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A small block slides down on a smooth inclined plane ....

A small block slides down on a smooth inclined plane, starting from rest at time t = 0. Let $S_n$ be the distance travelled by the block in the interval t = n-1 to t = n. Then, the ratio $\frac {S_n}{S_{n+1}}$ is:

(1) $\frac {2n-1}{2n}$
(2) $\frac {2n-1}{2n+1}$
(3) $\frac {2n+1}{2n-1}$
(4) $\frac {2n}{2n-1}$

Solution

The acceleration of the block $a=gsin\theta $ which is a constant and hence the following formula for distance in $n^{th}$ second would be applicable:

$S_n = u + (n-\frac {1}{2})a = 0 + (n-\frac {1}{2})a = (n-\frac {1}{2})a$

Now, $\frac {S_n}{S_{n+1}}=\frac {(n-\frac {1}{2})a}{(n+1-\frac {1}{2})a}=\frac {n-\frac {1}{2}}{n+\frac {1}{2}}=\frac {2n-1}{2n+1}$

Answer: (2)

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