Question

A block is moving down 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 $t=n$. The ratio $\frac{S^n}{S_n+1}$ is

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

Answer 1

The correct option is B $\frac{2n-1}{2n+1}$

REF. Image

Here on the cline plane

the down the incline

acceleration is

$a=\frac{force}{mass}=\frac{mgsin\theta }{m}$

or $a=gsin\theta$

distance after t second $s=ut+\frac{1}{2}a{t}^2$

initially at rest $\Rightarrow u=0\Rightarrow s=\frac{1}{2}a{t}^2$

$\Rightarrow$ distance in n second will be

$s=\frac{1}{2}a{n}^2$

distance in $n-1$ second is

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

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

or $s^n=\frac{1}{2}a\left\{n\right(n-1\left)\right(n-(n-1)\left)\right\}=\frac{1}{2}a(2n-1)\left(1\right)$

for getting $s^{n+1}$ just replace $n\to n+1$

we get $s^{n+1}=\frac{1}{2}a\left(2\right(n+1)-1)1=\frac{1}{2}a(2n+1)$

$\Rightarrow \frac{s^n}{s^{n+1}}=\frac{2n-1}{2n+1}$ Option - B