Question

In 0.678 s, a 14.2-kg block is pulled through a distance of 4.36 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 436 N/m. By how much does the spring stretch?

Answer 1

Here ,

m = 14.2 Kg

let the acceleration is a

Using second equation of motion

4.36 = 0.50 * a * 0.678^2

solving for a

a = 19 m/s^2

let the stretch in spring is x

m * a = k * x

19 * 14.2 = 436 * x

solving for x

x = 0.62 m

the elongation in the spring is 0.62 m