Question

A spring of natural lengthb19inches stretches 1.5 inches under a weight of pounds .find the work done in stretching the spring

Answer 1

Let, the equilibrium point is at origin i.e $x_i=0$

Information provided,

Natural length of spring $l_0=19$ inches.

Changes in the length of spring from natural length is $x=1.5\: inch=0.0381m$

Force acting on the spring is 1 pound weight $= 4.45N.$

As, magnitude of force is

$$F=kx\\\implies 4.5=k(0.0381)\implies k=116.67N/m$$

Since, work done in spring is the potential energy stored in the spring due to stretching or compressing it.

Since,

$$W=\Delta PE=\int_{x_i}^{x}kxdx=\frac{1}{2}kx^2-\frac{1}{2}kx_i^2$$

But in this case, $x_i=0$

Thus,

$$W=\frac{1}{2}kx^2$$

On plugin the value we get,

$$W=8.5\times 10^{-2}J$$