# An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 18 KJ to  4KJ over t in [0,12s]. What is the average speed of the object?

Feb 12, 2018

9.47 m/s

#### Explanation:

Based on the given values, the power loss (which is the constant here) is
$P = \frac{\setminus \Delta E}{t} = \frac{18 \setminus K J - 4 \setminus K J}{12 \setminus s} = - \frac{7}{6} \setminus k W$
so $E = {E}_{0} + P t$.

We can then find the velocity at any time based on that equation:
$E = \frac{1}{2} m {v}^{2} \to v \left(t\right) = \sqrt{2 \frac{E}{m}} = \sqrt{\frac{2 {E}_{0}}{m} + \frac{2 P}{m} t}$
From that equation, we can find its average:
${v}_{a v e} = \frac{1}{12 \setminus s} \cdot {\int}_{0}^{12 s} v \left(t\right) \mathrm{dt} = \frac{2 {\left(2 {E}_{0} / m + \left(2 \frac{P}{m}\right) \cdot \left(12 \setminus s\right)\right)}^{\frac{3}{2}}}{3 \cdot 2 \frac{P}{m}}$
${v}_{a v e} = \frac{m {v}_{f}^{3}}{3 P} = 9.47 \setminus \frac{m}{s}$

I hope the calculus is comfortable to you.