Question

11. A horizontal spring with a spring constant of 140 N/m is extended 0.20 m from the equilibrium position. How much elastic potential energy is stored in the spring?

Answer 1

`\huge\boxed{\sf PE=2.8 \ Joules}`

Step-by-step explanation:

Given data:

Spring constant = k = 140 N/m

Displacement = x = 0.2 m

Required:

Elastic Potential Energy = PE = ?

Formula:

`\displaystyle PE= (1)/(2) kx^2`

Solution:

`\displaystyle PE=(1)/(2) (140)(0.2)^2\\\\PE=(70)(0.04)\\\\PE=2.8 \ Joules\\\\\rule[225]{225}{2}`