Assuming you exclude the runaway solution in Prob. 11.19, calculate

(a) The work done by the external force,

(b) The final kinetic energy (assume the initial kinetic energy was zero),

(c) The total energy radiated.

Check that energy is conserved in this process.

Write the expression for the velocity of a particle.

$v\left(t\right)=\frac{F}{m}\left[t+t-t{e}^{\frac{(t-T)}{t}}\right]$ …… (1)

Here, F is the external force, m is the mass, and t is the time.

(a)

Write the expression for work done by an external force.

$$\begin{gathered} W_{ext}=\int F.dx \\ {W}_{ext}={\int }_0^{T}v?\left(t\right).dt \end{gathered}$$

Substitute the value of v ( t ) in the above expression.

$$\begin{gathered} W_{ext}=F{\int }_0^T\frac{F}{m}\left[t+t^{\text{'}}-t^{\text{'}}{e}^{\frac{(t-t^{\text{'}})}{t}}\right].dt \\ {W}_{ext}=\frac{F^2}{m}\left[{\int }_0^{T}tdt+\mathit{t}{\int }_0^{T}dt-\mathit{t}{e}^{-T/t}{\int }_0^t{e}^{t/t}dt\right] \\ {W}_{ext}=\frac{F^2}{m}{\left[\frac{t^2}{2}+\mathit{t}t-t{e}^{-T/t^{\text{'}}}t{e}^{t/t^{\text{'}}}\right]}_0^T \end{gathered}$$

On further solving,

$$\begin{gathered} W_{ext}=\frac{F^2}{m}\left[\frac{T^2}{2}+\mathit{t}T-{\mathit{t}}^2{e}^{-T/t}(e^{\mathbf{T}\mathbf{/}\mathbf{t}}-1)\right] \\ {W}_{ext}=\frac{F^2}{m}\left(\frac{T^2}{2}+\mathit{t}T-{\mathit{t}}^2+{\mathit{t}}^2{e}^{-T/t}\right) \end{gathered}$$

Therefore, the work done by an external force is $W_{ext}=\frac{F^2}{m}\left(\frac{T^2}{2}+\mathit{t}T-{\mathit{t}}^2+{\mathit{t}}^2{e}^{-T/t}\right)$.

(b)

Write the expression for final kinetic energy.

$K_f=\frac{1}{2}m{v}_f^2$ …… (2)

Here, $v_f$is the final kinetic energy.

Substitute t = T in equation (1) to calculate the final kinetic energy.

$$\begin{gathered} v\left(t\right)=\frac{F}{m}\left[T+t-t{e}^{\frac{(T-T)}{T}}\right] \\ {v}_f=\left(\frac{F}{m}\right)T \end{gathered}$$

Substitute the value of $v_f$in equation (2).

$$\begin{gathered} K_f=\frac{1}{2}{\left[\left(\frac{F}{m}\right)T\right]}^2 \\ {K}_f=\frac{F^2{T}^2}{2m} \end{gathered}$$

Therefore, the final kinetic energy is $K_f=\frac{F^2{T}^2}{2m}$.

(c)

Write the expression for the total energy radiated.

$W_{rad}=\int P.dt$ …… (3)

Using the Larmor formula, the value of P is given as:

$p=\frac{{\mu }_0{q}^2{a}^2}{6\mathrm{\pi c}}$

Here, a is the acceleration of the particle.

Write the expression for an acceleration of a particle.

$a\left(t\right)=\left\{\begin{array}{ll}\left(\frac{F}{m}\right)\left[1-e^{\_T/t}\right]& \left(t\underline{<}0\right)\\ \left(\frac{F}{m}\right)\left[1-e^{\frac{(t-T)}{t}}\right]& (0\underline{<}t\underline{<}T)\end{array}\right.$

Substitute all the value of P in equation (3).

$$\begin{gathered} W_{rad}=\frac{{\mu }_0{q}^2}{6\mathrm{\pi c}}\frac{F^2}{m^2}\left\{{(1-e^{-T/t})}^2{\int }_{-\infty }^0{e}^{2t/t}dt+{\int }_0^T{\left[1-e\frac{(t-T)}{t}\right]}^{2}dt\right\} \\ {W}_{rad}=t\frac{F^2}{m^2}\left\{\begin{array}{c}(1-e^{-T/t}){\left[\left(\frac{t}{2}{e}^{2t/t}\right)\right]}_{-\infty }^0+\\ {\int }_0^{T}dt-2e^{-T/t}{\int }_0^T{e}^{e/e}dt+e^{-2t/t}{\int }_0^T{e}^{et/t}dt\end{array}\right\} \\ {W}_{rad}=t\frac{F^2}{m^2}\left\{\frac{t}{2}{(1-e^{-T/t})}^2+T-2e^{-T/t}{\left[\left(t{e}^{e/e}\right)\right]}_0^T+e^{e^{-2t/t}}{\left[\left(\frac{t}{2}{e}^{e^{2t/t}}\right)\right]}_0^T\right\} \\ {W}_{rad}=t\frac{F^2}{m^2}\left(T-t+t{e}^{-\frac{T}{t}}\right) \end{gathered}$$

Energy conservation requires that the work done by an external force equal to the sum of the final kinetic energy and the radiated energy. Hence,

$$\begin{gathered} K_f+W_{rad}=\left[\frac{f^2{T}^2}{2m}\right]+\left[\mathit{t}\frac{F^2}{m^2}\left(T-\mathit{t}+\mathit{t}{e}^{-\frac{T}{t}}\right)\right] \\ {K}_f+W_{rad}=\frac{F^2}{m}\left(\frac{1}{2}{T}^2+\mathit{t}T-{\mathit{t}}^2+{\mathit{t}}^2{e}^{-\frac{T}{t}}\right) \\ {W}_{ext}=\frac{F^2}{m}\left(\frac{1}{2}{T}^2+\mathit{t}T-{\mathit{t}}^2+{\mathit{t}}^2{e}^{-\frac{T}{t}}\right) \end{gathered}$$

Therefore, the total energy radiated is $W_{rad}=\mathit{t}\frac{F^2}{m^2}\left(T-\mathit{t}+\mathit{t}{e}^{-\frac{T}{t}}\right)$.