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Homework Help: Block, Cart, and Pulley Question

  1. Oct 7, 2008 #1

    A cart of mass M1 = 2 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 = 0.7 m

    a) What is the work Wg done by gravity on the system?

    b) What is the increase in kinetic energy DK of the cart-plus-block system?

    c) What is the speed |v| of the cart-plus-weight system?

    d) What is the work Ws done on the cart (not the block!) by the string?

    e) What is the tension T in the string?

    I'm really Lost. I can't even figure out what I did wrong when solving for Wg. I did
    Wg = mgh = (5kg)(9.81)(0.7) = 34.335 and that wasn't right.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Oct 7, 2008
  2. jcsd
  3. Oct 7, 2008 #2


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    Welcome to PF!

    Hi r34racer01! Welcome to PF! :smile:

    Hint: work done by gravity = change in PE …

    so how much mass is having its PE changed? :wink:
  4. Oct 7, 2008 #3
    Re: Welcome to PF!

    Is it m2 or both masses?
  5. Oct 7, 2008 #4


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    Gravitational PE (potential energy) is caused by change in height.

    So how much mass is changing height? :wink:
  6. Oct 7, 2008 #5
    Oh so it is m2, I forgot also that forces perpendicular to the object perform 0 work, so gravity does no work on m1 right?
  7. Oct 7, 2008 #6


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    Bingo! :biggrin:
  8. Oct 12, 2008 #7
    Ok so I got Wg =20.601, But now I'm a little confused with finding the change in KE.
  9. Oct 12, 2008 #8


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    work done by gravity = change in PE = change in KE :smile:
  10. Oct 12, 2008 #9
    Wow I didn't even realize that. So Wg = 20.601. And with that I was able to figure out that |v|=2.87.

    So now I'm trying to figure out the work done by the string. We know that the speed is 2.87, so the KE of the cart is 8.32J. I thought I could do 20.601-8.32 to get Ws, but apparantly that's wrong. So what should I be doing?
  11. Oct 13, 2008 #10


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    HI r34racer01! :smile:

    You didn't realize that? But it's so important

    work and energy are the same ("work-energy theorem") …

    KE and PE are the same ("conservation of energy") …

    see how neatly it all fits together? :wink:

    It means that if you want to calculate the KE, say, it's sometimes easier to calculate either the PE or the work done, and use that instead! :biggrin:
    i] it's 8.24, not 8.32 :wink:

    ii] why do you want the work done … the question doesn't ask for it?

    but it's equal to the KE anyway, which you now have.

    Then you can find the tension by "reverse-engineering" from the work done. :smile:
  12. Oct 13, 2008 #11
    Wow thx, but how is 8.24 and not 8.32?
  13. Oct 13, 2008 #12
    part e.

    combine the two masses to find the acceleration using F=ma because acceleration is uniform through the system.

    Then use one of the masses and F=ma to determine tension.

    This equation would be set up like this:

    Winner force - loser force = ma
  14. Oct 14, 2008 #13


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    2.872 = 8.24 :wink:

    how did you get 8.32?
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