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Work Energy Theorem

  1. Oct 10, 2006 #1
    A 2kg block at rest on a frictionless surface is pulled for 2s by a 10N horizontal force (pull to the right).

    Use the Work Energy Theorem to determine the final velocity from the perspective of an observer moving to the left. Does the WET still work that observer?

    I did the calculations and got 14 m/s for the first part. However, I am stuck on the second part. Why does the Work Energy Theorem works for an observer who is moving at constant velocity?
  2. jcsd
  3. Oct 10, 2006 #2
    An observer moving to the left would have the effect of giving the block some additional initial and final velocity. Add this in to the problem and see how this does or does not affect the answer.
  4. Oct 10, 2006 #3
    I plugged it in the formula w = (1/2)mv^2 - (1/2)mv_o ^2

    w = (1/2)m (v^2-v_o^2)
    w = (1/2)(2)(14^2-4^2)
    w = 180 J.

    This is different from

    w = (1/2)(2) (10^2 - 0)
    w = 100J

    So the work in both cases are different. The answer key tells me that WET still works on the observer. But how? I don't know if I am misunderstanding something but if the work are different, how can WET still work?
  5. Oct 10, 2006 #4


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    Something else besides the velocities is different to the two observers. Think about the definition of work.
  6. Oct 11, 2006 #5
    The distance is also different. Thanks, I understand it now.
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