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B Force and potential energy

  1. Feb 22, 2017 #1
    I have just read chapter 7.4 FORCE AND POTENTIAL ENERGY in Sears and Zemansky's university physics 14th edition. There they show that a conservative force always acts to push the system toward lower potential energy in a one-dimensional motion with the equation Fx(x) = - dU(x)/dx. As I understand this it is saying that when a force is pushing in the same direction as its displacement then the potential energy is lowered. I felt like it made sense but after thinking of different scenarios I stumbled upon something that I couldn't understand.

    If positive x is to the right and I have a metal spring with it's right side attached to a wall. If I compress the spring with my thumb, my thumb experiences a force to the left from the metal spring while it is moving/pushing to the right. Since the force is in the other direction than the displacement Fx(x) = - dU(x)/dx gives an increase in potential energy.

    But what happens if we look at the spring? The spring is feeling the force from my thumb, a force to the right while it being compressed/displaced to the right. The force and displacement on the spring is in the same direction so according to my understanding of Fx(x) = - dU(x)/dx the potential energy should decrease? That doesn't make sense.

    According to all this my thumbs potential energy increases but the metal spring's decreases, what have I understood wrong? Writing this I'm thinking that it might be that my thumb is not a conservative force in the system but I'm not sure.

    (Was going to post this in the homework board but then I'd have to use the homework template and this doesn't feel like a homework template question, I'm guessing this is the right place to post?)
     
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  3. Feb 22, 2017 #2

    BvU

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    Yes, the potential energy of the spring-mass system ! Work is being done ON the system.

    It's like pushing a ball up the hill: you perform work to do that and your work can be converted to kinetic energy of the ball when you let it roll down again.
     
  4. Feb 22, 2017 #3

    PeroK

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    The equation you are using here is for the energy of a particle under a conservative force. The more the spring is compressed, the greater the potential energy of the particle.

    The potential energy in the spring is the potential energy of the particle.

    If you try to turn the system round and treat the spring as a particle under the force of your thumb, you will get a mess.
     
  5. Feb 22, 2017 #4
    So the equation is describing the change in potential energy for a mass that is affected by a conservative force? And since me pressing my thumb on a spring is not a conservative force I just get trash when trying to turn the system around, is this correct?

    Just to clarify, are you saying that if a particle affected by the force from a spring has 100 joules in potential energy, the spring also has 100 joules in potential energy?
     
  6. Feb 22, 2017 #5

    PeroK

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    Yes, all the PE is stored in the spring's mechanism, independent of the mass. If you hold the spring and remove the mass, then the mass loses all its PE!

    You might then ask where all the PE goes if there is no mass? That's where the assumption about a massless spring is exposed. It must go to KE of the spring.
     
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