Question

The mass of cyclists together with the bike is $90kg$. Calculate the increase in kinetic energy if the speed increase from $6.0kmph$ to $12kmph$.

Answer 1

Given:

Mass of the cyclist together with the bike

$m=90kg$

Initial Speed

$u=60kmph$

$\Rightarrow u=60\times \frac{5}{18}$ [Changing into $m{s}^{-1}$]

$\Rightarrow u=\frac{5}{3}m{s}^{-1}$

Final Speed

$v=12kmph$

$\Rightarrow v=12\times \frac{5}{18}$ [Changing into $m{s}^{-1}$]

$\Rightarrow v=\frac{10}{3}m{s}^{-1}$

Increases in Kinetic Energy($\Delta KE$);

$\Delta KE=\frac{1}{2}\times m\times (v^2-u^2)$

$\Rightarrow \Delta KE=\frac{1}{2}\times 90\times \left(\right(\frac{10}{3}{)}^2-\left(\frac{5}{3}{)}^2\right)$

$\Rightarrow \Delta KE=45\times \left(\right(\frac{100}{9})-(\frac{25}{9}\left)\right)$

$\Rightarrow \Delta KE=45\times \frac{75}{9}$

$\Rightarrow \Delta KE=5\times 75$

$\Rightarrow \Delta KE=375$

Hence, the increase in the kinetic energy of the cyclist is $375J$.