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Work done by lifting block attached to spring

  1. Feb 2, 2010 #1
    If one hangs a spring (of constant 10lbs/ft) vertically and attaches a 10lb. weight to the spring, how much work is done in raising the weight 6 inches from where it hangs naturally?

    i drew a free body diagram and deduced that when you are lifting the weight up, the forces of gravity and spring force act downward (since spring force always acts opposite to displacement). so my equation for force is F = mg + kx = 10 + 10x. so i took the integral of that from 0 to 1/2 and i got 6.25 ft lbs or 75 in lbs. but my book's answer is 240 in lbs. how did they get that?
  2. jcsd
  3. Feb 2, 2010 #2
    Spring force acts opposite to displacement from natural length.
    It would still act upward opposite to gravity when you pull the block up,at least upto a feet(that's what the question says).
  4. Feb 2, 2010 #3
    no the spring force has to be down since i'm displacing the spring upwards from its natural length. i think the problem just has a typo. it should be 10 lbs/in instead of 10 lbs/ft. then when i integrate 10+10x from 0 to 6, i get the answer as 240 in lbs.

    anyways thank you for your input!
  5. Feb 2, 2010 #4


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    I don't believe that is correct, you are displacing it upward from its natural hanging length when the weight is on it, in its equilibrium position (that's what the problem says). So when you move the spring up 6 inches, it is still in its stretched position, so the spring force acting on the weight is up. The answer is not 240 in-lb, even if there was a typo. Throw out the book answer, and toss out the question with it. Aim 1732's response was correct.
  6. Feb 4, 2010 #5
    oh i understand now. so would it be the integral from 0 to 1/2 of 10x-10?
  7. Feb 4, 2010 #6
    It would be 10-10x since the force is upwards.
    A better way to look at the problem is in terms of energy.Specifically,

    W(non-conservative) + W(external) = Change in mechanical energy of system
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