Question #87092

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Question #87092

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Answer 1 · 2018-02-01T17:51:15

I suspect you mean kinetic energy, not kinetic speed.
$K E = \frac{1}{2} m v^{2}$ $KE = 1/2 mv^2$

Explanation:

The equation for kinetic energy is $K E = \frac{1}{2} m v^{2}$ $KE = 1/2 mv^2$ .

So, we can plug in the values given to find the $K E$ $KE$ .

$K E = \frac{1}{2} (100 k g) {(12.5 \frac{m}{s})}^{2}$ $KE = 1/2(100 kg)(12.5m/s)^2$
$K E = \frac{1}{2} (100 k g) (156.25 \frac{m^{2}}{s^{2}})$ $KE=1/2(100 kg)(156.25m^2/s^2)$
$K E = 7812.5 k g \frac{m^{2}}{s^{2}}$ $KE = 7812.5 kgm^2/s^2$
$K E = 7812.5 J$ $KE = 7812.5 J$

If significant figures are considered, record answer as $7810 J$ $7810J$ because the speed had only three significant figures

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Question #87092

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