# What is the kinetic energy of an object with a mass of 2 kg that has been in freefall for 4 s?

Dec 25, 2016

The kinetic energy will be 1537 J.

#### Explanation:

After $4 s$ of freefall, the speed of a body will be

$v = \text{g} t = 9.8 \left(4\right) = 39.2 \frac{m}{s}$

Its kinetic energy will be

$K E = \frac{1}{2} m {v}^{2} = \frac{1}{2} \left(2\right) \left({39.2}^{2}\right) = 1537 J$

You can also find this by conservation of energy methods

During the 4s it will fall a distance of

$y = \frac{1}{2} \text{g} {t}^{2} = \frac{1}{2} \left(9.8\right) \left({4}^{2}\right) = 78.4 m$

The decrease in potential energy of the body will be
$D e l a t U = m g y = \left(2\right) \left(9.8\right) \left(78.4\right) = 1537 J$

This decrease in potential energy will equal its increase in kinetic energy.