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Kinetic energy ignoring friction

  1. Dec 8, 2015 #1
    1. The problem statement, all variables and given/known data
    2 trucks, one with twice as much mass than the other, are at rest on separate tracks. A man pushes each truck for 5 seconds. If you ignore friction and assume equal force is exerted on both trucks - which truck will have the most kinetic energy afterwards?

    2. Relevant equations

    3. The attempt at a solution
    Straight away I would say that the lighter truck would have more kinetic energy as it would take less force to move it, but as friction is being ignored and there is an equal force on both of them this stumps me, one of the options is that there is not enough information to answer the question but I feel like because friction is being ignored and there's equal force there is a really simple answer to this. I would appreciate someone steering me in the right direction so I can find the answer out for myself.

  2. jcsd
  3. Dec 8, 2015 #2

    Doc Al

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    Staff: Mentor

    How do the accelerations compare? The final velocities after 5 seconds?
  4. Dec 8, 2015 #3
    Thankyou for your prompt reply,

    Well since a = F/m and I know that the force is the same but that the masses are different, I can say that the acceleration of the heavy truck (x) would be half that of the lighter truck (y).
    For example:
    mass(x) = 2
    mass(y) = 1

    a(x) = 5/2 = 2.5
    a(y) = 5/1 = 5

    For the final velocity as both are stationary to begin with I can say that vf = at, meaning vf(x) would be half that of vf(y).

    Since KE = 1/2m*v^2 if velocity is doubled KE will then be quadrupled (as velocity is squared) however as the lighter truck (y) is half the mass of heavier truck (x) it means that the KE of (y) is only double that of (x).

    Therefore the lighter truck has more KE.

    Thanks! (if I have somehow managed to go completely wrong with my explanation please let me know)
  5. Dec 8, 2015 #4
    You can apply Newton's law too, F=dp/dt
  6. Dec 9, 2015 #5

    Doc Al

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    Staff: Mentor

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