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A work done comparison ( i find it confusing so a little help)

  1. Sep 7, 2014 #1
    [Mentor's note: this post does not use the homework template because it was originally posted in a non-homework forum. It was moved here because it had already received some help.]


    Weiping lifts a rock with a weight of 1.0 N through a height
    of 1.0 m, and then lowers it back down to the starting point. Bubba
    pushes a table 1.0 m across the
    oor at constant speed, requiring
    a force of 1.0 N, and then pushes it back to where it started. (a)
    Compare the total work done by Weiping and Bubba.

    I think Bubba did the greater work because he was pushing the object at a constant velocity where as weiping only lifted it . Is my assumption correct?
    Last edited by a moderator: Sep 7, 2014
  2. jcsd
  3. Sep 7, 2014 #2
    N is a unit of force, not weight, so this question makes no sense.
  4. Sep 7, 2014 #3
    The real assumptions you have here are not specified clearly. They are:

    A1. The weight is lifted and lowered in a constant gravity field.

    A2. The table is also subject to the gravity field, and it has a non-zero coefficient of friction with the floor.

    With these assumptions, the rest is simple. A constant gravity field is potential, so total work done on it or by it depends solely on the difference in the heights of the initial and final position. The work done on or by friction depends solely on the distance (not to be confused with displacement!) covered.
  5. Sep 7, 2014 #4
    Isn't weight of an object force directed towards the center of the earth?
  6. Sep 7, 2014 #5
    Weight is a force, so measuring it in Newtons is perfectly fine.
  7. Sep 7, 2014 #6
    ok so if we assume the Gravitational field is constant and there is no friction then who did the greater amount of work?
    But if we assume there is no friction then a forward force of 1 N would cause the table to accelerate forward but it moves at a constant velocity so lets assume there is 1 N frictional force then?
  8. Sep 7, 2014 #7


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    Weight is force, so you objection makes no sense.
  9. Sep 7, 2014 #8
    If there is no friction, why did Bubba need to apply 1 N to move the table?
  10. Sep 7, 2014 #9
    i just realised that a moment ago edit :P
  11. Sep 7, 2014 #10


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    Why don't you just apply the definition of work to both cases, and see what values for work done you get?

  12. Sep 7, 2014 #11
    Thats the first thing i did it gives weiping did 20 J and Bubba did 2 J but what confuses me is that the object is moving with a constant velocity so has some kinetic energy so shouldn't work done by bubba be the sum of the kinetic energy and the work done in moving the object?
  13. Sep 7, 2014 #12
    How did you get this?


    What we have here is an idealization. Of course it is not practically possible to get a stationary object, and move it back and forth with constant velocity - the velocity has to go from zero to some value, then it needs to change sign, and then go back to zero. We idealize it as follows:

    (1) The object is accelerated to a constant velocity in infinitesimal time and over infinitesimal distance. The requires work E.

    (1) The object is decelerated to zero velocity in infinitesimal time and over infinitesimal distance. The requires work -E.

    E + (-E) = 0, so we ignore it.
  14. Sep 7, 2014 #13


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    If the gravitational force on the rock is 1 N, what is its mass?

    If you push with a 1 N force against an object that is coming toward you through a 1 meter distance, how much work have you done?
  15. Sep 7, 2014 #14
    I got that by the work done in lifting the object plus the work done in lowering it back down mgΔh+mgΔh

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