The typical values of the voltage and the associated charge of a defibrillator are $V = 2240$ V and $Q = 0.392$ C. Half of these values means that we have $Q = 1120$ V and $Q = 0.196$ C, which gives the following stored energy

$$
begin{align*}
PE &= frac{1}{2}QV\
&= frac{1}{2}(0.196;mathrm{C})(1120;mathrm{V})\
&= 110;mathrm{J}
end{align*}
$$

The work done $W$ by the battery is stored as an electric potential energy $PE$ in the capacitor. Thus we have $W = PE = QV/2$. But remember that $Q = CV$, so we have

$$
W = frac{1}{2}(CV)V = frac{1}{2}CV^2
$$

Substituting the known values of $C$ and $V$, we find

$$
W = frac{1}{2}(7.8;times10^{-6};mathrm{F})(3.0;mathrm{V})^2 = 35;mumathrm{J}
$$

$35;mu$J

The potential energy stored in the capacitor is given by $PE = CV^2/2$ (see Problem 42). This amount of energy is converted into a kinetic energy of the go-kart motor, which is given by $KE = mv^2/2$, where $m$ is mass of the go-kart and $v$ is its final speed. According to the conservation of energy, we have $PE = KE$, or

$$
frac{1}{2}CV^2 = frac{1}{2}mv^2
$$

where the initial speed is zero (the go-kart starts at rest). Solving for $v$ and substituting the known values, we find

$$
begin{align*}
v &= Vsqrt{frac{C}{m}}\
&= (12;mathrm{V})sqrt{frac{0.31times10^{-6};mathrm{F}}{16;mathrm{kg}}}\
&= 0.17;mathrm{cm/s}
end{align*}
$$