Question

If a 66.0-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?

Answer 1

Answer:

U₁ = 129.4 J

Explanation:

The potential energy is

    U = mg y - m g y₀

Where I correspond to the initial position, with this it is an additive constant, we can make it zero with the placement of the reference system, in this case the system is placed on the floor where the ladder rests.

The power power for people on the floor is

     U₀ = 0 J

The potential energy for the person on the first step is

   U₁ = m g y₁

In general the steps are 20 cm high

     y₁ = 20 cm (1m / 100cm) = 0.20 m

     U₁ = 66 9.8 0.20

     U₁ = 129.4 J