# Question #5b15a

Feb 12, 2017

The kinetic energy increases by a factor of $4$.

#### Explanation:

As you know, the kinetic energy of an object is given by the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{K}_{\text{E}} = \frac{1}{2} \cdot m \cdot {v}^{2}}}}$

Here

• $m$ is the mass of the object
• $v$ is its velocity

This equation shows that the kinetic energy of an object varies with the square of its velocity.

Let's say that you have

${K}_{\text{E 1}} = \frac{1}{2} \cdot m \cdot {v}_{1}^{2}$

If you double the object's velocity

${v}_{2} = 2 \cdot {v}_{1}$

then its kinetic energy will increase by a factor of $4$, since

${K}_{\text{E 2}} = \frac{1}{2} \cdot m \cdot {v}_{2}^{2}$

${K}_{\text{E 2}} = \frac{1}{2} \cdot m \cdot {\left(2 \cdot {v}_{1}\right)}^{2}$

${K}_{\text{E 2}} = \frac{1}{2} \cdot m \cdot {v}_{1}^{2} \cdot {2}^{2}$

But since

${K}_{\text{E 1}} = \frac{1}{2} \cdot m \cdot {v}_{1}^{2}$

you can say that

${K}_{\text{E 2" = K_"E 1}} \cdot {2}^{2}$

which is equivalent to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{K}_{\text{E 2" = 4 * K_"E 1}}}}}$