Question

A spring with a spring constant of ${}^{180\mathrm{N}/\mathrm{cm}}$has a cage attached to its free end. (a) How much work does the spring force do on the cage when the spring is stretched from its relaxed length by${}^{7.60\mathrm{mm}}$? (b) How much additional work is done by the spring force when the spring is stretched by an additional${}^{7.60\mathrm{mm}}$ ?

Short answer

1. Work done by the spring force on the cage for displacement of 7.60m is${}^{-5.2\times {10}^2\mathrm{j}}$.

2. Additional work done by the spring force for additional displacement of 7.6${}^{-0.156\mathrm{j}}$

Step-by-step solution

Given information

It is given that,

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

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

Determining the concept

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.

(a) Determining the work done by the spring force on the cage for displacement of 7.60 m

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

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

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

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

$W_1=5.2\times {10}^{-2}\mathrm{J}$

(b) Determining the additional work done by the spring force for additional displacement of 7.60 m

Now, the total displacement is,

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

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

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