**Work and elastic potential energy**

In order to compress or stretch a spring, you have to do work. You must exert a force on the spring equal in magnitude to the force the spring exerts on you, but opposite in direction. The force you exert is in the direction of the displacement. You therefore do work.

The average force you exert as you change the displacement from 0 to x is (1/2)kx. The work you do when stretching or compressing a spring therefore is

**W = force*distance = (1/2)kx*x = (1/2)kx ^{2}.
**

As you do work, you transfer energy to the spring. The
**elastic
potential energy** stored in the spring is **U = (1/2)kx ^{2}**.

When a 4 kg mass is hung vertically on a certain light spring that obeys
Hooke's law, the spring stretches 2.5 cm. If the 4 kg mass is removed,

(a) how far will the spring stretch if a 1.5 kg mass is hung on it, and

(b) how much work must an external agent do to stretch the same spring 4 cm from
its unstretched position?

- Solution:
(a) We find the spring constant of the spring from the given data.

F = -kx.

F = -mg = -(4 kg)(9.8 m/s^{2) }= -39.2 N.

k = F/x = (39.2 N)/(0.025 m) = 1568 N/m.

Now we use x = -F/k to find the displacement of a 1.5 kg mass.

F = -(1.5kg)(9.8m.s^{2}) = -14.7 N.

x = (14.7 N)/(1568 N/m) = 0.009375 m = 0.975 cm.(b) W = (1/2)kx

^{2 }= (1/2)(1568 N/m)(0.04 m)^{2 }= 1.2544 Nm = 1.2544 J

**Problem:**

If it takes 4J of work to stretch a Hooke's law spring 10 cm from its unstretched length, determine the extra work required to stretch it an additional 10 cm.

- Solution:

The work done in stretching or compressing a spring is proportional to the square of the displacement. If we double the displacement, we do 4 times as much work. It takes 16 J to stretch the spring 20 cm from its unstretched length, so it takes 12 J to stretch it from 10 cm to 20 cm.Formally:

W = (1/2)kx^{2}. Given W and x we find k.

4 J = (1/2)k(0.1 m)^{2}; k =(8 J)/(0.1 m)^{2 }= 800 N/m.

Now x = 0.2 m. W = (1/2)(800 N/m)(0.2 m)^{2 }= 16 J.

Extra work: 16 J - 4 J = 12 J.