Question

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

Answer 1

According to the graphical second equation of motion for an n^th second we have;

S_n = u + a/2 (2n - 1)    ---(1)

where S_n = distance traveled by an object during the nth second.

u is the initial velocity.

a is the acceleration.

Suppose θ is the inclination of inclined plane acceleration along the inclined plane is written as;

a = g sin θ 

By using equation (1) of uniformly accelerated motion we have;

S_n = u + a/2 (2n - 1)

Putting the value of acceleration 'a' and initial velocity, u is zero, we have;

Distance traveled during (n + 1)^th second.

S_n+1 = 0 + \(\frac {g\sin \theta}2\) [2(n + 1) - 1]

S_n+1 = \(\frac {g\sin \theta}2\) (2n + 1)    -----(3)

Dividing equation (2) and (3) we have;

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