Question

A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let Sn be the distance traveled by the block in the interval t=n−1 to t=n. Then, the ratio \(\frac{S_n}{S_{(n+1)}}\)

• A. \(\frac{2n}{2n-1}\)
• B. \(\frac{2n-1}{2n}\)
• C. \(\frac{2n-1}{2n+1}\)
• D. \(\frac{2n+1}{2n-1}\)

Answer 1

The Correct Option is C

To solve this problem, we need to understand the motion of a block sliding down a smooth inclined plane. Since the inclined plane is smooth, there is no friction acting on the block, and it is under the influence of gravity. This scenario is a classic example of uniformly accelerated motion.

The block starts from rest, so its initial velocity \( u = 0 \). Let \( a \) be the acceleration of the block due to gravity along the inclined plane. The distance traveled by the block in the time interval \( t = 0 \) to \( t = n \) is given by the equation of motion:

\(S = ut + \frac{1}{2}at^2\) 

Since the initial velocity \( u = 0 \), this simplifies to:

\(S = \frac{1}{2}an^2\)

Now, let's calculate the distances \( S_n \) and \( S_{(n+1)} \) for the successive intervals:

The distance traveled in the interval from \( t = n-1 \) to \( t = n \) is:

\(S_n = \frac{1}{2}an^2 - \frac{1}{2}a(n-1)^2\)

\(= \frac{1}{2}a(n^2 - (n-1)^2)\)

\(= \frac{1}{2}a(n^2 - (n^2 - 2n + 1))\)

\(= \frac{1}{2}a(2n - 1)\)

Similarly, the distance traveled in the interval from \( t = n \) to \( t = n+1 \) is:

\(S_{(n+1)} = \frac{1}{2}a(n+1)^2 - \frac{1}{2}an^2\)

\(= \frac{1}{2}a((n^2 + 2n + 1) - n^2)\)

\(= \frac{1}{2}a(2n + 1)\)

Now, we find the ratio:

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

This matches the correct answer:

Option: \(\frac{2n-1}{2n+1}\)