Question

A spring with a force constant of 5.0 N/m has a relaxed length of 2.59 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 3.86 m. Calculate the elastic potential energy stored in the spring. Answer in units of J.

Answer 1

Answer:

Explanation:

Given

Spring force constant $k=5\ N/m$

Relaxed length $l_0=2.59\ m$

mass is attached to the end of spring and allowed to come at rest with relaxed length $l_2=3.86\ m$

Potential energy stored in the spring

$U=\frac{1}{2}k(\Delta x)^2$

$\Delta x=l_2-l_0$

$U=\frac{1}{2}\times 5\times (3.86-2.59)^2$

$U=4.03\ J$