Q34 of 146 Page 145

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)

Mass of scooter rider and the scooter, m = 150 kg

Initial velocity (v1) = 10 m/s


Final velocity (v2) = 5 m/s


K.E = 1/2 mv2


Initial kinetic energy = 1/2 mv12


= 1/2 x 150 x (10)2


= 7500J


Final kinetic energy = 1/2 mv22


= 1/2 x 150 x (5)2


= 1875 J


Work done by the brakes = change in kinetic energy


= (K.E) 2 - (K.E) 1


= 1875 - 7500


= - 5625J


(The negative sign indicates that the force applied by the breaks is opposite to the direction of motion of the body)


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