An angler hangs a 4.50 kgfish from a vertical steel wirelong and$\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{cm}}^{\mathbf{2}}$in a cross-sectional area. The upper end of the wire is securely fastened to a support.

1. Calculate the amount the wire is stretched by the hanging fish.

The angler now applies a varying forceat the lower end of the wire, pulling it very slowly downward byfrom its equilibrium position. For this downward motion, calculate

1. The work done by gravity;
2. The work done by the force$\overrightarrow{\mathbf{F}}$,
3. The work done by the force the wire exerts on the fish;
4. The change in the elastic potential energy (the potential energy associated with the tensile stress in the wire). Compare the answers in parts (d) and (e).

Here we have, mass of fish is${\mathrm{m}}_{\mathrm{f}}=4.50\mathrm{kg}$

Length of vertical steel wire is${\mathrm{l}}_0=1.50\mathrm{m}$

Cross sectional area of wire is$\mathrm{A}=5.00\times {10}^{-3}{\mathrm{cm}}^2$

Young’s modulus for steel wire is$Y=20\times {10}^{10}Pa$

The idea of completed work W is the product of the force and the distance. Which is given by,

W = F.d (1)

Where, W is work done, F is force and d is distance.

(a)

Now, Young’ modulus we have formula which is given by

$\begin{array}{r}\mathrm{Y}=\frac{{\mathrm{F}}_{\perp }{\mathrm{I}}_0}{\mathrm{A\Delta I}}\\ \mathrm{\Delta I}=\frac{{\mathrm{F}}_{\perp }{\mathrm{I}}_0}{\mathrm{AY}}\\ \mathrm{\Delta I}=\frac{\left({\mathrm{m}}_{\mathrm{f}}\cdot \mathrm{g}\right){\mathrm{I}}_0}{\mathrm{AY}}\\ \mathrm{Now},\mathrm{substitute}\mathrm{all}\mathrm{numerical}\mathrm{values}\mathrm{in}\mathrm{above}\mathrm{equation}\mathrm{we}\mathrm{get},\\ \mathrm{\Delta I}=\frac{(4.50\mathrm{kg})\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)(1.50\mathrm{m})}{\left(5.00\times {10}^{-3}{\mathrm{cm}}^2\right)\left(20\times {10}^{10}\mathrm{Pa}\right)}\\ =6.63\times {10}^{-4}\mathrm{m}\end{array}$

Hence. the amount the wire is stretched by the hanging fish is$6.63\times {10}^{-4}m$ .

(b)

Here we have additional acting force of the angler F, so the additional elongation of the wire is $∆{\mathrm{l}}_{\mathrm{F}}=0.5\times {10}^{-4}\mathrm{m}$.

Now, the workdone by gravity for that additional elongation is given by equation (1) which is says that,

$\begin{array}{r}{W}_g=F_g\cdot \mathrm{\Delta }{I}_F\\ W_g=\left(m_g\cdot g\right)\mathrm{\Delta }{l}_F\\ W_g=\left(4.50\mathrm{kg}\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(0.5\times {10}^{-4}\mathrm{m}\right)\\ =0.0219\mathrm{J}\end{array}$

Hence, the work done by gravity is 0.0219J .

(c)

Here we have additional acting force of the angler F, so the additional elongation of the wire is $∆l_F=0.5\times {10}^{-4}m$.

Now, we know that,

$\begin{array}{r}\mathrm{Y}=\frac{{\mathrm{FI}}_0}{\mathrm{Al}}\\ \mathrm{F}=\frac{\mathrm{YAI}}{{\mathrm{I}}_0}\\ \mathrm{Also},\mathrm{The}\mathrm{product}\mathrm{of}\mathrm{displacement}\mathrm{and}\mathrm{the}\mathrm{force}\mathrm{acting}\mathrm{on}\mathrm{the}\mathrm{body}\mathrm{as}\mathrm{a}\mathrm{result}\mathrm{of}\mathrm{that}\mathrm{displacement}\mathrm{is}\mathrm{given}\mathrm{as}\mathrm{work}.\mathrm{So}\mathrm{we}\mathrm{have},\\ {\mathrm{W}}_{\mathrm{F}}={\int }_{{\mathrm{\Delta }}_{\mathrm{f}}} \mathrm{F}\cdot \mathrm{dl}\\ ={\int }_0^{{\mathrm{\Delta }}_{\mathrm{f}}} \frac{\mathrm{YA}}{{\mathrm{I}}_0+\mathrm{\Delta l}}\cdot \mathrm{I}\cdot \mathrm{dl}\\ ={\left[\frac{\mathrm{YA}}{{\mathrm{I}}_0+\mathrm{\Delta l}}\cdot \frac{{\mathrm{l}}^2}{2}\right]}_0^{{\mathrm{\Delta }}_{\mathrm{f}}}\\ =\frac{\mathrm{YA}}{{\mathrm{I}}_0+\mathrm{\Delta l}}\left(\frac{{\left({\mathrm{\Delta I}}_{\mathrm{f}}\right)}^2}{2}\right)\end{array}$

By substituting all the numerical values in above equation we get,

$\begin{array}{r}W=\frac{\left(5.00\times {10}^{-3}{\mathrm{cm}}^2\right)\left(20\times {10}^{10}\mathrm{Pa}\right)}{1.5\mathrm{m}+6.63\times {10}^{-4}\mathrm{m}}\left(\frac{{\left(0.5\times {10}^{-3}\mathrm{m}\right)}^2}{2}\right)\\ =8.33\times {10}^{-3}\mathrm{J}\end{array}$

(d)

Here we have additional acting force of the angler , so the additional elongation of the wire is$∆l_F=0.5\times {10}^{-4}m$

Now, the force of the wire acting on the fish is given by:

$T=-\left(F_g+F\right)$

Now, the similar way from step 5, we have

$\begin{array}{r}{\mathrm{W}}_{\mathrm{T}}=\int \mathrm{T}\cdot \mathrm{dl}\\ =\int -\left({\mathrm{F}}_{\mathrm{g}}+\mathrm{F}\right)\cdot \mathrm{dl}\\ ={\int }_0^{{\mathrm{\Delta l}}_{\mathrm{f}}} -\left({\mathrm{m}}_{\mathrm{fish}}\cdot \mathrm{g}+\frac{\mathrm{YA}}{{\mathrm{I}}_0+\mathrm{\Delta l}}\cdot \mathrm{l}\right)\cdot \mathrm{dl}\\ =-{\mathrm{m}}_{\mathrm{fish}}\cdot \mathrm{g}\cdot {\mathrm{\Delta l}}_{\mathrm{f}}-{\left[\frac{\mathrm{YA}}{{\mathrm{I}}_0+\mathrm{\Delta l}}\cdot \frac{{\mathrm{I}}^2}{2}\right]}_0^{{\mathrm{\Delta l}}_{\mathrm{f}}}\\ =-{\mathrm{m}}_{\mathrm{fish}}\cdot \mathrm{g}\cdot {\mathrm{\Delta l}}_{\mathrm{f}}+\frac{\mathrm{YA}}{{\mathrm{I}}_0+\mathrm{\Delta l}}\left(\frac{{\left({\mathrm{\Delta l}}_{\mathrm{f}}\right)}^2}{2}\right)\\ \mathrm{Substitute}\mathrm{all}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{above}\mathrm{equation}\mathrm{we}\mathrm{get},\\ {\mathrm{W}}_{\mathrm{T}}=-(4.5\mathrm{kg})\cdot \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\cdot \left(0.5\times {10}^{-3}\mathrm{m}\right)+\frac{\left(5.00\times {10}^{-3}{\mathrm{cm}}^2\right)\left(20\times {10}^{10}\mathrm{Pa}\right)}{1.5\mathrm{m}+6.63\times {10}^{-4}\mathrm{m}}\left(\frac{{\left(0.5\times {10}^{-3}\mathrm{m}\right)}^2}{2}\right)\\ =-3\times {10}^{-2}\mathrm{J}\end{array}$

Hence, the work done by the force the wire exerts on the fishis$-3\times {10}^{-2}J$ .

(e)

Now, change in elastic potential energy is

$\begin{array}{r}\mathrm{\Delta U}={\mathrm{U}}_2-{\mathrm{U}}_1\\ =\frac{1}{2}{\mathrm{kl}}_2^2-\frac{1}{2}{\mathrm{kl}}_1^2\\ =\frac{1}{2}\mathrm{k}{\left(\mathrm{\Delta l}+{\mathrm{\Delta l}}_{\mathrm{F}}\right)}^2-\frac{1}{2}\mathrm{k}(\mathrm{\Delta l}{)}^2\\ =\frac{1}{2}\mathrm{k}\left({\left({\mathrm{\Delta I}}_{\mathrm{F}}\right)}^2+2{\mathrm{\Delta l\Delta I}}_{\mathrm{F}}\right)\\ \mathrm{Now},\mathrm{we}\mathrm{know}\mathrm{from}\mathrm{Hooke}’\mathrm{s}\mathrm{law},\\ \mathrm{k}=\frac{\mathrm{YA}}{{\mathrm{I}}_0}\end{array}$

Mow, put the value of in above equation we get,

$\mathrm{\Delta }U=\frac{1}{2}\left(\frac{YA}{I_0}\right)\left({\left(\mathrm{\Delta }{I}_F\right)}^2+2{\mathrm{\Delta \Delta }}_F\right)$

Put all the numerical values in above equation we get,

$\begin{array}{r}\mathrm{\Delta }U=\frac{1}{2}\left(\frac{\left(5.00\times {10}^{-3}{\mathrm{cm}}^2\right)\left(20\times {10}^{10}\mathrm{Pa}\right)}{1.5\mathrm{m}}\right)\left({\left(5\times {10}^{-4}\mathrm{m}\right)}^2+2\left(5\times {10}^{-4}\mathrm{m}\right)\left(6.63\times {10}^{-4}\mathrm{m}\right)\right)\\ =3\times {10}^{-2}\mathrm{J}\end{array}$

Hence, the change in the elastic potential energyis$3\times {10}^{-2}\mathrm{J}$ .