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Work done related to spring

  1. Jul 27, 2009 #1
    1. The problem statement, all variables and given/known data
    A weight of mass m is connected to a spring of spring constant k. Initially the weight hangs down and is at rest. Support the weight by hand and lift it up slowly until the length of the spring is equal to its natural length. What is the work done by the hand during the process? Let the gravitational constant be g.


    2. Relevant equations
    F = kx
    Weight = mg
    Work done on spring = 1/2 kx^2

    3. The attempt at a solution
    When the object is at rest :

    [tex] F = W [/tex]

    [tex]mg = kx[/tex]

    [tex]x=\frac{mg}{k}[/tex]

    When the mass is lifted up, the work needed = gain in potential energy + work done to compress the spring

    [tex]
    W total = mgx + 1/2 kx^2

    =\frac{m^2 g^2}{k} + 1/2 \frac{m^2 g^2}{k}

    =\frac{3 m^2 g^2}{2k}
    [/tex]

    But there is no such choice for my answer
    Where is my mistake?

    thx
     
    Last edited: Jul 27, 2009
  2. jcsd
  3. Jul 27, 2009 #2

    rl.bhat

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    Homework Helper

    Hand does work on the mass against the gravity. So mgx should be negative.
     
  4. Jul 27, 2009 #3
    If mgx is negative, then :

    [tex] W total = - \frac{m^2 g^2}{2k} [/tex]

    There is choice [tex] W total = \frac{m^2 g^2}{2k} [/tex]

    What is the meaning of the negative sign? The work done by the hand is negative so the hand received work?

    thx
     
  5. Jul 27, 2009 #4

    Doc Al

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    Staff: Mentor

    The work done by the hand is positive: The hand pushes up and the mass moves up.

    But what about the work done by the spring?
     
  6. Jul 27, 2009 #5
    The work done by the spring is negative because it's being compressed? So the work done by the hand to the spring is positive ??

    I get confused with the concept...

    thx
     
  7. Jul 28, 2009 #6

    rl.bhat

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    Attach a weight to the spring when it has its natural length. The potential energy in the spring is zero. The weight must have some potential energy with respect to some reference point. Let the reference point be the lowest point of the un stretched spring.
    So the total energy of the system = 0 When the weight is allowed to fall through a height x, the rise in potential energy of the spring = 1.2*k*x^2 and the potential energy of weight is - mgx. So the total energy of the system = 1/2*k*x^2 - mgx.
    When the work is done on the weight by the hand, increase in the PE of the weight is mgx and PE of spring is zero. So final energy of the system is mgx. Hence the net work done on the system = the change in the energy of the system
    = mgx - ( 1/2*k*x^2 - mgx) = 2mgx - 1/2*k*x^2
     
  8. Jul 28, 2009 #7
    So the work done by the hand = 2mgx - 1/2*k*x^2 = 3/2 m^2 g^2/k but there is no such choice...

    The choices are :
    a. m^2 g^2 /k
    b. 2 m^2 g^2 /k
    c. m^2 g^2 / 2k
    d. mg/k
    e. 2mg/k
    f. mg/2k
    g. kmg^2
    h. 2kmg^2
    i. kmg^2/2

    Though i think 3/2 m^2 g^2/k is the right answer...
     
  9. Jul 28, 2009 #8

    rl.bhat

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    Since in the problem work done by the hand is asked
    W = mgx - 1/2*k*x^2.
    It is obvious, because if you compare the work done by the hand to lift a mass directly with a mass attached to a stretched spring, ii is easier in the second case.Again the work done = change in potential energy.
    Potential energy is positive if the work is done against repulsive force.
    Potential energy is negetive if the work is done against attractive force.
     
  10. Jul 28, 2009 #9

    Doc Al

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    Staff: Mentor

    As the spring compresses, it pulls up on the mass. Thus the work done by the spring is positive.
    Yes.

    If the force and displacement are in the same direction, the work done is positive.
     
  11. Jul 28, 2009 #10

    Doc Al

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    Staff: Mentor

    The final energy of the system, when returned to the unstretched position, is 0 (since you measure PE from that point).
     
  12. Jul 28, 2009 #11
    In this problem, how to determine if it is repulsive or attractive?

    thx
     
  13. Jul 28, 2009 #12

    rl.bhat

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    Homework Helper

    It is a general statement. In fact it is nothing to do with this problem.
    A peculiar situation arises in this problem.
    Initially the block is rest, because the gravity tries to pull it down and stretched spring pulls it up. Since these two force are equal and opposite net force is zero.
    When you hold the block in your hand, the gravitational force is neutralized by the normal reaction of the hand.
    When you try to push the block slowly up, the stretched spring readily assist you to lift the block.
    The force is exerted by your hand. Work done = F*x. But you don't know much force is exerted by your hand.
    But while stretching,work done by the gravity = mgx.
    Since you are returning to the unstretched position, F*x = mg*x
    Part of the wark is done by the spring. So the net work done by the hand is equal to W = mgx - 1/2k*x^2.
     
  14. Jul 28, 2009 #13
    That's plenty of explanation

    thx a lot rl.bhat and doc al ^^
     
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