Question:
The mass of cyclist together with the bike is $90 \mathrm{~kg}$. Calculate the increase in kinetic energy if the speed increases from 6 ' $0 \mathrm{~km} / \mathrm{h}$ to $12 \mathrm{~km} / \mathrm{h}$.
Solution:
$K . E_{x}=\frac{1}{2} m v^{2}$
Change in K.E. $=\frac{1}{2} \mathrm{~m}\left[\mathrm{v}^{2}-\mathrm{u}^{2}\right]$
$=\frac{1}{2} \times 90 \times\left[12^{2}-6^{2}\right]$
$=375 \mathrm{~J}$
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