Question

What is the power output for a 65-kg woman who runs up a 3.5 m high flight of stairs in 2 s, starting from rest but having a final speed of 2 m/s?

P = units =

Answer 1

The work-energy theorem states that the total work done on an object is the change in its kinetic energy. Now, the total work done is comprised of the work done by the woman and the work done by the gravitational force.

W_total = ∆K ⟹ W_gravity +W_woman = ∆K

Now, W_gravity = -∆U, where ∆U is the change in potential energy of the woman.

Thus, W_woman = ∆K - W_gravity = ∆K - (-∆U) = ∆K + ∆U = 1/2(mv_f²) - 1/2(mv_i²) + mgh_f - mgh_i

⟹ W_woman = (m/2)(v_f²-v_i²) + mg(h_f - h_i)

Now, m = 65 Kg, v_f = 2m/s, v_i=0m/s, h_f - h_i=3.5m.

Therefore: W_woman = (65/2)(2²-0²) + 65×9.8×3.5 = (65/2)×4 + 2229.5 = 65×2 + 2229.5 = 130 + 2229.5 = 2359.5 Joule

The power output from the woman is thus : P = W_woman / time = 2359.5 Joule/2s = 1179.75 W

Therefore, the power output from the woman is 1179.75W.