Question

In 0.521 s, a 8.15-kg block is pulled through a distance of 3.62 m on a frictionless horizontal surface, starting from rest. The block has a constant acceleration and is pulled by means of a horizontal spring that is attached to the block. The spring constant of the spring is 415 N/m. By how much does the spring stretch

Answer 1

Step-by-step explanation:

It is given that,

Mass of the block, m = 8.15 kg

Time, t = 0.521 s

It is pulled through a distance of 3.62 m, s = 3.62 m

Initially, u = 0

Spring constant of the spring, k = 415 N/m

We need to find the stretching in the spring. Firstly, we will find the acceleration of the block. It can be calculated using second equation of motion as :

`s=ut+(1)/(2)at^2`

`a=(2s)/(t^2)`

`a=(2* 3.62\ m)/((0.521)^2)`

`a=26.67\ m/s^2`

Now, the force due to this acceleration is balanced by the force in spring as :

`ma=kx`

`x=(ma)/(k)`

`x=(8.15\ kg* 26.67\ m/s^2)/(415\ N/m)`

x = 0.52 meters

So, the spring is stretched by 0.52 meters. Hence, this is the required solution.