Question

It takes 20 s for a girl A to climb up the stairs while girl B takes 15 s for the same job. Compare: (i) the work done and (ii) the power spent by then.

Answer 1

The above problem can be resolved using the concepts and fundamentals of the work done and the power generated by any moving object. When any object is said to be lifted through some vertical distance, then the entire energy of the object is stored in potential energy. Therefore, the mathematical expression for the work is equivalent to the expression for the potential energy.

Complete step by step solution:
Given:
The time taken by girl A to climb the stairs is, ${T_1} = 20\;{\rm{s}}$.
The time taken by girl B to climb is, ${T_2} = 15\;{\rm{s}}$.
(i) The expression for the work done is,
$W = mgh$
Here, h is the height.
Clearly it shows that the work done is dependent on the height. And, as girls covered equal distance as the height. Then the work done for both the girls A and B are the same and in the ratio 1:1.
(ii) The expression for the power spent by them is,
$P = \dfrac{W}{T}$
Taking the ratio for power for both girls as,

\Rightarrow \dfrac{{{P_A}}}{{{P_B}}} = \dfrac{{{W_A}/{T_1}}}{{{W_B}/{T_2}}}\\\ \Rightarrow \dfrac{{{P_A}}}{{{P_B}}} = \dfrac{{{W_A}}}{{{W_B}}} \times \dfrac{{{T_2}}}{{{T_1}}}\\\ \Rightarrow \dfrac{{{P_A}}}{{{P_B}}} = \left( {\dfrac{1}{1}} \right) \times \dfrac{{15\;{\rm{s}}}}{{20\;{\rm{s}}}}\\\ \Rightarrow \dfrac{{{P_A}}}{{{P_B}}} = \dfrac{3}{4} \end{array}$$ **Therefore, the power spent by them is in the ratio of 3:4 and as girl B took less time, so power spent by her is more than girl A. ** **Note:** To resolve the above problems, one must understand the concept and application of the work done and the power developed by any object. Moreover, there are significant work and power applications, and each of these is related to each other wisely, as the direct relation in each other's context.