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Homework Help: Work-Kinetic Energy Theorem applied to a Spring Force

  1. Apr 18, 2010 #1
    If I apply the Work-Kinetic Energy theorem to a situation in which an object is lifted or lowered then I can form the equation K(f)-K(i)=W(net)=W(applied)+W(gravity)

    This equation shows that if K(f)=K(i) then the above equation reduces to:
    W(applied)= -W(gravity)

    Now in the situation in which a force is applied to an object attached to a spring we can form a similar equation:

    Now my textbook says that this equation reduces to W(applied)= -W(spring) if and only if the object to which the force was applied to is stationary before and after the displacement.

    Why is this so.If the object has some value K(i)>0 at the start of the displacement and at the end of the displacement this value is the same ie. K(f)=K(i), then surely the work done by the applied force to maintain the kinetic energy must be equal in magnitude and opposite in sign to the work done by the spring.

    Am I missing something here?I don't see how my reasoning is incorrect.
  2. jcsd
  3. Apr 18, 2010 #2


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    You are correct. As long as K(f) = K(i), i.e. the applied force is such that the final kinetic energy equals the initial kinetic energy, W(applied) = -W(gravity). The object may be stationary with respect to the room it is in, but with respect to the Moon or Sun or the Milky Way it is not. Like any energy, kinetic energy has an arbitrary zero point, which your textbook does not seem to recognize.

    *** Edit ***
    I meant to say W(applied) = -W(spring)
    As PhantomJay suggests, this would be the case for a horizontal spring.

    To illustrate: If the applied force has magnitude kx in a direction opposite to the spring force and is applied over some fraction of the amplitude, then the system will lose energy and oscillate with a smaller amplitude.
    Last edited: Apr 19, 2010
  4. Apr 18, 2010 #3


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    I guess you are assuming a horizontal spring such that the workdone by gravity is excluded. Your reasoning is incorrect because the kinetic energies before and after cannot be the same because the applied force will accelerate the object if it is not fixed (stationary), per Newton 2, implying a change in velocity, unless friction between the object and surface balances the appied force, so that it moves at constant velocity, and in which case you must consider work done by friction. The book answer is correct.
  5. Apr 19, 2010 #4
    Good point,I forgot to mention that the system is isolated,horizontal and on a frictionless surface.And now I realise that this question relates to more generally to Conservation of Energy rather than to the work-kinetic energy theorem.

    OK,with that in mind,say that the object compresses the spring from an arbitrary point [tex]x_i[/tex] to a point [tex]x_f[/tex].The object at point [tex]x_i[/tex] has a Kinetic energy [tex]K_i>0[/tex].Now as the object moves to a greater displacement from the equilibrium point of the spring ie. to the point [tex]x_f[/tex], the spring exerts a force on the object according to the relation [tex]F_{spring}=-kx[/tex].Now suppose that an applied force acts on the object during the displacement from [tex]x_i[/tex] to [tex]x_f[/tex] and is exactly equal in magnitude but opposite in direction to the spring force ie. [tex]F_{app}=kx[/tex].Then the work done by this applied force during the displacement from [tex]x_i[/tex] to [tex]x_f[/tex] is:

    [tex]W_{app} = \int_{x_i}^{x_f}F.dx[/tex]

    which is equal to

    [tex]W_{app} = \int_{x_i}^{x_f}kx.dx[/tex]

    As the object moves from [tex]x_i[/tex] to [tex]x_f[/tex] there is a change in the elastic potential energy of the spring.The change in potential energy can be related to the work done by the spring force by the equation [tex]\Delta{U}=-W_{spring}[/tex]

    So [tex]\Delta{U} = \int_{x_i}^{x_f}kx.dx[/tex]

    So the work of the applied force is equal to the change in potential energy of the system and equal in magnitude but opposite in sign to the work of the spring force.This means that the conservation of energy equation [tex]W_{app}=\Delta{K}+\Delta{U}[/tex] simplifies to [tex]\Delta{K}=0[/tex] or [tex]K_f=K_i[/tex].By reverse reasoning if [tex]K_f=K_i \ge 0[/tex] then [tex]W_{app}=\Delta{U}[/tex] , therefore [tex]W_{app}=-W_{spring}[/tex]

    Still,my textbook says,"If the block is not stationary before and after the displacement,then this statement is NOT true." Fundamentals of Physics.

    Go figure...I certainly can't.

    PS If the symbols and equations are screwed up then apologies.I'm not really too sure what's going wrong.When I preview my post most equations and symbols which were meant to appear are wrong,even though the source code is correct in the context.I'm a newbie so hopefully I'll figure it out.
    Hopefully you gan gather the meaning regardless.
    Last edited: Apr 19, 2010
  6. Apr 19, 2010 #5


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    You're right. All you need is Ki=Kf. I'll hesitate to say the book is wrong since it's possible the statement is correct in context.
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