Two children $A$ and $B$ each weighing $20 \, kg$ climb a rope up to a height of $10 \, m$. Child $A$ takes $10 \, s$ and child $B$ takes $20 \, s$ to climb. State whether the work done by both is equal or unequal. Who has more power?
(A) Work done by each child is calculated using the formula $W = mgh$.
Given $m = 20 \, kg$,$g = 9.8 \, m/s^2$,and $h = 10 \, m$.
Work done $= 20 \times 9.8 \times 10 = 1960 \, J$.
Since $m$,$g$,and $h$ are the same for both children,the work done is equal.
Power is defined as the rate of doing work,$P = W / t$.
For child $A$: $P_A = 1960 / 10 = 196 \, W$.
For child $B$: $P_B = 1960 / 20 = 98 \, W$.
Since $196 \, W > 98 \, W$,child $A$ has more power.
Given $m = 20 \, kg$,$g = 9.8 \, m/s^2$,and $h = 10 \, m$.
Work done $= 20 \times 9.8 \times 10 = 1960 \, J$.
Since $m$,$g$,and $h$ are the same for both children,the work done is equal.
Power is defined as the rate of doing work,$P = W / t$.
For child $A$: $P_A = 1960 / 10 = 196 \, W$.
For child $B$: $P_B = 1960 / 20 = 98 \, W$.
Since $196 \, W > 98 \, W$,child $A$ has more power.
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