# An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 225 KJ to 150 KJ over t in [0, 8 s]. What is the average speed of the object?

Mar 13, 2018

The average speed is $= 249.6 m {s}^{-} 1$

#### Explanation:

The kinetic energy is

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

The mass is $m = 6 k g$

The initial velocity is $= {u}_{1} m {s}^{-} 1$

The final velocity is $= {u}_{2} m {s}^{-} 1$

The initial kinetic energy is $\frac{1}{2} m {u}_{1}^{2} = 225000 J$

The final kinetic energy is $\frac{1}{2} m {u}_{2}^{2} = 150000 J$

Therefore,

${u}_{1}^{2} = \frac{2}{6} \cdot 225000 = 75000 {m}^{2} {s}^{-} 2$

and,

${u}_{2}^{2} = \frac{2}{6} \cdot 150000 = 50000 {m}^{2} {s}^{-} 2$

The graph of ${v}^{2} = f \left(t\right)$ is a straight line

The points are $\left(0 , 75000\right)$ and $\left(8 , 50000\right)$

The equation of the line is

${v}^{2} - 75000 = \frac{50000 - 75000}{8} t$

${v}^{2} = - 3125 t + 75000$

So,

$v = \sqrt{- 3125 t + 75000}$

We need to calculate the average value of $v$ over $t \in \left[0 , 8\right]$

$\left(8 - 0\right) \overline{v} = {\int}_{0}^{8} \left(\sqrt{- 3125 t + 75000}\right) \mathrm{dt}$

$8 \overline{v} = {\left[{\left(- 3125 t + 75000\right)}^{\frac{3}{2}} / \left(- \frac{3}{2} \cdot 3125\right)\right]}_{0}^{8}$

$= \left({\left(- 3125 \cdot 8 + 75000\right)}^{\frac{3}{2}} / \left(- 4687.5\right)\right) - \left({\left(- 3125 \cdot 0 + 75000\right)}^{\frac{3}{2}} / \left(- 4687.5\right)\right)$

$= {75000}^{\frac{3}{2}} / 4687.5 - {50000}^{\frac{3}{2}} / 4687.5$

$= 1996.6$

So,

$\overline{v} = \frac{1996.6}{8} = 249.6 m {s}^{-} 1$

The average speed is $= 249.6 m {s}^{-} 1$

Mar 14, 2018

$248.73 \frac{m}{s}$

#### Explanation:

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

$v = \sqrt{\frac{2 K E}{m}}$

So, v initial =(sqrt((2*225,000J)/(6 kg))=273.86 m/s

and v final=(sqrt((2*150,000J)/(6 kg))=223.607 m/s

$v a v g = \left(\frac{v f + v i}{2}\right) = \left(\frac{273.86 + 223.607 \frac{m}{s}}{2}\right) = 248.73 \frac{m}{s}$