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### Theory:

In this section, we will derive the formula to find the kinetic energy of an object.
Kinetic energy is the energy contained by an object due to its motion. The kinetic energy of an object increases with its speed.
The kinetic energy of a body moving with a certain velocity is equal to the work done on it to make it obtain that velocity.
Let us now represent the kinetic energy of an object in the form of an equation.

Consider an object of mass ($$m$$) moving with a uniform velocity ($$u$$.

Let it now be displaced through a distance $$s$$ when a constant force ($$F$$, acts on it in the direction of its displacement.

From the equation of  work done, We know that,

$\mathit{Work}\phantom{\rule{0.147em}{0ex}}\mathit{done}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}F\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}s$

The work done on the object will cause a change in its velocity. Let its velocity change from initial velocity $$u$$ to final velocity $$v$$.

Let $$a$$ be the acceleration produced.

We studied three equations of motion. The relation connecting the initial velocity ($$u$$) and final velocity ($$v$$) of an object moving with a uniform acceleration $$a$$, and the displacement ($$s$$) is,

$\begin{array}{l}{v}^{2}\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}{u}^{2}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}2\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}a\phantom{\rule{0.147em}{0ex}}×s\\ \\ \mathit{Rearranging}\phantom{\rule{0.147em}{0ex}}\mathit{it}\phantom{\rule{0.147em}{0ex}}\mathit{to}\phantom{\rule{0.147em}{0ex}}\mathit{get}\phantom{\rule{0.147em}{0ex}}s,\\ \\ s\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{{v}^{2}\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}{u}^{2}\phantom{\rule{0.147em}{0ex}}}{2\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}a}\end{array}$

We also know that,

$\mathit{Force}\phantom{\rule{0.147em}{0ex}}\left(F\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\mathit{Mass}\phantom{\rule{0.147em}{0ex}}\left(m\right)\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}\mathit{Acceleration}\phantom{\rule{0.147em}{0ex}}\left(a\right)$

Substitute the equation $$s$$ and $$F$$ in work done,

$\begin{array}{l}\mathit{Work}\phantom{\rule{0.147em}{0ex}}\mathit{done}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}F\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}s\\ \\ =\phantom{\rule{0.147em}{0ex}}F\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}\frac{{v}^{2}\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}{u}^{2}\phantom{\rule{0.147em}{0ex}}}{2\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}a}\\ \\ =\phantom{\rule{0.147em}{0ex}}\left(m\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}a\right)\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}\frac{{v}^{2}\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}{u}^{2}\phantom{\rule{0.147em}{0ex}}}{2\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}a}\\ \\ \mathit{Simplifying}\phantom{\rule{0.147em}{0ex}}\mathit{it}\phantom{\rule{0.147em}{0ex}}\mathit{we}\phantom{\rule{0.147em}{0ex}}\mathit{get},\\ \\ =\phantom{\rule{0.147em}{0ex}}\frac{1}{2}\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}m\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}\left({v}^{2}\phantom{\rule{0.147em}{0ex}}-\phantom{\rule{0.147em}{0ex}}{u}^{2}\right)\phantom{\rule{0.147em}{0ex}}\end{array}$

If the object starts from its stationary position, that is, $$u = 0$$, then,

$\mathit{Work}\phantom{\rule{0.147em}{0ex}}\mathit{done}\phantom{\rule{0.147em}{0ex}}\left(W\right)\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{1}{2}\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}m\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}{v}^{2}$

It is clear that the work done is equal to the change in the kinetic energy of an object.

If $$u = 0$$, the work done will be $\frac{1}{2}\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}m\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}{v}^{2}$.

Thus, the kinetic energy possessed by an object of mass ($$m$$) and moving with a uniform velocity ($$v$$ is,

${E}_{k}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{1}{2}\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}m\phantom{\rule{0.147em}{0ex}}×\phantom{\rule{0.147em}{0ex}}{v}^{2}$