Question

A boy takes 3 minutes to lift a 20 litre water bucket from a $20\,{\text{m}}$ deep well, while his father does it in $2$ minutes.
(a) Compare:
(i) The work and
(ii)Power developed by them
(b)How much work each does?
(Take density of water is ${10^3}\,{\text{kg}}\,{{\text{m}}^{ - 3}}$ and $g$ is $9.8\,{\text{N}}\,{\text{k}}{{\text{g}}^{ - 1}}$.)

Answer 1

First of all, we will convert the density in a way such that we can get the mass contained in a litre of water. Then we will find out work done and then power, since power means rate of work done. We will manipulate accordingly and obtain the result.

Complete step by step answer: In the given problem, we are supplied with the following data:
Time taken by the boy is $3$ minutes.
The boy lifts a total of $20$ litre water bucket.
Timer taken by his father is $2$ minutes.
His father also lifts a total of $20$ litre water buckets.
The density of water is given as ${10^3}\,{\text{kg}}\,{{\text{m}}^{ - 3}}$ .

To begin with, first we will convert the density of water to mass per litre.
Since, the density is given as ${10^3}\,{\text{kg}}\,{{\text{m}}^{ - 3}}$ . So, we can elaborate as follows:
We know,
$1\,{\text{l}} = 1000\,{\text{c}}{{\text{m}}^3}$
Again,

$$\begin{aligned} {10^3}\,{\text{kg}}\,{{\text{m}}^{ - 3}} \\\ \Rightarrow {10^3}\,{\text{kg}}\,{{\text{m}}^{ - 3}} &= \dfrac{{1000\,{\text{kg}}}}{{1\,{{\text{m}}^3}}} \\\ \Rightarrow {10^3}\,{\text{kg}}\,{{\text{m}}^{ - 3} } &= \dfrac{{1000\,{\text{kg}}}}{{1\,{\text{m}} \times 1\,{\text{m}} \times 1\,{\text{m}}}} \\\ \Rightarrow {10^3}\,{\text{kg}}\,{{\text{m}}^{ - 3}} &= \dfrac{{1000\,{\text{kg}}}}{{100\,{\text{cm}} \times 100\,{\text{cm}} \times 100\,{\text{cm}}}} \\\ \end{aligned}$$

Further manipulating, we get:

$$\begin{aligned} \Rightarrow {10^3}\,{\text{kg}}\,{{\text{m}}^{ - 3}} &= \dfrac{{1000\,{\text{kg}}}}{{1000 \times 1000\,{\text{c}}{{\text{m}}^3}}} \\\ \Rightarrow {10^3}\,{\text{kg}}\,{{\text{m}}^{ - 3}} &= \dfrac{{1\,{\text{kg}}}}{{1000\,{\text{c}}{{\text{m}}^3}}} \\\ \Rightarrow {10^3}\,{\text{kg}}\,{{\text{m}}^{ - 3}} &= \dfrac{{1\,{\text{kg}}}}{{1\,{\text{l}}}} \\\ \end{aligned}$$

So, we can say that $20\,{\text{l}}$ of water contains $20\,{\text{kg}}$ by mass.
(a)
(i) Since, the work done is dependent on only the mass, acceleration due to gravity and height, so, the work done for both the boy and his father will be the same.
It is given by the formula:

$$w = mgh \\\ \Rightarrow w = 20 \times 9.8 \times 20\,{\text{J}} \\\ \Rightarrow w = 3920\,{\text{J}} \\\$$

Hence, the work done is found to be $3920\,{\text{J}}$ .

(ii) Power is given by the formula, which is given below:
$P = \dfrac{w}{t}$ …… (1)
Where,
$P$ indicates power.
$w$ indicates work done.
$t$ indicates time taken by each.

So, the power developed by the boy is:

$$P = \dfrac{{3920\,{\text{J}}}}{{3 \times 60\,{\text{s}}}} \\\ \Rightarrow P = 21.78\,{\text{W}} \\\$$

Hence, the power developed by the boy is $21.78\,{\text{W}}$ .

The power developed by his father is:

$$P = \dfrac{{3920\,{\text{J}}}}{{2 \times 60\,{\text{s}}}} \\\ \Rightarrow P = 32.67\,{\text{W}} \\\$$

Hence, the power developed by his father is $32.67\,{\text{W}}$.

(b)
The work done for both the boy and his father will be the same.
It is given by the formula:

$$w = mgh \\\ \Rightarrow w = 20 \times 9.8 \times 20\,{\text{J}} \\\ \Rightarrow w = 3920\,{\text{J}} \\\$$

Hence, the work done for each case is found to be $3920\,{\text{J}}$.

Note: It is important to note the time duration as given in the question has nothing to do with the work done. Work done is simply the product of mass, acceleration and height. Work done is the same for each case. However, power will be different depending on the duration of time taken by each person. Power is precisely the rate of work done.