Q51E - University Physics with Modern Physics

Question

How many joules of energy does a $\mathbf{100}\mathbf{\text{ watt}}$ light bulb use per hour? How fast would a $\mathbf{70}\mathbf{\text{ kg}}$  person have to run to have that amount of kinetic energy?

Step-by-Step Solution

Answer

The energy of the bulb is $3.6\times {10}^5\text{ J}$ and the velocity of the person is $101.42 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Step 1: A definitions of energy and kinetic energy:

Energy is defined as the amount of work that can be done by a force.

Kinetic energy is the form of energy that an object or particle has because of its motion.

Step 2: Given data and required formulas:

The energy of a light bulb is,

$E=Pt$

Here, $E$ is the energy of the light bulb, $P$ is the power of the light bulb, and $t$ is the time.

 

The energy of a running person is equal to the kinetic energy of the person. Therefore,

$E=\frac{1}{2}m{v}^2$

Here, $E$ is the energy of the person, m is the mass of the person, and $v$ is the velocity of the person.

Consider the given data as below.

The power, \(P = 100{\rm{ }}W\)

The time, \(t = 1{\rm{ }}h = 3600{\rm{ }}s\)

Mass of the person, \(m = 70{\rm{ }}kg\)

Step 3: The energy of the bulb:

Calculate the energy of the light bulb by using following formula.

$E=Pt$

 

Substitute known values in the above equation.

$\begin{array}{rcl}E& =& 100\text{ W}\times 3600\text{ s}\\ & =& 3.6\times {10}^5\text{ J}\end{array}$

Hence, the energy of the bulb is  $\begin{array}{rcl}& & 3.6\times {10}^5\text{ J}\end{array}$.

Step 4: The velocity of the person:

The velocity of the person is define by,

$$\begin{gathered} dE=\frac{1}{2}m{v}^2 \\ {v}^2=\frac{2E}{m} \\ v=\sqrt{\frac{2E}{m}} \end{gathered}$$

Putting known numerical values in the above equation.

$\begin{array}{rcl}v& =& \sqrt{\frac{2\times 3.6\times {10}^5\text{ J}}{70\text{ kg}}}\\ & =& \sqrt{10285.714 \raisebox{1ex}{$m^2$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}\\ & =& 101.42 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

 

Hence, the velocity of the person is  $101.42 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.