91Ó°ÊÓ

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.

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

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

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

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.

$W_{rad}=∫P.dt$ …… (3)

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

$p=\frac{{\mathrm{μ}}_0{q}^2{a}^2}{6\mathrm{Ï€³¦}}$

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).

$$\begin{gathered} W_{rad}=\frac{{\mathrm{μ}}_0{q}^2}{6\mathrm{Ï€³¦}}\frac{F^2}{m^2}\left\{{(1-e^{-T/t})}^2{∫}_{-∞}^0{e}^{2t/t}dt+{∫}_0^T{\left[1-e\frac{(t-T)}{t}\right]}^{2}dt\right\} \\ {W}_{rad}=t\frac{F^2}{m^2}\left\{\begin{array}{c}(1-e^{-T/t}){\left[\left(\frac{t}{2}{e}^{2t/t}\right)\right]}_{-∞}^0+\\ {∫}_0^{T}dt-2e^{-T/t}{∫}_0^T{e}^{e/e}dt+e^{-2t/t}{∫}_0^T{e}^{et/t}dt\end{array}\right\} \\ {W}_{rad}=t\frac{F^2}{m^2}\left\{\frac{t}{2}{(1-e^{-T/t})}^2+T-2e^{-T/t}{\left[\left(t{e}^{e/e}\right)\right]}_0^T+e^{e^{-2t/t}}{\left[\left(\frac{t}{2}{e}^{e^{2t/t}}\right)\right]}_0^T\right\} \\ {W}_{rad}=t\frac{F^2}{m^2}\left(T-t+t{e}^{-\frac{T}{t}}\right) \end{gathered}$$

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 $W_{rad}=\mathit{t}\frac{F^2}{m^2}\left(T-\mathit{t}+\mathit{t}{e}^{-\frac{T}{t}}\right)$.