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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)
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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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