Work, Energy and Power — Question 16

(a)

(i) Work done = force x displacement

Work done is same for boy and father as both lift 20 litre water from 20 m deep well.

Work done by boy : Work done by father = 1:1

(ii) Comparing Power developed:

$\text{Power} = \dfrac{\text{work done}}{\text{time}} \\[0.5em] {\text {Power of boy}} : {\text {Power of father }} = \dfrac{\dfrac{W}{3}}{\dfrac{W}{2}} \\[0.5em] \Rightarrow {\text {Power developed by boy}} : {\text {Power developed by father}} = \dfrac{2}{3} \\[0.5em]$

Power developed by boy : Power developed by father = 2:3

(b) Work done = mgh

Given,

density of water = 10³kg m ^-3

To convert, density of water to mass per litre

We know,

1 litre = 1000 cm³
and
1 m³ = 10⁶ cm³

So,

$10^3\text{ kgm}^{-3} = \dfrac{1000\text{ kg}}{1\text{ m}^3} \\[1em] = \dfrac{1000\text{ kg}}{10^6 \text{ cm}^3} \\[1em] = \dfrac{1\text{ kg}}{1000\text{ cm}^3} \\[1em] = \dfrac{1\text{ kg}}{1 \text{ litre}}$

Hence,

20L of water = 20 Kg by mass

height = 20m

Now, substituting the values in the formula for work done we get

W = 20 x 9.8 x 20
= 3.92 kJ