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About the concept of work done (High school student)

  1. Apr 15, 2014 #1
    Hi I am kind of Confused about an example in work done:

    It says that a man applies 250 N parallel to the slope to push a box up a slope for 3 m, Now I know that the formula for work done if Force by Perpendicular distance which is 250 x 3 = 750 (which is the correct answer in my book). The thing is, isn't the object "Gaining" Gravitational potential energy ie change in its energy ie transfer of energy and therefor we should add that energy to the 750??
    This is where I am confused.
    Can any one help please :)

    Edit: Or is the Force x Distance formula includes all changes in energy together?
  2. jcsd
  3. Apr 15, 2014 #2


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    Staff: Mentor

    Welcome to the PF.

    Since you are explicitly given the force needed to push the box up the ramp, that includes both the force needed to push against the friction of the box on the ramp, and the force needed to lift the box up the delta-height that it gains. Does that make more sense?
  4. Apr 16, 2014 #3
    So what you mean is that if we get the component of each force and see the distance moved by each to get the total energy transfered it would add up to the Force x distance? so this formula should really be
    Resultant force by displacement right?
  5. Apr 16, 2014 #4

    Doc Al

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    Staff: Mentor

    It depends on what you want to compute.

    If you want the work done by the man, then that is simply the force he exerts times the displacement. But to determine the final energy of the box, you'll need to consider all the forces acting, which includes gravity and friction.
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