# What is the kinetic energy of an object with a mass of  16 kg that has been in freefall for  2 s?

May 15, 2016

$3000 J$

#### Explanation:

Recall that the formula for kinetic energy is:

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

where:
$\Delta {E}_{k} =$change in kinetic energy
$m =$mass
$\Delta v =$change in velocity

Step 1
In order to use the kinetic energy formula, you first need to determine the initial velocity, which can be found by using the rearranging the formula for acceleration.

$a = \frac{\Delta v}{\Delta t}$

$a = \frac{{v}_{f} - {v}_{i}}{\Delta t}$

$a \Delta t = {v}_{f} - {v}_{i}$

${v}_{i} = {v}_{f} - a \Delta t$

Step 2
Substitute ${v}_{i} = {v}_{f} - a \Delta t$ into the kinetic energy formula.

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

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

$\Delta {E}_{k} = \frac{1}{2} m {\left({v}_{f} - \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left({v}_{f} - a \Delta t\right)}\right)}^{2}$

$\Delta {E}_{k} = \frac{1}{2} m {\left(a \Delta t\right)}^{2}$

Plug in the values.

$\Delta {E}_{k} = \frac{1}{2} \left(16 k g\right) {\left(\left(9.81 \frac{m}{s} ^ 2\right) \left(2 s\right)\right)}^{2}$

Solve.

$\Delta {E}_{k} = \frac{1}{2} \left(16 k g\right) \left(384.9444 {m}^{2} / {s}^{2}\right)$

$\Delta {E}_{k} = 3079.5552 J$

$\Delta {E}_{k} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{3000 J} \textcolor{w h i t e}{\frac{a}{a}} |}}}$