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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)
Solution
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 = 1/2 mv^2`
Therefore ,
`(K.E)_1 = 1/2(150)(10)^2` J
= 7500 J
Similarly , final kinetic energy ,
`(K.E)_2 = 1/2(150)(5)^2` J
= 1875 J
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 , = -5625 J
Negative sign shows that the force applied by brakes is opposite to the direction of motion of the body.
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