Question

The mass of the cyclist together with the bike is $90kg$. Calculate the increase in kinetic energy if the speed increases from $6km/h$ to $12km/h$.

Answer 1

We know that, $1km/h=\frac{5}{18}m/s$

Here, initial velocity is $u=6km/h=6\times \frac{5}{18}=\frac{5}{3}m/s$

Final velocity is $v=12km/h=12\times \frac{5}{18}=\frac{10}{3}m/s$

Change in Kinetic energy is given by $\Delta K=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=\frac{1}{2}m(v^2-u^2)$

$\Delta K=\frac{90}{2}(\frac{100}{9}-\frac{25}{9})$

$⟹\Delta K=\frac{90}{2}\times \frac{75}{9}$

$⟹\Delta K=375J$