How to Calculate Kinetic Energy

In this article, we will look at how to calculate kinetic energy. Kinetic Energy is the energy that an object has due to its motion and it depends on both the object’s speed and mass. The direction of the body’s motion has no effect on kinetic energy. For a moving body, kinetic energy is defined as the net work that needs to be done in order to accelerate the body to its speed from rest. Let us assume that an object is accelerated from rest by a constant net force. In this situation, the acceleration is also constant and we can use our ‘suvat’ equations of motion.

If the kinetic energy of the body after acceleration is $K$, we can find it by

$K=W=Fs=mas$

where $F$ is the size of the constant force, $m$ is the mass of the object, $a$ isthe constant acceleration and $s$ is the displacement.

From the equation $v^2=u^2+2as$, we have $as=\frac{v^2-u^2}{2}$. Since our initial velocity is 0, then we have $as=\frac{v^2}{2}$. Then,

$K=\frac{1}{2}mv^2$

This defines the object’s kinetic energy.

Suppose the object was not at rest initially. Then, the net work done:

$W=m\frac{v^2-u^2}{2}$.

I.e., the work done is equal to the final kinetic energy – the initial kinetic energy, or the net work done on the object is equal to the change in the object’s kinetic energy.

$W=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

But, what if the force was not constant? For this case, we need to use calculus. We use the , with $W$ as our net work done on the body, and $F$ as our net force:

$W=\int_{x_1}^{x_2}F(x)dx$

Now,

$W=\int_{x_1}^{x_2}Fdx=m\int_{x_1}^{x_2}adx=m\int_{x_1}^{x_2}\frac{\mathrm{d}v}{\mathrm{d}t}dx$.

Applying chain rule,

$W=m\int_{x_1}^{x_2}\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}dx=m\int_{v_1}^{v_2}\frac{\mathrm{d}x}{\mathrm{d}t}dv=m\int_{v_1}^{v_2}vdv=\frac{m}{2}{\left[ v^2\right]}_{v_1}^{v_2}$

Then we get,

$W=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$

If we again think of the case where the object’s initial speed was 0, we can define kinetic energy to be $K=\frac{1}{2}mv^2$ when the speed of the object is $v$. We also end up with the situation where the work done has been used to change the kinetic energy of the body.

The result $W=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$ is often referred to as the work-kinetic energy theorem. This states that the net work done on an object is equal to the object’s change in kinetic energy. Note that if the net work done on the body $W<0$, then the speed of the object reduces. In this case, the net work is done by the object.

Kinetic energy is a scalar quantity: since there is a $v^2$ term, the sign of the velocity does not matter to the kinetic energy. Like work, kinetic energy is also measured in joules (J).

How to Calculate Kinetic Energy – Examples

Example 1

Find the kinetic energy of a horse and a rider, having masses 450 kg and 70 kg respectively, moving at a speed of 18 m s-1.

$K=\frac{1}{2}mv^2=\frac{1}{2}\times (450+70)\times 18^2=84\mathrm{\:kJ}$

The horse and the rider in this example have about 84 kJ of kinetic energy

Example 2

An object of mass 20 kg is pulled forward by a constant force of 300 N while a constant resistive force of 400 N acts on it in the opposite direction. If the object is traveling at a speed of 15 m s-1  forwards at a certain time, find how much the object’s kinetic energy would be after it travels a further 2 m.

The resultant force is $\sum F=300-400=-100\mathrm{\:N}$. The net work done is then $W=Fs=-100\times 2=-200\mathrm{\:N}$.

From the work-kinetic energy theorem,

$W=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2\Rightarrow v^2-u^2=\frac{2W}{m}$.

Then,

$v=\sqrt{u^2+\frac{2W}{m}}=\sqrt{15^2+\frac{2\times (-200)}{20}}=\sqrt{225-20}=\sqrt{205}=14\mathrm{\:m\:s^{-1}}$

This is expected: since the net work done is in the opposite direction to the object’s motion, we should expect the kinetic energy to decrease.

Example 3

Show that for an object with momentum $p$, its kinetic energy $K$ could be given by $K=\frac{p^2}{2m}$

$K=\frac{1}{2}mv^2=\frac{1}{2m}m^2v^2=\frac{p^2}{2m}$

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