Question

A body slides down a frictionless inclined plans starting from rest if sn and sn+1 be the distance travelled by the body during nth and n+1 th seconds then the ratio sn+1 upon sn is...

Answer 1

Let $\theta$ is the angle made by inclined plane with horizontal surface.

so, net force on body , F = ma = $mgsin\theta$

$\implies a=gsin\theta$....(1)

body slides down the inclined plane starting from rest .

so, initial velocity of body, u = 0

so, distance covered by body in nth second , S = u + 1/2 a(2n - 1)

= 0 + $\frac{1}{2}gsin\theta(2n-1)$

similarly, distance covered by body in (n +1)th second, $S_{(n+1)}=\frac{1}{2}gsin\theta(2n+2-1)=\frac{1}{2}gsin\theta(2n+1)$

now, $\frac{s_{(n+1)}}{s_n}=\frac{(2n+1)}{(2n-1)}$

Answer 2

hey mate here is Ur answer

please mark as brainlist if it's helpful