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Question involving work and springs

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data

    When a 4.00-kg object is hung vertically on a certain light
    spring that obeys Hooke’s law, the spring stretches 2.50 cm.
    If the 4.00-kg object is removed, (a) how far will the spring
    stretch if a 1.50-kg block is hung on it? (b) How much
    work must an external agent do to stretch the same spring
    4.00 cm from its unstretched position?

    2. Relevant equations

    F = -k*x
    W = F*x

    3. The attempt at a solution

    **I get a) correct so you can skip to b) if you want

    a) First using the 4kg block I get a force of 39.2N. Then using 39.2 = -k*-0.025 and solving for k I get k= 1568 N/m. .
    Then inputting k for the 1.5 kg block(F=14.7 N) I get 14.7=1568x. x=0.00938 m or x= 0.938 cm

    b)This is what I tried to do: F=-1568*-0.04=62.8 N So the force needed is 62.8 N. So using W=F*x, I get W= 62.8*0.04 = 2.51 J

    I can see in the back of the textbook the answer is 1.25 J, so I'm tempted to just divide my answer by 2 but then I won't understand why. Can anyone out there explain this to me please??
     
    Last edited: Nov 3, 2012
  2. jcsd
  3. Nov 3, 2012 #2
    The force is not constant as the spring is stretched, so you need to integrate. Alternatively, if you know the formula for spring's potential energy, you could just use that.
     
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