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 .^{24}
\text { (a) } W_{ ext }=\int F d x=F \int_{0}^{T} v(t) d t . From Prob. 11.19, v(t)=\frac{F}{m}\left[t+\tau-\tau e^{(t-T) / \tau}\right] . So
W_{ ext }=\frac{F^{2}}{m}\left[\int_{0}^{T} t d t+\tau \int_{0}^{T} d t-\tau e^{-T / \tau} \int_{0}^{T} e^{t / \tau} d t\right]=\left.\frac{F^{2}}{m}\left[\frac{t^{2}}{2}+\tau t-\tau e^{-T / \tau} \tau e^{t / \tau}\right]\right|_{0} ^{T}
=\frac{F^{2}}{m}\left[\frac{1}{2} T^{2}+\tau T-\tau^{2} e^{-T / \tau}\left(e^{T / \tau}-1\right)\right]=\frac{F^{2}}{m}\left(\frac{1}{2} T^{2}+\tau T-\tau^{2}+\tau^{2} e^{-T / \tau}\right).
(b) From Prob. 11.19, the final velocity is v_{f}=(F / m) T, \text { so } W_{\text {kin }}=\frac{1}{2} m v_{f}^{2}=\frac{1}{2} m \frac{F^{2}}{m^{2}} T^{2}=\frac{F^{2} T^{2}}{2 m}.
\text { (c) } W_{ rad }=\int P d . According to the Larmor formula, P=\frac{\mu_{0} q^{2} a^{2}}{6 \pi c} , and (again from Prob. 11.19)
a(t)=\left\{\begin{array}{l}(F / m)\left[1-e^{-T / \tau}\right] e^{t / \tau}, \quad(t \leq 0) \\(F / m)\left[1-e^{(t-T) / \tau}\right], \quad(0 \leq t \leq T)\end{array}\right.W_{ rad }=\frac{\mu_{0} q^{2}}{6 \pi c} \frac{F^{2}}{m^{2}}\left\{\left(1-e^{-T / \tau}\right)^{2} \int_{-\infty}^{0} e^{2 t / \tau} d t+\int_{0}^{T}\left[1-e^{(t-T) / \tau}\right]^{2} d t\right\}
=\tau \frac{F^{2}}{m}\left\{\left.\left(1-e^{-T / \tau}\right)^{2}\left(\frac{\tau}{2} e^{2 t / \tau}\right)\right|_{-\infty} ^{0}+\int_{0}^{T} d t-2 e^{-T / \tau} \int_{0}^{T} e^{t / \tau} d t+e^{-2 T / \tau} \int_{0}^{T} e^{2 t / \tau} d t\right\}
=\frac{\tau F^{2}}{m}\left[\frac{\tau}{2}\left(1-e^{-T / \tau}\right)^{2}+T-\left.2 e^{-T / \tau}\left(\tau e^{t / \tau}\right)\right|_{0} ^{T}+\left.e^{-2 T / \tau}\left(\frac{\tau}{2} e^{2 t / \tau}\right)\right|_{0} ^{T}\right\}
=\frac{\tau F^{2}}{m}\left[\frac{\tau}{2}\left(1-2 e^{-T / \tau}+e^{-2 T / \tau}\right)+T-2 \tau e^{-T / \tau}\left(e^{T / \tau}-1\right)+\frac{\tau}{2} e^{-2 T / \tau}\left(e^{2 T / \tau}-1\right)\right]
=\frac{\tau F^{2}}{m}\left[\frac{\tau}{2}-\tau e^{-T / \tau}+\frac{\tau}{2} e^{-2 T / \tau}+T-2 \tau+2 \tau e^{-T / \tau}+\frac{\tau}{2}-\frac{\tau}{2} e^{-2 T / \tau}\right]=\frac{\tau F^{2}}{m}\left(T-\tau+\tau e^{-T / \tau}\right).
Energy conservation requires that the work done by the external force equal the final kinetic energy plus the energy radiated:
W_{\text {kin }}+W_{ rad }=\frac{F^{2} T^{2}}{2 m}+\frac{\tau F^{2}}{m}\left(T-\tau+\tau e^{-T / \tau}\right)=\frac{F^{2}}{m}\left(\frac{1}{2} T^{2}+\tau T-\tau^{2}+\tau^{2} e^{-T / \tau}\right)=W_{ ext } .