The steps of a flight of stairs are 21.0 cm high (vertically). If a 63.0-kg pers

Question

The steps of a flight of stairs are 21.0 cm high (vertically). If a 63.0-kg person stands with both feet on the same step, what is the gravitational potential energy of this person on the first step of the flight of stairs relative to the same person standing at the bottom of the stairs? If a 63.0-kg person stands with both feet on the same step, what is the gravitational potential energy of this person on the second step of the flight of stairs relative to the same person standing at the bottom of the stairs? If a 63.0-kg person stands with both feet on the same step, what is the gravitational potential energy of this person on the third step of the flight of stairs relative to the same person standing at the bottom of the stairs? What is the change in energy as the person descends from step 7 to step 3?

Explanation / Answer

Part A:

P.E. = mgh

P.E. = 63.0*9.81*(21.0*10-2) = 129.786 J

Part B:

P.E. = mg*2h

P.E. = 63.0*9.81*(2*21.0*10-2) = 259.573 J

Part C:

P.E. = mg*3h

P.E. = 63.0*9.81*(3*21.0*10-2) = 389.359 J

Part D:

P.E. = mg*(hf - ho)

where (hf - ho) is the difference of the height, for this case

P.E. = 63.0*9.81*(2h - 7h) = -3090.15 J

where 'h' is the height between each floor in meters.

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