91Ó°ÊÓ

• The potential difference between a cloud and the ground is $∆V=1.0×{10}^{9}V$.
• The quantity of charge transferred is q = 30 C.
• The mass of the car is m = 1000 kg.

Using the equation of potential energy U = qV, you can find change in potential energy, you can find change in energy of the transferred charge.

Using the conservation of energy, you can find the car's final speed.

The change in energy of the transferred charge is.

$\begin{array}{rcl}∆U& =& q∆V\\ & =& \left(30C\right)\left(1.0×{10}^{9}V\right)\\ & =& 3.0×{10}^{10}J\end{array}$

Hence, the change in energy is $\begin{array}{rcl}& & 3.0×{10}^{10}J\end{array}$.

If all this energy is used to accelerate a 1000 kg car from rest, then the energy will be,

$∆U=K=\frac{1}{2}m{v}^2$

And you can find the car’s final speed to be

$\begin{array}{rcl}v& =& \sqrt{\frac{2K}{m}}\\ & =& \sqrt{\frac{2∆U}{m}}\end{array}$

Substitute known values in the above equation.

role="math" localid="1662533780803" $\begin{array}{rcl}V& =& \sqrt{\frac{2\left(3.0×{10}^{10}J\right)}{1000kg}}\\ & =& \sqrt{60×{10}^6{m}^2/s^2}\\ & =& 7.7×{10}^{3}m/s\end{array}$

Hence, the final speed of the car is $\begin{array}{rcl}& & 7.7×{10}^{3}m/s\end{array}$.