Question

An elastic linear spring of spring constant 144 N/cm is compressed from an initial unconstrained length to a final of 6cm. If the work required on the spring is 648J, determine the initial length of the spring in centimeters:

Answer 1

Elastic potential energy is directly proportional to the square of the change in length and the spring constant.

$$U=\dfrac{kx^2}{2}$$

Then the work required is

$$W=U=\dfrac{kx^2}{2}$$

$$x=|l-l_0|=\sqrt{\dfrac{2W}{k}}$$

Given

$$l_0>l, l=6\ cm=0.06\ m,$$

$$k=144\ N/cm=14400\ N/m, W=648 J$$

Substitute

$$l_0=0.06\ m+\sqrt{\dfrac{2(648\ J)}{14400\ N/m}}=0.06\ m+0.3\ m=0.36\ m$$

$$=0.06\ m+0.3\ m=0.36\ m$$

The initial length of the spring is 36 cm.