Kinetic energy is defined as the energy possessed by a body as a result of its movement.
For an object to set itself in motion, something has to be done to it: a force has to be exerted on it so that it can set itself in motion.
The kinetic energy possessed by an object is determined by the following equation:
||E_{k} = \dfrac {1}{2} \times m \times v^{2}||
where
|E_{k}| is kinetic energy |\text {(J)}|
|m| is the object's mass |\text {(kg)}|
|v| is the object's velocity |\text {(m/s)}|
The amount of kinetic energy possessed by an object depends on two factors: the mass of the moving object and its velocity. So if you double the mass of an object, its kinetic energy will also double. However, if you double its velocity, its kinetic energy will be four times greater.
To convert velocity in metres per second into kilometres per hour, follow the procedure below.
||\dfrac{\text {m}}{\text {s}} \times \dfrac {1 \text { km}}{1\:000 \text { m}} \times \dfrac {3\:600 \text { s}}{1 \text {h}}||
To make it easier, simply do |\dfrac{\text {m}}{\text {s}} \times 3{,}6 = \dfrac{\text {km}}{\text {h}}.|
It's also possible to convert velocity in kilometres per hour into metres per second.
||\dfrac{\text {km}}{\text {h}} \times \dfrac {1\:000 \text { m}}{1 \text { km}} \times \dfrac {1 \text { h}}{3\:600 \text {s}}||
To make it easier, simply do |\dfrac{\text {km}}{\text {h}} \div 3{,}6 = \dfrac{\text {m}}{\text {s}}.|
The driver of a |\text {2 000 kg}| car accelerates from |\text {20 m/s}| to |\text {30 m/s}.| How much kinetic energy does the car need?
In this problem, it is not possible to take only the variation in velocity of |\text {10 m/s}|, since the energy is not proportional to the velocity, but to the square of the velocity. We need to calculate the energy at each of the two velocities and then differentiate between them.
First, we need to determine the initial kinetic energy.
||\begin{align}
E_{k_{i}} = \dfrac {1}{2} \times m \times v^{2} \quad \Rightarrow \quad
E_{k_{i}}&= \dfrac {1}{2} \times \text {2 000 kg} \times \text {(20 m/s)}^{2} \\
&= \text {400 000 J} \\
\end{align}||
The final kinetic energy must then be determined.
||\begin{align}
E_{k_{f}} = \dfrac {1}{2} \times m \times v^{2} \quad \Rightarrow \quad
E_{k_{f}}&= \dfrac {1}{2} \times \text {2 000 kg} \times \text {(30 m/s)}^{2} \\
&= \text {900 000 J} \\
\end{align}||
By determining the difference between the final kinetic energy and the initial kinetic energy, it is possible to determine the energy required to produce the change in velocity.
||\begin{align}
\triangle E = E_{k_{f}} - E_{k_{i}} \quad \Rightarrow \quad
\triangle E&= \text {900 000 J} - \text {400 000 J} \\
&= \text {500 000 J} \\
\end{align}||
A work of |\text {500 000 J}| must therefore be carried out for the car to go from |\text {20 m/s}| to |\text {30 m/s}.|
As mentioned in the previous example, work must be done to observe a change in kinetic energy. In a frictionless environment, all the energy produced by work can be converted into kinetic energy.
|W = \triangle E_{k}|
where
|W| is the work |\text {(J)}|
|\triangle E_{k}| is the variation in kinetic energy |\text {(J)}|