A ball (initially at rest) is released from t | vedclass

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(A) The work done by the force of gravity is given by $W = F \cdot S_n = mg \cdot S_n$,where $S_n$ is the distance traveled in the $n$th second.Since

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$1: 3: 5$

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(A) The work done by the force of gravity is given by $W = F \cdot S_n = mg \cdot S_n$,where $S_n$ is the distance traveled in the $n$th second.Since $m$,$g$,and $F$ are constant,$W \propto S_n$.The distance traveled in the $n$th second for an object starting from rest is given by $S_n = u + \frac{a}{2}(2n - 1)$.Here,$u = 0$ and $a = g$,so $S_n = \frac{g}{2}(2n - 1)$.For $n = 1, 2, 3$:$S_1 = \frac{g}{2}(2(1) - 1) = \frac{g}{2}(1)$$S_2 = \frac{g}{2}(2(2) - 1) = \frac{g}{2}(3)$$S_3 = \frac{g}{2}(2(3) - 1) = \frac{g}{2}(5)$The ratio of distances is $S_1 : S_2 : S_3 = 1 : 3 : 5$.Therefore,the ratio of work done is $1 : 3 : 5$.

$A$ ball (initially at rest) is released from the top of a tower. The ratio of work done by the force of gravity,in the first,second,and third seconds is

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