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Alex works at a legal firm which produces a lot of documents. Often Alex has to carry heavy stacks of documents from one office to another. On this

particular day Alex,who weighs 75 kilograms,has to transport a stack of documents which weighs 10 kilograms. He has to lift them 5 meters from the floor and then carry them to an office 100 meters down the hall,where he will drop them on the floor. Alex walks swiftly at a steady rate of 1 meter per second. What two pieces of information given in the above description will help you compute the amount of work Alex does when he lifts the papers?



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Pieces of information:
   1. Alex has to lift the documents by 5 meters.  So he has to climb that height against gravity and so he has to do work to do that.
   2. Alex drops the books onto the floor, but not slowly putting them down.
   3. Alex walks at a velocity.  So he gains speed from rest.  So there is a change in Kinetic energy.
   4. Nothing is specified regarding his slowing down or coming to rest, before dropping the books at the end.
amount of work done = change in potential energy while lifting the books and change in potential energy while putting books on floor again

work done for lifting the books and himself by 5 meters = m g h
         =  85 kg * 9.81 m/sec² *  5 meters   = 4, 169.25 Joules
work done in carrying the documents and himself across the hall by 100 meters
         = 0  as the change in potential energy is zero, as there is no change in the height of the books or the person.

The amount of work done in walking at a uniform pace = 1 m/sec is zero.  But to gain the kinetic energy of 1/2 * m * v² =  1/2 * 85 kg * (1 m/sec) ²
         = 42.5 Joules

While the person walks at the steady rate there is no work done, if the friction is ignored.

If he drops the documents directly without putting them gently on the floor, there is no work done :negative or positive.

Total work done :  4,169.25 + 42.5 =  4, 211.75 Joules

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