Question

A force of 10 lb is required to hold a spring stretched 4 in.beyond its naturallength. How much work is done instretching it from its natural length to 6 in. beyond its natural length?

Short answer

The work done in stretching the spring from natural length to 6in beyond its natural length is \(\frac{{15}}{4}lb \cdot ft\).

Step-by-step solution

Work done in elasticity

If an object moves from a to b and\({\rm{f(x)}}\)is the force applied; work done is given as:\({\rm{W = }}\int_{\rm{a}}^{\rm{b}} {\rm{f}} {\rm{(x)dx}}\)

Hooke’s law gives us the force\({\rm{f(x)}}\)by formula:\({\rm{f(x) = kx}}\)

Given parameters

force required is\(10lb\)to hold a spring stretched 4 in beyond its natural length.

i.e., For \(x = 4in,{\rm{ }}f(x) = 10lb\)

convert x into ft. as- \(4in \times \frac{{1ft}}{{12in}} = \frac{1}{3}in\)

Calculatingproportionality constant k

Using Hooke’s law,

\(\begin{aligned}{}f(x) = kx\\10 = k\left( {\frac{1}{3}} \right)\\k = 30\end{aligned}\)

Calculating work done

Convert 6 in into ft as- \(6in \times \frac{{1ft}}{{12in}} = \frac{1}{2}in\)

The work done in stretching the spring from \(x = 0\) to \(x = \frac{1}{2}\)is,

\(\begin{aligned}{}W = \int_a^b f (x)dx\\ = \int_0^{\frac{1}{2}} 3 0xdx\\ = 30\int_0^{\frac{1}{2}} x dx\\ = 30\left( {\left. {\frac{{{x^2}}}{2}} \right|_0^{\frac{1}{2}}} \right)\\ = 30\left( {\frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{2} - \frac{{{0^2}}}{2}} \right)\\ = 30\left( {\frac{1}{8} - 0} \right)\\ = \frac{{30}}{8}\\ = \frac{{15}}{4}\end{aligned}\)

Therefore, the work done in stretching the spring from natural length to 6in beyond its natural length is \(\frac{{15}}{4}lb \cdot ft\).