Question

A small block is released from rest at the top of a frictionless incline. The block travels a distance 0.633 m in the first second after it is released. How far does it travel in the next second

Answer 1

Answer:1.89 m

Explanation:

Given

Block travels $0.63\ m$ in first second

It is released from rest i.e. initial speed is zero (u=0)

using

$s=ut+\frac{1}{2}at^2$

where a=acceleration

here acceleration is the component of gravity on incline plane (say $\theta$)

so

$s_1=\frac{1}{2}\times g\sin \theta (1)^2$

$0.633\times 2=9.8\sin \theta \times 1^2$

$\sin\theta =0.1291$

$\theta =7.41^{\circ}$

So distance traveled in $2\ sec$

$s=\frac{1}{2}\times g\sin \theta (2)^2$

$s=0.5\times 9.8\times \sin (7.41)\times 4$

$s=2.52\ m$

So distance traveled in $2^{nd}\ sec$ is

$s-s_1=2.52-0.633=1.89\ m$