Question: An automobile traveling on a highway has an average kinetic energy of . Its mass is . What is its average speed? Convert your answer to miles per hour to see whether it makes sense. If you could use all of themc²rest energy of some amount of fuel to provide the car with its kinetic energy of , What mass of fuel would you need?

The kinetic energy is given by:

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

Here$m$is mass of the object, and $v$is the velocity of the object.

A simple approximate expression for the kinetic energy of the car can be obtained by comparing the speed v of a particle to the speed of light $v<<c$.

Let vbe the automobile's speed.

In the given scenario, the kinetic energy is negligible in comparison to the rest energy, with the kinetic energy being approximately zero at low speeds.

Solve the formula of kinetic energy for v.

$v=\sqrt{\frac{2K}{m}}$

Substitute kinetic energy of the automobile$k=1.1\times {10}^{5}J$and its mass$m=1500kg$ into the obtained formula.

$$\begin{gathered} v=\sqrt{\frac{2\left(1.1\times {10}^{5}J\right)}{1500kg}} \\ =12.0m/s \end{gathered}$$

Convert the units of the speed into mile/h by using the conversion $1mile=1609m$and $1h=3600s$.

$$\begin{gathered} v=12.0\frac{m}{s}\left[\frac{1mile}{1609m}\right]\left[\frac{3600s}{1h}\right] \\ =26.85miles/h \end{gathered}$$

Thus, the average speed is$12.0m/s$ and in mile per hour the speed is $28.85miles/h$.

The rest energy of the fuel is given by:

$E_{rest}=m{c}^2$

Substitute $E_{rest}=1.1\times {10}^{5}J$and $c=3\times {10}^{8}m/s$into the formula and solve for $m$.

$$\begin{gathered} 1.1\times {10}^{5}J=m{\left(3\times {10}^{8}m/s\right)}^2 \\ m=\frac{1.1\times {10}^{5}J}{{\left(3\times {10}^{8}m/s\right)}^2} \\ =1.22\times {10}^{-12}kg \end{gathered}$$

Thus, the mass of fuel need is$1.22\times {10}^{-12}kg$.

As illustrated, a small amount of fuel is sufficient to propel the vehicle at a speed of 26.85 miles/h.