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Work and Energy

  1. Jan 24, 2007 #1
    1. The problem statement, all variables and given/known data
    A 0.010-kg positively-charged particle is accelerated by a uniform electric field directed to the right. The object accelerates from 5.0 to 15.0 m/s while traveling a distance of 0.10 m.

    Suppose the particle is moved by an external force a distance of 0.10 m against the field. What is the change in electric potential energy of the particle? Is the change an increase or a decrease?

    2. Relevant equations
    DeltaU = -W

    3. The attempt at a solution
    I know that Fext must be greater than Fel, but how much greater?
    W=fd, distance is .10m. How do I determine the change in EPE, specifically calculating the amount of work neccesary?
  2. jcsd
  3. Jan 24, 2007 #2

    Chi Meson

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    What kind of energy does this particle gain?
  4. Jan 24, 2007 #3
    It gains kinetic energy, but how do I determine the appropiate amount of energy to overtake the electric field?
  5. Jan 24, 2007 #4

    Chi Meson

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    This problem is exactly like lifting a mass in a gravitational field.

    It takes "W" work to lift it to height "h." How much PE does it have?

    The object is let go, and it falls the same distance it was lifted. How much PE does it lose/ How much KE does it gain? Compare these quantities to the quantity W.
  6. Jan 24, 2007 #5
    Okay, the PE at the drop height is mgh while KE is 0. They reverse numbers as it falls, so KE picks up while PE decreases. The entire time Work equals the change in mechanical energy, so W = PE at it's greatest point. I still don't see how that equates to the electric field since the particle just doesn't stop moving when it hits the "ground." I already know Wel, so Wext would have to be greater - but I still don't get it.
  7. Jan 24, 2007 #6
    Okay, I've tried to rethink it. The work will be my force*distance. I know the distance needed, .10m, but I don't know what I should do for the force. It has to be greater than the Force of the electric field. What really confuses me is the follow up question which asks what the work done by the external force would be if the movement was constant.

    So ... the movement from the first question I posted has an accelleration - so wouldn't that change it's force since F=MA, in turn changing the work? AHH!
  8. Jan 24, 2007 #7

    Chi Meson

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    no, it has to be equal to (on average) if it moves .10 m. The applied force has to be a little bit more at first (in order to accelerate from zero) but then, to come to a stop, the applied force has to be a little bit less. Over the entire distance, the average applied force = the electric force, the net change in KE is zero, so W = change in PE.

    This is a little unclear; are we talking "from rest to a constant motion against the field"?
  9. Jan 24, 2007 #8
    So, I need to determine the force of the electric field then set it equal to the force of the external object? I know then that W=fd, change in KE is 0 since it begins and finishes at rest, so like you said so W = my DeltaU.

    The question for the next part is: For the situation of part c, suppose the particle is moved at constant velocity. Determine the work done by the external force. Begin with the generalized conservation of energy equation: Wext = DeltaK + DeltaUel

    I'd assume for that situation that Delta K would be 0 since it would be at constant speed, since there is no acceleration. The external force's work would then directly equal my DeltaUel - is that a correct assumption?
  10. Jan 24, 2007 #9

    Chi Meson

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    Yes you are correct, but since your position in the field will be continuously increasing, the change in PE and thus your work will also be continuously increasing. So there won't be "work done" but "the rate at which work is done," which is also known as power.

    Are these questions from a book? Or did a teacher write them?
  11. Jan 24, 2007 #10
    It's an online course, that's why I've been utilizing this board lately! I'm not sure whether they're from a book or his own, but he's pretty good with asking questions that the book covers. I just don't always get them on the first couple of tries :).
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