A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm.

(a) How much elastic potential energy does the spring contribute when its length is 23 cm?

(b) How much more potential energy does it contribute if its length increases to 26 cm?

a) When its length is 23 cm, the elastic potential energy of the spring is

0.18 J

b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

Explanation:

Hi there!

a) The elastic potential energy (EPE) is calculated using the following equation:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretched lenght.

Let's calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).

First, let's convert the spring constant units into N/m:

4 N/cm · 100 cm/m = 400 N/m

EPE = 1/2 · 400 N/m · (0.03 m) ²

EPE = 0.18 J

When its length is 23 cm, the elastic potential energy of the spring is 0.18 J

b) Now let's calculate the elastic potential energy when the spring is stretched 0.06 m:

EPE = 1/2 · 400 N/m · (0.06 m) ²

EPE = 0.72 J

When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J