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

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 :

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

$a=\dfrac{2s}{t^2}$

$a=\dfrac{2\times 3.62\ m}{(0.521)^2}$

$a=26.67\ m/s^2$

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

$ma=kx$

$x=\dfrac{ma}{k}$

$x=\dfrac{8.15\ kg\times 26.67\ m/s^2}{415\ N/m}$

x = 0.52 meters

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