# If an object with a mass of 2 kg  changes speed from 3 m/s to  m/s, by how much does its kinetic energy change?

May 14, 2016

$4 J$

#### Explanation:

Recall that the formula for kinetic energy is:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \Delta {E}_{k} = \frac{1}{2} m {v}^{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where:
$\Delta {E}_{k} =$change in kinetic energy
$m =$mass (kilograms)
$v =$velocity (metres per second)

Start by breaking down $\Delta {E}_{k}$ into final kinetic energy minus the initial kinetic energy.

$\Delta {E}_{k} = \frac{1}{2} m {v}_{f}^{2} - \frac{1}{2} m {v}_{i}^{2}$

Simplify.

$\Delta {E}_{k} = \frac{1}{2} m {\left({v}_{f} - {v}_{i}\right)}^{2}$

Substitute the given values.

$\Delta {E}_{k} = \frac{1}{2} \left(2 k g\right) {\left(1 \frac{m}{s} - 3 \frac{m}{s}\right)}^{2}$

Solve.

$\Delta {E}_{k} = \frac{1}{2} \left(2 k g\right) \left(4 {m}^{2} / {s}^{2}\right)$

$\Delta {E}_{k} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{4 J} \textcolor{w h i t e}{\frac{a}{a}} |}}}$