91Ó°ÊÓ

Hooke’s law states that the restoring force is always directly proportional to the amount of stretch/deformation in the spring, such that if deformation in the spring is x, then restoring force in the spring will be,

$\mathbf{F}\mathbf{=}\mathbf{-}\mathbf{kx}$

where k is a positive constant.

$$\begin{gathered} {\mathrm{F}}_{\mathrm{x}}=\mathrm{kx}-{\mathrm{bx}}^2+{\mathrm{cx}}^3 \\ \mathrm{k}=100 \mathrm{N}/\mathrm{m} \\ \mathrm{b}=700 \mathrm{N}/{\mathrm{m}}^2 \\ \mathrm{c}=12,000 \mathrm{N}/{\mathrm{m}}^3 \end{gathered}$$

Use the concept of work done by a variable force that uses the method of integration.

The work done by the variable force is given as follows:

$\mathrm{w}={∫}_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}{\mathrm{F}}_{\mathrm{x}}\mathrm{dx}$

Substitute 0 for $x_1$, 0.050 m for $x_2$, and $kx-b{x}^2+c{x}^3$for $F_x$.

$$\begin{gathered} {\mathrm{W}}_{\mathrm{stretching}}={∫}_0^{0.050}\left(\mathrm{kx}-{\mathrm{bx}}^2+{\mathrm{cx}}^3\right)\mathrm{dx} \\ ={\left[\mathrm{k}\left(\frac{{\mathrm{x}}^2}{2}\right)-\mathrm{b}\left(\frac{{\mathrm{x}}^3}{3}\right)+\mathrm{c}\left(\frac{{\mathrm{x}}^4}{4}\right)\right]}_0^{0.050\mathrm{m}} \\ =\mathrm{k}\left(\frac{{\left(0.050\mathrm{m}\right)}^2}{2}-0\right)-\mathrm{b}\left(\frac{\left(0.050\mathrm{m}\right)}{3}-0\right)+\mathrm{c}\left(\frac{\left(0.050\mathrm{m}\right)}{4}-0\right) \\ =\left(100\mathrm{N}/\mathrm{m}\right)\frac{{\left(0.050\mathrm{m}\right)}^2}{2}-\left(700\mathrm{N}/{\mathrm{m}}^2\right)\frac{\left(0.050\mathrm{m}\right)}{3}+\left(12,000\mathrm{N}/{\mathrm{m}}^3\right)\frac{{\left(0.050\right)}^4}{4} \\ =0.115\mathrm{JA} \end{gathered}$$

Therefore, the work done in stretching the spring from zero to

Is 0.115 J .

The work done by the variable force is given as follows:

Substitute 0 m for $x_1$, $-0.050\mathrm{m}$for $x_2$, and $\mathrm{kx}-{\mathrm{bx}}^2+{\mathrm{cx}}^3$for ${\mathrm{F}}_{\mathrm{x}}$.

$$\begin{gathered} {\mathrm{w}}_{\mathrm{compress}}={∫}_0^{−0.050\mathrm{m}} \left(\mathrm{kx}−{\mathrm{bx}}^2+{\mathrm{cx}}^3\right)\mathrm{dx} \\ ={\left[\mathrm{k}\left(\frac{{\mathrm{x}}^2}{2}\right)−\mathrm{b}\left(\frac{{\mathrm{x}}^3}{3}\right)+\mathrm{c}\left(\frac{{\mathrm{x}}^4}{4}\right)\right]}_0^{−0.050\mathrm{m}} \\ =\mathrm{k}\left(\frac{(−0.050\mathrm{m}{)}^2}{2}−0\right)−\mathrm{b}\left(\frac{(−0.050\mathrm{m})}{3}−0\right)+\mathrm{c}\left(\frac{(−0.050\mathrm{m}{)}^4}{4}−0\right) \\ =(100\mathrm{N}/\mathrm{m})\frac{(0.050\mathrm{m}{)}^2}{2}+\left(700\mathrm{N}/{\mathrm{m}}^2\right)\frac{(0.050\mathrm{m}{)}^3}{3}+\left(12,000\mathrm{N}/{\mathrm{m}}^3\right)\frac{(0.050\mathrm{m}{)}^4}{4} \\ =0.173\mathrm{J} \end{gathered}$$

Therefore, the work done in compressing the spring from zero to 0.050 m

is 0.173 J.

More work is required to compress the spring than stretching for the same distance.

This is because of the reason that the work is the function of polynomials of x and one term $-{\mathrm{bx}}^2$is negative which makes compression tougher.