91Ó°ÊÓ

It is given that,

1. Spring constant is${}^{\mathrm{K}=18.0\mathrm{N}/\mathrm{cm}}$

2. Displacement is${}^{{\mathrm{x}}_{\mathrm{i}}=7.60\mathrm{mm}}$

Here the concept of spring’s potential energy and kinetic energy of the object can be applied. spring’s potential energy is the stored energy in a compressible or stretchable objects such as spring, rubber. Plugging the values of displacement and the spring constant, the work done by the spring on the cage can be calculated. For additional displacement, find the work done. Finally, the difference between the previous work done and this work done will give us the additional work done.

Formula is as follow:

$\mathrm{w}=\frac{1}{2}{{\mathrm{kx}}^2}_{\mathrm{f}}-\frac{1}{2}{{\mathrm{Kx}}^2}_{\mathrm{i}}$

Where, W is the work done, k is the spring constant,localid="1654068523998" ${}^{{{\mathrm{x}}^2}_{\mathrm{f}},{{\mathrm{x}}^2}_{\mathrm{i}}}$are the final and initial displacements.

Firstly, convert the given values in the SI unit system,

$$\begin{gathered} \mathrm{K}=\frac{18.0\mathrm{N}}{\mathrm{cm}}×\frac{100\mathrm{cm}}{1\mathrm{m}} \\ =1800\mathrm{N}/\mathrm{m} \\ {\mathrm{x}}_{\mathrm{i}}=7.60\mathrm{mm}×\frac{1\mathrm{m}}{1000\mathrm{mm}} \\ =7.60×{10}^{-3}\mathrm{m} \end{gathered}$$

The work done by the spring is the change in the spring’s potential energy,

$$\begin{gathered} {\mathrm{W}}_1=\frac{1}{2}{{\mathrm{kx}}^2}_{\mathrm{t}}-\frac{1}{2}{{\mathrm{kx}}^2}_{\mathrm{i}} \\ {\mathrm{W}}_1=\frac{1}{2}\left(1800\right)\left(0\right)-\frac{1}{2}\left(1800\right){(7.60×{10}^{-3})}^2{\mathrm{w}}_1=-5.2×{10}^{-2}\mathrm{J} \\ {\mathrm{W}}_1=-5.2×{10}^{-2}\mathrm{J} \\ \mathrm{Hence},\mathrm{work}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{spring}\mathrm{force}\mathrm{on}\mathrm{the}\mathrm{cage}\mathrm{for}\mathrm{displacement}\mathrm{of}7.60\mathrm{m}\mathrm{is} \end{gathered}$$

$W_1=5.2×{10}^{-2}\mathrm{J}$

Now, the total displacement is,

$$\begin{gathered} {\mathrm{x}}_{\mathrm{i}}=7.60×{10}^{-3}+7.60×{10}^{-3} \\ =0.0152\mathrm{m} \\ {\mathrm{w}}_2=\frac{1}{2}{{\mathrm{kx}}^2}_{\mathrm{i}}-\frac{1}{2}{{\mathrm{kx}}^2}_{\mathrm{i}} \\ {\mathrm{w}}_2=\frac{1}{2}\left(1800\right)\left(0\right)-\frac{1}{2}\left(1800\right){(0.0152)}^2{\mathrm{w}}_2=-0.2079\mathrm{j} \\ \mathrm{Additional}\mathrm{work}\mathrm{done}\mathrm{by}\mathrm{the}\mathrm{spring}\mathrm{is}, \\ \mathrm{w}-{\mathrm{w}}_2-{\mathrm{w}}_1 \\ \mathrm{w}=-0.2079-(-0.052) \\ =-0.1559\mathrm{j} \\ \mathrm{w}=-0.156\mathrm{j} \end{gathered}$$

Hence, additional work done by the spring force for additional displacement of 7.60is

$\mathrm{w}=-0.156\mathrm{j}$