An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 72 KJ to  120 KJ over t in [0, 3 s]. What is the average speed of the object?

Mar 16, 2016

${v}_{a} = \frac{8}{3} \frac{m}{s}$

Explanation:

$\Delta {E}_{k} = {E}_{\text{k l"-E_"k i"" " "changing of the kinetic energy of object }}$
$\frac{1}{2} m \left({v}_{l}^{2} - {v}_{i}^{2}\right) = 120 - 72$
$\frac{1}{2} \cdot \cancel{6} \left({v}_{l}^{2} - {v}_{i}^{2}\right) = \cancel{48}$
$\left({v}_{l}^{2} - {v}_{i}^{2}\right) = 16$
$\left({v}_{l} - {v}_{i}\right) \left({v}_{l} + {v}_{i}\right) = 16$
${v}_{l} - {v}_{i} = 2$
${v}_{l} + {v}_{i} = 8$
$2 \cdot {v}_{l} = 10 \text{ } {v}_{l} = 5 \frac{m}{s}$
$5 - {v}_{i} = 2$
${v}_{i} = 3 \frac{m}{2}$
$a = \frac{\Delta v}{\Delta t} = \frac{{v}_{l} - {v}_{i}}{\Delta t} = \frac{5 - 3}{3 - 0} = 1 \frac{m}{s} ^ 2 \text{ (acceleration of object)}$
${v}_{l}^{2} = {v}_{i}^{2} + 2 \cdot a \cdot \Delta x$
${5}^{2} = {3}^{2} + 2 \cdot 1 \cdot \Delta x$
$25 = 9 + 2 \cdot \Delta x$
$16 = 2 \cdot \Delta x$
$\Delta x = 8 m \text{ displacement for (0-3)}$
$\text{Average velocity of an object is given by:}$
${v}_{a} = \left(\text{Total displacement")/("Total time}\right)$
${v}_{a} = \frac{8}{3} \frac{m}{s}$