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Hooke's law Potential Energy

  1. Jul 7, 2013 #1

    I am self studying physics and am a bit confused with Hooke's law.

    Using the conservation of energy equation:

    (PE+KE)_i = (PE+KE)_f

    I don't see why the springs potential energy (½ kx^2) goes in the final part.
    Could someone please explain?

    EDIT: Also, (I'm a new user) can you type in LaTeX on here?
    Last edited: Jul 7, 2013
  2. jcsd
  3. Jul 7, 2013 #2
    You can type in latex: just write
    Code (Text):
    your formula
    Code (Text):
    Alternatively, use the [itex]\Sigma[/itex] button inside the editor.
    What do you mean by "the spring potential energy going in the final part"?
  4. Jul 7, 2013 #3
    Thank you for your reply.
    Some questions which I see in the textbook I am using have the spring potential, [itex]$\frac{1}{2}kx^2[/itex]
    listed in the 'final' part of the Conservation of Energy formula.
    Problems like these usually involve a stationary block connected to a spring, and another block which collides with the first block.
  5. Jul 7, 2013 #4
    It would be helpful if you gave an example of such a problem, however, there can be many reasons why the connected block still holds some potential energy in the "final" instant: friction, for example, could prevent it from going back to the rest position.
    Please, post one of such problems so that it's easier for me to help you :-)
  6. Jul 7, 2013 #5
    Example Problem:

    A 12.0 g bullet is fired horizontally into a 100 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 150 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 80.0 cm, what was the speed of the bullet at impact with the block?
  7. Jul 7, 2013 #6
    Ok, then the "final" instant is not the "equilibrium" instant and it makes sense to have potential energy without friction.

    When you call time instants as "initial"/"final" it doesn't mean they need to be the starting/ending point of the process, but simply two time insants such that [itex]t_i<t_f[/itex]. In the case of your problem, you don't need to know what happens when your system is at rest, you just need to know how compressed the spring was, because the kinetic energy of the bullet is completely converted into potential energy of the spring.

    Before the impact, the block is at rest in the equilibrium position of the spring, so it's potential energy is 0. Then there is the impact, which compresses the spring. At some point, the spring "rebels" to the compression and elongates again. To solve your problem, though, we are not interested in this last stage of the process: we just need to know about the first compression.

    So we call "initial" any moment before the impact and "final" the istant of maximum compression of the spring.
    Then you have
    Kinetic energy of the bullet before the impact = potential energy of the spring when it is first compressed after the impact.

    Is it clear?
  8. Jul 7, 2013 #7
    Thanks! Just one more thing since i'm not 100% getting it:
    Can you write an equation which models the energy? Like which side does the PE/KE go?
  9. Jul 8, 2013 #8


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    Your equation can be arranged (or re-arranged) any way you like as long as you initially put the components on the appropriate side - so that the total energy you start with, equates to the total energy you end up with.
    You have to be aware that some of the 'energy out' may not be in Mechanical form (i.e. it may be regarded as 'lost' to the system under examination).
  10. Jul 8, 2013 #9
    The energy conservation equation (which holds if there is no friction) tells you that the total mechanical energy is always the same: you could also write it as
    KE + PE = constant for all t

    The sum of KE and PE is always the same for every time instant, whatever you wish to label it ("initial", "final", "t1", "t2" ecc.).
    As sophiecentaur says, as long as you are equating the TOTAL energy taken at different times, you are then free to move the kinetic (or potential) energy which side you want.
    For example:


    so [itex]KE(t_3)=KE(t_1)+PE(t_1)-PE(t_3)[/itex] or [itex]KE(t_1)=KE(t_2)+PE(t_2)-PE(t_1)[/itex] etc.
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