Question

Two children say A and B each weighing 20 kg climb a rope up to the height of 10 m. Child A takes 10 s and child B takes 20 s to climb. Out of these two children, who have more power.

Answer 1

Step 1: Given,

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{A}\left({\mathrm{m}}_{\mathrm{A}}\right)=\mathrm{Mass}\mathrm{of}\mathrm{B}\left({\mathrm{m}}_{\mathrm{B}}\right)=20\mathrm{kg} \\ \mathrm{Height}\left(\mathrm{h}\right)=10\mathrm{m} \\ \mathrm{Time}\mathrm{taken}\mathrm{for}\mathrm{child}\mathrm{A}\left({\mathrm{t}}_{\mathrm{A}}\right)=10\mathrm{s} \\ \mathrm{Time}\mathrm{taken}\mathrm{for}\mathrm{child}\mathrm{B}\left({\mathrm{t}}_{\mathrm{B}}\right)=20\mathrm{s} \end{gathered}$$

Step 2: Formula used,

$$\begin{gathered} \mathrm{Work}\mathrm{done}=\mathrm{force}\times \mathrm{displacement},\mathrm{and} \\ \mathrm{Power}=\frac{\mathrm{work}\mathrm{done}}{\mathrm{time}} \end{gathered}$$

Step 3: Calculation,

$$\begin{gathered} \mathrm{Work}\mathrm{done}=\mathrm{force}\times \mathrm{displacement} \\ =\mathrm{mg}\times \mathrm{h} \\ =20\times 10\times 10 \\ =2000\mathrm{joule} \\ \mathrm{Power}\mathrm{of}\mathrm{A}\left({\mathrm{P}}_{\mathrm{A}}\right)=\frac{\mathrm{W}}{{\mathrm{t}}_{\mathrm{A}}}=\frac{2000}{10}=200\mathrm{W} \\ \mathrm{Power}\mathrm{of}\mathrm{B}\left({\mathrm{P}}_{\mathrm{B}}\right)=\frac{\mathrm{W}}{{\mathrm{t}}_{\mathrm{B}}}=\frac{2000}{20}=100\mathrm{W} \end{gathered}$$

Hence, the power of child A is maximum.