Question

A spring of natural length 10 in. stretches 1.5 in. under a weight of 8 lb. Find the work done in stretching the spring

(a) from its natural length to a length of 14 in.

(b) from a length of 11 in. to a length of 13 in.

Answer 1

The external force that stretches a spring a distance x from its equilibrium position is 

$F = kx$ ,

$F=mg$,

so $k = \frac{F}{x}=\frac{mg}{x}$

$m=8lb\approx8\cdot0.4536kg=3.6288kg$

$1.5in=1.5\cdot2.54\cdot10^{-2}m=3.81\cdot10^{-2}m$

$k = \frac{3.6288kg\cdot 9.81 m/s^2}{3.81\cdot10^{-2}m}\approx934.34N/m$ .

The work that must be done to stretch spring a distance x from its equilibrium position is

$W=\frac12kx^2$

(a) Find the work done in stretching the spring from its natural length to a length of 14 in.

$x=14in-10in=4in=4\cdot2.54\cdot10^{-2}m$=0.1008m;

$W=\frac12\cdot934.34N/m\cdot (0.1008m)^2$$\approx4.75J$.

(b) Find the work done in stretching the spring from a length of 11 in. to a length of 13 in.

To find this work we should subtract work that must be done to stretch spring a distance 11 in from the work that must be done to stretch spring a distance 13 in.

$\Delta W=\frac12k(x_2)^2-\frac12k(x_1)^2$ , where

$x_2=13in-10in=3in=7.62\cdot10^{-2}m$ ;

$x_1=11in-10in=1in=2.54\cdot10^{-2}m$

$\Delta W=\frac12\cdot934.34N/m((7.62\cdot10^{-2}m)^2-$$(2.54\cdot10^{-2}m)^2)\approx2.41J$