A weightlifter carries 1000kg across campus. He does this in steps: He lifts the

Question

A weightlifter carries 1000kg across campus. He does this in steps: He lifts the weight to I m above the ground He accelerates the weight to I m/s He negatively accelerates the weigh to 0m/s He lowers the weight back to I m above the ground. What is the potential energy and kinetic energy of the weight at the end of each step? How much cumulative work did the weightlifter do on the weights at each step? If the campus was on a hill, such that the destination was 10m above the starting point, how much cumulative work would he have done?

Explanation / Answer

PE = mgh, KE = mv2/2

Potentail energy = 1000*9.8*1 = 980 J, KE =0,( at the end v=0 hence KE=0)

PE = 980 J, KE = 0.5*1000*12 = 500 J ( no change in height, hence pE remains same, vel= 1 m/s)

PE = 980 J, KE = 0 ( height remianed same vel=0 at the end)

PE =0, KE=0 ( height is 0 with reference point ground and vel=0)

b)

work done = 980 J, equal to change in PE

work done = 500J, equal to change in KE

work done = -500 J, equal to change in KE

work done = -980 J, equal to change in PE

he will do any additional work of 1000*9.8*10 = 9800J, this is the energy required to lift the height by naother 10 m against gravity.

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