Aug 8, 2007 #17

    learningphysics

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    It should be:

    T - M2g = M2a
    T = M1a
     
  19. Aug 8, 2007 #18

    learningphysics

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    You mixed up M1 and M2...
     
  20. Aug 8, 2007 #19

    M2a + M2g = M1a
    M1a - M2a = M2g
    a(M1-M2) = M2g

    a= M2g / (M1-M2)

    a = 9.8m/s/s

    F=ma
    58.8N = 6*9.8 (of M1)

    W=Fx

    58.8N * 2m = 117.6J

    doesnt add up though????
     
  21. Aug 8, 2007 #20

    learningphysics

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    Your original description says 1m, so it should be 58.8N * 1m = 58.8J. Is the answer 58.8J?
     
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