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Homework Help: Energy conservative (Kinetic Energy and Potential Energy)

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data
    May i know how do i compute an energy conservative equation for the scenario below, which i don't understand.

    If a book is lifted up by a force in order to make it moving upward with a constant velocity.
    For conservation of total energy, W1-2 = U1-U2 = K2 - K1
    If the object is moving in a constant velocity, isn't K2 - K1 = 0
    However, U1 is not equal to U2.

    It seems not correct obviously, hope someone could enlighten me.

    I guess W1-2 = U1-U2 = K2 - K1 is only valid for isolated system, which should be a system without any external force involves except gravitational force.

    I'm very confuse now, hope someone could help me.
  2. jcsd
  3. Nov 19, 2009 #2
    d=distance moved
    earth does -ve work , you do equal +ve.


    +ve change in potiential energy always needs external energy source.

    so W=U2-U1=m.g.d
  4. Nov 19, 2009 #3


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    The work done by non conservative forces (such as the lifting force on the book) is
    [tex] W_{nc} = (U_2 - U_1) + (K_2 - K_1)[/tex] or in its alternate form, [tex]W_{nc} = \Delta U + \Delta K [/tex]

    For a system subject to conservative forces only (like gravity or spring forces), then [tex] (U_2 - U_1) + (K_2 -K_1) = 0 [/tex], or [tex]\Delta U + \Delta K =0 [/tex], and the work done by such conservative forces is [tex] W_c = U_1- U_2[/tex], or [tex] W_c = - (\Delta U)[/tex] in its alternate form.

    For a system subjct to both conservative and non conservative forces, then the total work done by those forces is [tex]W_T = (K_2- K_1)[/tex], or [tex]W_T= \Delta K[/tex]

    staraet's answer for the work done by the lifting force is also correct.

    This can get confusing for sure, but actually, I'm not exactly sure what your question is. Mechanical Energy (U + K) is NOT conserved when non-conservative forces that do work are acting.
    Last edited: Nov 19, 2009
  5. Nov 20, 2009 #4
    why we no need to include the [tex]\Delta U[/tex]? What if the object is in a falling motion with air resistive force acting on it.

    Could i write it as: mg([tex]\Delta h[/tex]) - Ffriction([tex]\Delta h[/tex]) = [tex]\Delta K[/tex] + [tex]\Delta U[/tex]

    or i should write it as without taking the work done by gravitational force: - Ffriction([tex]\Delta h[/tex]) = [tex]\Delta K[/tex] +[tex]\Delta U[/tex]
  6. Nov 20, 2009 #5
    always think in terms of gain and loss

    we some thing goes along only acting force , it gains k.e
    so decreases its p.e

    when ball falls, so it means, going along a force, so increases k.e (say after dt);
    then if u find k.e is constant-- it means body has spent that k.e in either heat, vibrational,mass or something in that dt time.
    then write equations. donot write before.............

    here friction is heat. so E spent in heat= mgh
  7. Nov 20, 2009 #6
    I think my problem is I don't understand between work done, kinetic energy, and potential energy.
  8. Nov 20, 2009 #7


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    I suggest sticking with PhanthomJay's first equation:
    As you realize for your example of lifting a book, K2=K1=0. Also, Wnc is the work W1-2 done by lifting, so in that case

    W1-2 = U2 - U1

    If instead we have an isolated system with only conservative forces such as gravity, we have Wnc = 0 and conservation of mechanical energy applies. PhanthomJay's equation then becomes:

    0 = U2 - U1 + K2 - K1

    this is often rewritten in the form

    U1 + K1 = U2 + K2

    and is simply a statement that the total energy (potential + kinetic) does not change.
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