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  • 1 Post By MarkFL
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Math Help - Work

  1. #1
    Aug 2011


    I'm definitely no physics major, and I'm still trying to grasp the concept of Work = Force x Distance. Scenario: Let's say I'm dragging a huge rock with a rope, and after two minutes of pulling (non-variable force) I manage to drag it 20 feet. Now, let's say I pull it another 20 feet, but this time I pull twice as hard, and get the job done in only a minute. Did I do twice as much work the second time? I guess that fits the formula. But intuitively I'm thinking that I did the same amount of work: I was pulling twice as hard, but only half as long.

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  2. #2
    MHF Contributor MarkFL's Avatar
    Dec 2011
    St. Augustine, FL.

    Re: Work

    You are right, it is the distance over which the force is applied, not the duration of time, which is relevant in the computation of work done. However, you cannot say that if you double the force, the time will be cut in half. Displacement varies quadratically with acceleration, which is proportional to the (net) applied force. Since there is a force of kinetic friction to contend with, doubling the applied force will not double the net force.
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  3. #3
    MHF Contributor ebaines's Avatar
    Jun 2008

    Re: Work

    If you double the force and keep the distance the same the amount of work done does indeed double. That work has to result in an increase in energy, and there are two places where that energy resides:

    1. Kinetic friction between stone and ground, which generates heat. In both cases the work done to overcome friction is the same.
    2. Kinetic energy of the stone. If in the first case you pull the stone with just enough force to equal friction then it doen't accelerate while you are pulling on it. Then if in the second case you double the amount of force the result is that the stone accelerates, moving faster and faster, so that after 20 feet it is moving at a much faster pace than in the first case. So your work has gone into increasing the kinetic energy of the stone.

    Mathematically what you have is this:

    Work done = Force x Distance = Work to Overcome Friction + Work turned into KE:

    Here the coefficient mu is the coefficient of kinetic friction.
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