Question

On a level road, a scooterist applies brakes to slow down from a speed of 10 m/s to 5 m/s. If the mass of the scooterist and the scooter be 150 kg, calculate the work done by the brakes. (Neglect air resistance and friction)

Answer 1

Mass of the scooterist and the scooter, (m) = 150 kg
Initial velocity, (v₁) = 10 m/s
Final velocity, (v₂) = 5 m/s
So, initial kinetic energy can be calculated as,
$K.E=\frac{1}{2}m{v}^2$
Therefore,
$$\begin{gathered} (K.E{)}_1=\frac{1}{2}(150\left)\right(10{)}^2\mathrm{J} \\ =7500\mathrm{J} \end{gathered}$$
Similarly, final kinetic energy,
$$\begin{gathered} (K.E{)}_2=\frac{1}{2}(150\left)\right(5{)}^2\mathrm{J} \\ =1875\mathrm{J} \end{gathered}$$
So, Work done by the brakes = Change in kinetic energy
Therefore, work done by the brakes,
Work done by the brakes = (K.E)₂ – (K.E)₁
= (1875 – 7500) J, $=\boxed{-5625\mathrm{J}}$
Negative sign shows that the force applied by brakes is opposite to the direction of motion of the body.