# An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 84 KJ to  12 KJ over t in [0, 6 s]. What is the average speed of the object?

Feb 27, 2017

$115.3 m {s}^{-} 1$

#### Explanation:

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

This is the equation for kinetic energy. It can be rearranged to give $v$ as the subject:

$v = \sqrt{\frac{2 {E}_{k}}{m}}$

We know the mass $m = 6$ and the energy at the beginning is ${E}_{k} = 84 k J = 84000 J$, so

$v = \sqrt{\frac{2 \cdot 84000}{6}} = \sqrt{28000} = 167.332 m {s}^{-} 1$

We know that the energy at the end is $12 k J = 12000 J$, so

$v = \sqrt{\frac{2 \cdot 12000}{6}} = \sqrt{4000} = 63.25 m {s}^{-} 1$

Now we have the initial and the final velocities.

Work out the average (of anything) by adding them all together and dividing by the total number of things. In this case, we only have two velocities, so:

$\overline{v} = \frac{167.332 + 63.25}{2}$

$= 115.291 m {s}^{-} 1$