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Homework Help: Calculate the work as the MINIMUM amount of work

  1. Sep 11, 2006 #1


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    Hey. Heres the question:

    "A weightlifter lifts a 100kg mass 2.4m from the floor. He then holds the mass steady for 3 seconds before dropping it. What is the work done by the weightlifer in raising the mass?"

    Well. I know mass is applying a force downward of 980N because of gravity, so for this question, since an acceleration of the mass is not stipulated, do i just calculate the work as the MINIMUM amount of work he would do? Which would be around 2352J, so for the answer, would i just say that he must do MORE THAN 2352J of work to lift the weight?
  2. jcsd
  3. Sep 11, 2006 #2
    It doesn't matter what the acceleration of the mass is. If here were to provide a drastic acceleration for 1 meter, he would no longer need to push upwards because the mass would have been thrown up and going up at this moment.
  4. Sep 11, 2006 #3


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    If an object is applying a weight force of 980N downwards, and i applied a force back on it of 980N upwards, the object would be stationary right?

    So to make the mass move upwards, wouldnt it require that the weightlifer apply a force of MORE than the weight force of 980N? OR am i completely missing the point here
  5. Sep 11, 2006 #4
    No you do not need to apply a force more that 980N, just 980N will do.

    Any force that is more than 980N will result in an aceleration of the weight, since there is net force. ( X - 980 = Net force, where X >980)

    The minimum amount of work that is required, requires the weight to be travelling at constant velocity.(Net force = zero, force applied=weight)

    Hope it helps. :)
  6. Sep 11, 2006 #5


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    Yes, you are completely missing the point! Work= force*distance. Any force, above 980 N, applied to accelerate the the mass so that it will move upward, will have to be "negated" by a force under 980 N to stop it. The work done by those will cancel. The only force that counts is the 980 N that must be applied upward to counteract gravity and keep the weight moving at constant speed.

    Or, you could argue that the work done is the difference in potential energies at the two points- again, weight*distance.
  7. Sep 11, 2006 #6


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    Ahhh ok i see now :P

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