Chapter 4.2, Problem 17PP

To determine

To calculate: the increase in kinetic energy when the car doubles its speed.

Answer to Problem 17PP

The kinetic energy of the car increases four times when the car doubles its speed.

Explanation of Solution

Given data:

Mass of the car, $m=1500\text{ kg}$

Speed of the car, $v_1=50\text{ km/h, }{v}_2=100\text{ km/h}$

Formula used:

Kinetic Energy Equation:

  $KE=\frac{1}{2}m{v}^2$

Where $KE$ is the kinetic energy of the object in joules, $m$ is the mass of the object in $\text{kg}$ and $v$ is the speed of the object in $\text{m/s}$ .

Calculation:

As per the problem,

  $m=1500\text{kg}$

To calculate the kinetic energy when the speed of the car is $v_1=50\text{ km/h}$ , first convert kilometer per hour into meter per second.

  $\begin{array}{l}{v}_1=50\text{ km/h}\\ \text{1 km}=1000\text{ m}\\ \text{1 h}=3600\text{ s}\\ v_1=50\times \frac{1000\text{ m}}{3600\text{ s}}=13.89\text{ m/s}\end{array}$

Substitute the values of mass and speed in the kinetic energy equation:

  $\begin{array}{l}K{E}_1=\frac{1}{2}m{v}_1{}^2\\ K{E}_1=\frac{1}{2}(1500\text{ kg)(13}{\text{.89 m/s)}}^2\\ K{E}_1=144699.01\text{ J}\simeq 144700\text{ J}\end{array}$

To calculate the kinetic energy when the speed of the car is $v_2=100\text{ km/h}$ , first convert kilometer per hour into meter per second.

  $\begin{array}{l}{v}_1=100\text{ km/h}\\ \text{1 km}=1000\text{ m}\\ \text{1 h}=3600\text{ s}\\ v_2=100\times \frac{1000\text{ m}}{3600\text{ s}}=27.78\text{ m/s}\end{array}$

Substitute the values of mass and speed in the kinetic energy equation:

  $\begin{array}{l}K{E}_2=\frac{1}{2}m{v}_2{}^2\\ K{E}_2=\frac{1}{2}(1500\text{ kg)(27}{\text{.78 m/s)}}^2\\ K{E}_2=578796.3\text{ J}\approx \text{578,800 J}\end{array}$

The increase in kinetic energy = $\frac{K{E}_2}{K{E}_1}=\frac{578,800\text{ J}}{144,700\text{ J}}=4\text{ times}$

Conclusion:

The kinetic energy of the car increases four times when the car doubles its speed.