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

Step-by-step 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+(1)/(2)at^2`

where a=acceleration

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

so

`s_1=(1)/(2)* g\sin \theta (1)^2`

`0.633* 2=9.8\sin \theta * 1^2`

`\sin\theta =0.1291`

`\theta =7.41^(\circ)`

So distance traveled in
`2\ sec`

`s=(1)/(2)* g\sin \theta (2)^2`

`s=0.5* 9.8* \sin (7.41)* 4`

`s=2.52\ m`

So distance traveled in
`2^(nd)\ sec` is

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