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Confused conceptually about work done with pulleys

  1. Oct 28, 2014 #1
    Hello,

    I am confused about a problem regarding work done and pulleys. I am confused mainly by the last two, but not 100% confident in the first two either.

    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 (Figure 2). The 20.0 N block moves 75.0 cm to the right and the 12.0 N block moves
    75.0 cm downward.

    (image attached)

    Find the work done on
    (a) the 20.0 N block if there is no friction between the table and the 20.0 N block.
    (b) the 12.0 N block if there is no friction between the table and the 20.0 N block.
    (c) the 20.0 N block if µs=0.500 and µk=0.325 between the table and the 20.0 N block.
    (d) the 12.0 N block if µs=0.500 and µk=0.325 between the table and the 20.0 N block.

    2. Relevant equations


    3. The attempt at a solution

    For case (a) the force that is moving the [itex]20.0N[/itex] block is the weight of the [itex]12.0N[/itex] block. The work done on the [itex]20.0N[/itex] block is therefore [itex]F x d = 12 x 0.75 = 9 Nm[/itex]

    For case (b) I think I can find the work done by finding the change in gravitational potential energy of the [itex]12.0N[/itex] block which is [itex]mgΔh = -9 Nm[/itex].

    I think this is correct, to expect the net work done to be [itex]0[/itex], since there is no friction which is non-conservative.

    However, for question (c), I am confused about using the coefficient of static friction.

    The [itex]20N[/itex] box will move if the static friction force is less than [itex]12N[/itex]. [itex]μn= 10N[/itex] so the net force in the [itex]x[/itex] direction is [itex]2N[/itex].

    Now I know the box will move, but is this information relevant to the rest of the question?

    The kinetic friction force comes to be [itex] f_{k} = \mu_{k} N = 6.5N[/itex], so the net force when the box is moving is [itex]5.5N[/itex], and thus the work done on the [itex]20N[/itex] box is [itex] w = f x d 4.125N[/itex].

    I think the work done on the [itex]12N[/itex] box is still [itex]-9Nm[/itex] because it's change in GPE is still the same, it just changes more slowly.

    I know this is a bit of a ramble, but i'd really appreciate it if you could confirm whether my thoughts make sense.

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Oct 28, 2014 #2
    Both the parts are wrong .I haven't checked other parts yet .

    In a) force acting on 20 N block is the tension in the string attached to it ,not the weight of the hanging block .

    In b) the change in GPE is not equal to the work done .

    Draw FBD of the two blocks and write Newton's 2nd Law separately.
     
  4. Oct 28, 2014 #3
    Thanks for the response.

    I have calculated the tension in the string to be [itex]0.76 N[/itex], by using newton's second law and simultaneous equations. The work done on the 20N block is therefore [itex]w = f*d = 0.57Nm[/itex]

    The net force on the [itex]12N[/itex] block is equal to [itex]12 - T = 11.24N[/itex], thus the work done is [itex]w = f*d = 8.43Nm[/itex]

    If this is correct I shall try and apply it to the other parts of the question.

    Thanks for your help.
     
    Last edited: Oct 28, 2014
  5. Oct 28, 2014 #4
    This is incorrect .Recheck your calculations.

    I see you have edited your response . Your initial response in post#3 was correct .
     
    Last edited: Oct 28, 2014
  6. Oct 28, 2014 #5
    Are you sure?

    I changed my answer because I used the weight of the block rather than the mass. My new results were found by using 20/g and 12/g in exactly the same equations.
     
  7. Oct 28, 2014 #6
    Using 20/g and 12/g as masses should give you T = 7.5N .
     
  8. Oct 28, 2014 #7
    Fx = 20/g * a = T

    Fy = 12 - T = 12/g * a

    12 - 12/g * a = 20/g * a

    a = 0.375 ms^-2 Sub this value into top equation.

    20/g * a = T = 0.76N
     
  9. Oct 28, 2014 #8
    I think it should be a = 3.75 ms-2 .
     
  10. Oct 28, 2014 #9
    Ah - you are correct, though i'm getting 3.68ms^-2, using g=9.81ms^-2.

    Algebra error. Thanks!
     
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