Question

A spring 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?

Answer 1

0.20 J

Step-by-step explanation:

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

`U=(1)/(2)k(x-x_0)^2`

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:

`U=(1)/(2)(150 N/m)(0.11 m-0.08 m)^2=0.0675 J`

- when x = 14.0 cm = 0.14 m:

`U=(1)/(2)(150 N/m)(0.14 m-0.08 m)^2=0.27 J`

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

`\Delta U=0.27 J-0.0675 J=0.20 J`