1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work of spring vs force of spring?

  1. Nov 7, 2011 #1
    Let's say you have a hydraulic lever. On the end with the smaller area, you want to slowly pour sand, so as to compress a spring on the other end, which has the bigger area. If you want to find how much mass of sand is needed to compress this spring a certain distance, would you use Hooke's law or W=kx^2/2 ?

    My reasoning is that you would say kx^2/2=mgy, where y is the vertical displacement of the smaller piston, so mgy is the work done by the sand, and since "the work done on the input piston by the applied force is equal to the work done by the output piston in lifting the load placed on it" (according to my book), then mgy must equal kx^2/2.

    Is this right?
  2. jcsd
  3. Nov 7, 2011 #2
    I believe you are correct.

    Instead of work, you can also think of it in terms of potentials (which are very similar). The potential energy lost by the sand moving down in the gravitational field must go somewhere, and if the spring is compressing, the energy is being converted from gravitational potential energy to spring potential. So I think you are correct when you use kx^2/2, because this is the potential of the spring (The potential of a spring is the force integrated over x)
  4. Nov 7, 2011 #3
    That's what I thought too but then I saw several solved problems like this one from various sources that used only Hooke's law, not integrating it...? So our reasoning must be wrong?
  5. Nov 7, 2011 #4

    If we were to drop sand on a spring directly, the amount it would compress would come from the force diagrams I think... So if I drew my spring on a table, and I put sand on top, it would compress an amount mg/k. However, one might be able to use energy. If I have a mass at the top of the spring, and I drop it, and the mass drops a distance x, than the gravitational energy lost by the mass is mgx, while the energy gained in the spring would be kx^2/x.

    I think you can use both methods.
  6. Nov 7, 2011 #5
    Im quite sure you cannot just use both methods since they do not lead to the same answers...

    and I think in the example you mentioned you would use work, not Hooke's law. If you put sand on top of your spring, you would use mgy=kx^2/2, because the force from the sand stays constant, but the force needed to compress the spring varies. So at first, the spring will compress more easily than towards the end..

    I mean obviously you cannot use both ways since one way leads to x=[itex]\sqrt{2mgy/k}[/itex]
    while the other leads to x=mg/k

    Hooke's Law I think would apply to something like the following: Lets say we know that once the sand fully compresses the spring, the force the spring exerts upwards towards the sand is 10 N. Then we can find how much the spring has been compressed from its equilibrium point by x=10/k ...
  7. Nov 7, 2011 #6
    You might be right. I know that if we want to determine the speed of a mass at the end of a spring we would need to use conservation of energy, and the kx^2/2, but I guess if you wanted to find the amount of mass that will compress the spring a particular distance you could use F=ma equations.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook