A boy of weighing 40 kgf climbs up a stair of 30 steps, each 20 cm high in 4 minutes and a girl weighing 30 kgf does the same in 3 minutes.

Compare :

(i) the work done by them, and

(ii) the power developed by them. Take g = 10 N kg^-1.

(i) We know that work done is a product of force and displacement

W = F x S = mgh

Given,

m₁ = 40 kgf
m₂ = 30 kgf
t₁ = 4 min = 4 x 60 =240 sec
t₂ = 3 min = 3 x 60 = 180 sec
h₁ = h₂ = 30 x 20 = 600 cm = 6 m

$\dfrac{\text {Work done by boy}}{\text {Work done by girl}} = \dfrac{m$1gh1}{m2gh2} \\[0.5em] = \dfrac{m1}{m2} = \dfrac{40}{30} \\[0.5em] = \dfrac{4}{3} \\[0.5em]

Therefore,

Work done by boy : Work done by girl = 4 : 3

(ii) The power developed —

We know that,

$\text{Power} = \dfrac{\text{work done}}{\text{time}} \\[0.5em] \dfrac{\text{Power developed by boy}}{\text{Power developed by girl}} \\[1em] = \dfrac{\dfrac{\text{work done by boy}}{\text{time taken by boy}}}{\dfrac{\text{work done by girl}}{\text{time taken by girl}}} \\[1em] = \dfrac{\dfrac{40 \times 10 \times 6}{240}}{\dfrac{30 \times 10 \times 6}{180}} \\[1em] = \dfrac{\dfrac{2400}{240}}{\dfrac{1800}{180}} \\[1em] = \dfrac{10}{10} \\[1em] = \dfrac{1}{1}$

Therefore,

Power developed by boy : Power developed by girl = 1:1