Aspring with a spring constant of 150 n/m has an equilibrium length of 8.0 cm. what is the change in spring potential energy when the spring is stretched from a length of 11.0 cm to 14.0 cm?

0.20 J

Explanation:

The elastic potential energy stored by a spring is given by:

where

k is the spring constant

x is the length of the compressed/stretched spring

x0 is the equilibrium length of the spring

In this problem, we have:

k = 150 N/m

x0 = 8.0 cm = 0.08 cm

We need to calculate the elastic potential energy in two situations:

- when x = 11.0 cm = 0.11 m:

- when x = 14.0 cm = 0.14 m:

So, the change in elastic potential energy between the two situations is

a.) travels faster in solids because the particles are closer together.

air is a gas water is a liquid

no

sodium is solid at rt