37-43

OK, no diagrams for these. The kinetic energy of a particle in SR is given by

$$K=(\gamma-1)mc^2.$$

I will briefly give a reason as to why the definition should be different. Suppose we apply a constant force to a particle. Then the work done is

$$\Delta K = W = F\Delta x.$$

The problem we have is that, since $F$ is constant, the kinetic energy should increase proportionally to $\Delta x$. But the speed of the particle is bounded above by $c$. So the definition of kinetic energy as $mv^2/2$ gives a maximum value of $mc^2/2$ for the kinetic energy. Thus we need a new definition that has value 0 when $v=0$ and increases to infinity as $v\to c$. The definition above does exactly that since $\gamma\to 1$ as $v\to 0$ and $\gamma\to\infty$ as $v\to c$.

In any case, a change in kinetic energy can be expressed as

$$\begin{eqnarray*} \Delta K &=& K_f-K_i\\ &=&(\gamma_f-1)mc^2-(\gamma_i-1)mc^2\\ &=& (\gamma_f-\gamma_i)mc^2, \end{eqnarray*}$$

but since $\gamma$ increases non-linearly with $v$, the answer will depend on the starting value of $v$, not just the change in $v$. We have

$$\begin{eqnarray*} \gamma_i &=& (1-.18^2)^{-1/2}=1.0166\\ \gamma_f &=& (1-.19^2)^{-1/2}=1.0185, \end{eqnarray*}$$

so that

$$W=(1.0185-1.0166)(511\;{\rm keV/c^2})c^2=971\;{\rm eV}\approx1\;{\rm keV},$$

since the mass of an electron is $511\;{\rm keV/c^2}$.

for the second part we have

$$\begin{eqnarray*} \gamma_i &=& (1-.98^2)^{-1/2}=5.025\\ \gamma_f &=& (1-.99^2)^{-1/2}=7.088, \end{eqnarray*}$$

so that

$$W=(7.088-5.025)(511\;{\rm keV/c^2})c^2=1.05\;{\rm MeV},$$

which is about $10^6/10^3=1000$ times larger!