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Does work account for direction when considering only KE?

  1. Nov 1, 2015 #1
    1. The problem statement, all variables and given/known data
    A 60 g particle is moving to the left at 20 m/s . How much net work must be done on the particle to cause it to move to the right at 40 m/s ?

    2. Relevant equations
    1/2 mv^2 = KE, W = ΔE

    3. The attempt at a solution

    My initial method of solving this problem was to take the initial kinetic energy: 1/2 (60/1000) 20^2, and the final kinetic energy: 1/2 (60/1000) 40^2, then to find the work I would add the two, since they are opposite directions. This nets 60 J of work, but the answer is 36 J.

    I'm aware that the correct way to find the solution to an equation like this is to take the final kinetic energy of the particle subtracted by the initial, but that simple method doesn't make conceptual sense to me. It seems like the solution to the problem would be the same if the particle was moving to the right at 20 m/s initially. But why? To give a simpler scenario: if a particle moves 1m/s, and another particle moves -1m/s, would the work it takes to move each particle to 2m/s be the same?

    Now with a question like that comes other considerations. Work is the change of energy, but in a scenario like this, it might also be Fd. So how much force would it require to move a particle moving at -1 m/s to 1 m/s? over any distance, the work, from my understanding to the solution from this problem, -seems- to be 0. Meaning it takes 0 N of force to do this. Which just doesn't make any logical sense to me.
     
  2. jcsd
  3. Nov 1, 2015 #2

    haruspex

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    The trouble is that work is not the same in all frames of reference, even if they are all inertial.
    In the frame of reference of the original motion of the particle, you would calculate the work to accelerate the particle from rest to 60m/s.
    In the ground frame, however, consider: is it possible for a force to change the velocity of the particle without doing work?
    Note I wrote velocity, not speed.
     
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