The work done by the brakes of a car of mass $1000 k g$ when its speed is reduced from $20 m / s$ to $10 m / s$

100 k J

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120 k J

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150 k J

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180 k J

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Solution

The correct option is C $150 k J$

Given,
$m = 1000 k g, v_{f} = 10 m / s, v_{i} = 20 m / s$

We know,

Work done= $Δ K E$

Work done $= \frac{1}{2} \times 1000 \times (20^{2} - 10^{2}) = 500 \times 300 = 150000 = 150 k J$

Option $C$ is the correct answer

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