Chapter 33, Problem 44A

The potential difference between a storm cloud and the ground is $5.0\times {10}^7$ volts. During a lightning flash, 3.0 coulombs of charge are transferred to the ground.

a. How much energy is transferred to the ground in this lightning flash?

b. If this much energy were used to accelerate a 3500-kg truck from rest, how fast would the truck end up going?

To determine

The energy transferred to the ground during lightning.

Answer to Problem 44A

The energy transferred = $1.5\times {10}^8\text{ J}$ .

Explanation of Solution

Given data:

Potential difference between storm cloud and ground, $\text{U}=5\times {10}^7\text{ volts}$ .

Charge transferred $q=3\text{ C}$ .,

Formula used:

The energy transferred is also known as the work done, and the work done is given as,

  $W=qU$

Where,

W = work done.

q = Charge.

U = potential energy

Calculation:

The energy transferred will be:

  $\begin{array}{l}\Delta E=W=qU\\ \Delta E=3\times 5\times {10}^7\\ \Delta E=1.5\times {10}^8\text{ J}\end{array}$

Conclusion:

The energy transferred = $1.5\times {10}^8\text{ J}$ .

To determine

The speed of truck got accelerated due to this energy transferred.

Answer to Problem 44A

The velocity of truck = $292.77\text{ m/s}\text{.}$

Explanation of Solution

Given data:

Potential difference between storm cloud and ground, U = $5\times {10}^7\text{volts}\text{.}$

Charge transferred, q $=3\text{ C}$ .

Mass of truck, m $=3500\text{ kg}\text{.}$ .

Formula used:

The work done will be the change in kinetic energy.

  $\Delta E=K.E.=0.5m{v}^2$

Where,

K.E. = kinetic energy.

m = mass of truck.

v = velocity of truck.

Calculation:

The final velocity of truck will be.

  $\begin{array}{l}\Delta E=K.E.\\ K.E.=0.5m{v}^2\\ 1.5\times {10}^8=0.5\times 3500\times v^2\\ v^2=\frac{1.5\times {10}^8}{0.5\times 3500}\\ v=\sqrt{\frac{1.5\times {10}^8}{0.5\times 3500}}\\ v=292.77\text{ m/s}\text{.}\end{array}$

Conclusion:

The velocity of truck = 292.77 m/s.