Question

Johnny stretches a spring 40 cm from its resting position. The spring’s constant of proportionality is 20 N/m. What is the spring potential energy?
J=

Answer 1

The spring potential energy is 1.6 J

Explanation:

The elastic potential energy of a spring is given by the equation:

$U = \frac{1}{2}kx^2$

where

U is the potential energy

k is the spring constant

x is the compression/stretching of the spring

For the spring in this problem, we have:

x = 40 cm = 0.40 m is the stretching

k = 20 N/m is the spring constant

Substituting into the equation, we can find the spring potential energy:

$U=\frac{1}{2}(20)(0.40)^2=1.6 J$

#LearnwithBrainly