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Q.

Position (x) of a particle of mass 2kg varies with time as $x = \frac{t^{2}}{3}$ . (‘x’ in meter and ‘t’ in seconds) the work done by the net force acting on the particle in first two seconds is

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a

$\frac{4}{3} J$

b

$\frac{4}{9} J$

c

$\frac{2}{3} J$

d

$\frac{16}{9} J$

answer is D.

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Detailed Solution

Work done by net force = $K_{f} - K_{i}$ (WE theorem)

$x = \frac{t^{2}}{3} \Rightarrow V = \frac{d x}{d t} = \frac{2 t}{3}$
at $t = 0 V_{i} = 0$
at t = 2sec $V_{f} = \frac{2}{3} \times 2 = \frac{4}{3}$

$W = K_{f} - K_{i} = \frac{1}{2} \times 2 \times {(\frac{4}{3})}^{2}$

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Position (x) of a particle of mass 2kg varies with time as x=t23. (‘x’ in meter and ‘t’ in seconds) the work done by the net force acting on the particle in first two seconds is