Question

A spring with a force constant of 5.2 N/m has a relaxed length of 2.45 m. When a mass is attached to the end of the spring and allowed to come to rest, the vertical length of the spring is 3.57 m. Calculate the elastic potential energy stored in the spring.

Answer 1

Answer:

Elastic potential energy, E = 3.26 J

Explanation:

It is given that,

Force constant of the spring, k = 5.2 N/m

Relaxed length of the spring, X = 2.45 m

When the mass is attached to the end of the spring, the vertical length of the spring is, x' = 3.57 m

To find,

The elastic potential energy stored in the spring.

Solution,

The extension in the length of the spring is given by :

$x=x'-X$

$x=3.57\ m-2.45\ m$

x = 1.12 m

The elastic potential energy of the spring is given by :

$E=\dfrac{1}{2}kx^2$

$E=\dfrac{1}{2}\times 5.2\times (1.12)^2$

E = 3.26 J

So, the elastic potential energy stored in the spring is 3.26 joules.