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Homework Help: Work problem-blocks,pulley

  1. Feb 12, 2009 #1
    Work problem--blocks,pulley

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

    Two blocks are connected by a very light string passing over a massless and frictionless pulley . The 20.0 N block moves 75.0 cm to the right and the 12.0 N block moves 75.0 cm downward.

    Find the total work done on 20.0N block if there is no friction between the table and the 20.0N block.

    2. Relevant equations
    [tex]w=f\cdot s[/tex]
    [tex]f_k=\mu_k\cdot n[/tex]


    3. The attempt at a solution
    w=12N(.75m)=9J--mastering physics says it's wrong.
     
  2. jcsd
  3. Feb 12, 2009 #2
    Re: Work problem--blocks,pulley

    Ya, your attempt is not correct. What you need to do is find out the force the hanging block exerts on the sliding block. Then you can do W=F*d using this force over a distance of 75cm.
     
  4. Feb 12, 2009 #3

    Delphi51

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    Homework Helper

    Re: Work problem--blocks,pulley

    First, are you sure the blocks are in Newtons? Very odd - would expect kg.
    Second, I think the whole thing is more complicated because the blocks are accelerating. A force of 12 N down on the 2nd block will be opposed by the Tension in the string, so
    sum of forces = ma
    mg - T = ma
    T = mg-ma and it will not be the full 12N pulling on the other block.
    I think the first step will be to calculate the acceleration of the whole system.
     
  5. Feb 12, 2009 #4
    Re: Work problem--blocks,pulley

    That's probably it. There is a problem nearly identical to this in the book, but it says the blocks are moving at a constant speed.
     
  6. Feb 12, 2009 #5
    Re: Work problem--blocks,pulley

    I believe Delphi51 is correct.
     
  7. Feb 12, 2009 #6
    Re: Work problem--blocks,pulley

    T would give me the force on the 20N block and thus the work, yes?
    a=12N/total weight?
     
  8. Feb 12, 2009 #7
    Re: Work problem--blocks,pulley

    I think you mean that a=(Force)/(mass) not weight. For the force, use the T from your first statement.
     
  9. Feb 12, 2009 #8
    Re: Work problem--blocks,pulley

    What first statement?
     
    Last edited: Feb 12, 2009
  10. Feb 12, 2009 #9
    Re: Work problem--blocks,pulley

    This statement. You say that solving for T would give you the force you need in the equation to find work.
     
  11. Feb 13, 2009 #10
    Re: Work problem--blocks,pulley

    What about the work done on the other block? Is it the same?
     
    Last edited: Feb 13, 2009
  12. Feb 13, 2009 #11
    Re: Work problem--blocks,pulley

    The work will not be the same for the other block. In this case the only force acting on it is mg. Therefore, your work will just be mg*.75m and I think it will be negative.
     
  13. Feb 13, 2009 #12
    Re: Work problem--blocks,pulley

    I had already found that out, but thanks anyway.
     
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