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Work = force * displacement

  1. Nov 28, 2005 #1
    Hello Everyone,

    I am sure that I am missing something obvious here. It might have to do with the fact that force is more explicitly defined as the derivative of momemtum. Anyway, can anyone explain why the following example seems to violate one of the two equations for work? (thanks in advance)

    eq 1. Work = force * displacement
    eq 2. Work = change in kinetic engery

    Here is the example:

    I have a 747 and a matchbox car in a frictionless environment. Let's say the 747 weighs about 1 million pounds, and the toy car weighs about 1 ounce. (Exact weights, masses, conversions, etc are not necessary for this qualitative example.)

    If I apply a force of 1 Newton to each of them for 1 second, the change in kinetic energy of each of them will be the same, but the change in displacement will be vastly different. If I then divide by 1 to get power, I am left to believe that the power to create a 1 newton force varies dramatically in these two cases. Is this becase it is more difficult to maintain a 1 newton force on the toy because it is accelerating so much faster?

    I know if I turn the problem around and try to determine how much power is required to accelerate each of them at a certain rate or propel them across a given displacement over a given function of time, the power required works out exactly as I would expect.

    Also, I know that if I apply a given force to an object that ends up moving in the opposite direction of my applied force, that I actually do negative work on the object, and so a negative amount of power is required (in other words, I can extract power from the system). I have a feeling that something along these lines has caused me to missunderstand "work = force * displacement"; however, the exact problem with my example is not clear to me.

    I am just trying to get an intuitive feel for the concept.


  2. jcsd
  3. Nov 28, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    If you apply the same force for the same time the impulse (and thus the final momentum) will be the same, not the work done or the final kinetic energy. The kinetic energy of the matchbox car will be much greater (by a factor equal to the ratio of their masses).
    Power is not required to create a force; it's required to apply a force over a distance. Power is energy per time. The energy gained by the car is much greater.
    The power delivered to the car is much greater because it travels much farther in that one second.
  4. Nov 28, 2005 #3
    Thank you!
  5. Nov 29, 2005 #4
    I should have done this before I asked, but in case it might be useful to someone else, here is an easy example:


    1. bigbox has a mass of 5 kg.
    2. littlebox has a mass of 1 kg.
    3. we apply a constant horizontal force of 1 Newton to each of them for a duration of 1 second.
    4. things are slippery (no friction)

    1 Newton ---> bigbox (5 kg)
    1 Newton ---> littlebox (1 kg)

    The acceleration of littlebox is 1 m/s/s (f=ma --> 1 N = 1 kg * a --> a = 1/1 = 1)
    The acceleration of bigbox is likewise .2 m/s/s

    The total displacement of littlebox is .5 m (x = .5at^2 --> x = .5 * 1 * 1) The total displacement of bigbox is likewise .1 m

    The final velocity of littlebox is 1 m/s (v = v0 + at --> v = 0 + 1 * 1) The final velocity of bigbox is likewise .2 m/s

    The final momentum of littlebox is 1 kg m/s (momentum = velocity * mass )
    The final momentum of bigbox is likewise 1 kg m/s

    The final kinetic energy of littlebox is .5 kg m^2/s^2 (Joules) (ke = .5mv^2)
    The final kinetic energy of bigbox is likewise .1 Joules

    So as you can see although the momentums are the same, the kinetic energies are quite different. Intuitively, this makes sense to me since it is much more difficult to continue to apply a force to something as it moves away from you.

    This link has another good way of looking at it:


    I hope these are useful.


  6. Nov 29, 2005 #5
    I then started thinking about the case of a rocket engine, and decided that in that case change kinetic energy was relative to the rocket and not some other frame of reference. As it turns out, the kinetic energy is relative to any frame of reference. Here is an excellent treatment of the subject:


    I hope this is useful.


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