A particle of mass 0.5 kg is moving along x-axis and at time t, its velocity is V = 5 t m/s where t is time in second. Then work done by the applied force in the time interval t = 1 sec to t = 2 sec is

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20.5 J

25 J

6.25 J

18.75 J

answer is D.

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Detailed Solution

At $t = 1 \sec, V_{1} = 5 \times 1 m / s = 5 m / s$

At $t = 2 \sec, V_{2} = 5 \times 2 m / s = 10 m / s$

By work - energy theorem

Work done $= \frac{1}{2} {mV}_{2}^{2} - \frac{1}{2} {mV}_{1}^{2} = \frac{1}{2} \times 0 .5 \times (10^{2} - 5^{2}) J$

$\Rightarrow W = 18 .75 J$

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