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Work done by spring vs hookes law?

  1. Nov 7, 2011 #1
    what is wrong with this problem?


    In Fig. 14-36, a spring of spring constant 3.60x104 N/mis between a rigid beam and the output piston of a hydraulic lever.An empty container with negligible mass sits on the input piston.The input piston has area Ai, and the outputpiston has area 15.5Ai. Initially the spring isat its rest length. How many kilograms of sand must be (slowly)poured into the container to compress the spring by 5.80 cm?

    See above link if you want to see image.
    Basically I solved this same as was solved in the above link, but I do not understand why Hooke's law is used instead of the kx^2/2. What I did was:

    work to compress spring=kx^2/2
    work applied to piston 1=mgy1, where m is mass of sand
    the two works are equal by Pascal's Law, so kx^2/2=mgy1

    the volume of water displaced same at both pistons:
    Solving for yi and x and plugging into my work equations gives:


    So the V does not cancel out. But I cannot see how Hooke's Law can be used. Yes, the spring will feel force of kx once it is compressed 5cm, but the work to acheive this compression is a variable force, so we must integrate or deal with the work...?
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Nov 7, 2011 #2


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    From what I can gather, if you use conservation of energy, then you'd have to take into account the fact that the beam might move up some distance y2 while the spring will compress a distance x.

    Whereas if you were using Hooke's law, the pressure transmitted throughout the fluid is constant. So the force on the beam is the force that the spring is seeing.
  4. Nov 8, 2011 #3

    The distance the spring compresses is obviously equal to the distance the beam moves up...so that doesnt solve it..
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