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 the problem, we need to calculate and understand the motion of a block sliding down a smooth inclined plane. Given that the block starts from rest at \( t=0 \), it accelerates under the influence of gravity along the incline.

The motion of the block on the inclined plane can be considered a uniformly accelerated motion, where the acceleration \( a \) is constant. The formula for the distance \( S \) traveled by an object under uniform acceleration from rest is given by:

\(S = \frac{1}{2} a t^2\)

Let us understand what the problem asks:

• \( S_n \) is the distance traveled during the interval \( t = n-1 \) to \( t = n \).
• \( S_{n+1} \) is the distance traveled during the interval \( t = n \) to \( t = n+1 \).

 

Calculate \( S_n \):

• The distance at time \( t = n \) is \(S_n = \frac{1}{2} a n^2\).
• The distance at time \( t = n-1 \) is \(S_{n-1} = \frac{1}{2} a (n-1)^2\).
• Thus, \(S_n = \frac{1}{2} a n^2 - \frac{1}{2} a (n-1)^2\).
• Expanding, \(S_n = \frac{1}{2} a (n^2 - (n^2 - 2n + 1)) = \frac{1}{2} a (2n - 1)\).

 

Calculate \( S_{n+1} \):

• The distance at time \( t = n+1 \) is \(S_{n+1} = \frac{1}{2} a (n+1)^2\).
• The distance at time \( t = n \) is already calculated as \(\frac{1}{2} a n^2\).
• Thus, \(S_{n+1} = \frac{1}{2} a ((n+1)^2 - n^2)\).
• Expanding, \(S_{n+1} = \frac{1}{2} a ((n^2 + 2n + 1) - n^2) = \frac{1}{2} a (2n + 1)\).

 

Now, the ratio \(\frac{S_n}{S_{(n+1)}}\) is:

• \(\frac{S_n}{S_{(n+1)}} = \frac{\frac{1}{2} a (2n - 1)}{\frac{1}{2} a (2n + 1)}\)
• This simplifies to \(\frac{2n - 1}{2n + 1}\).

 

Therefore, the correct answer is \(\frac{2n-1}{2n+1}\).