under the action of a force, a 2 kg body moves such that its position x as a function of time is given by x= t33 , where x is in metre and t is in second. The work done by the force in the first two seconds is

displacement x=t33,differentiate with respect to time to get Velocity=dxdt=t2Work done=change in KEW=12mv2substitute v=t2work done=12m(t2)2work done=12×2×34work done=81J

under the action of a force, a 2 kg body moves such that its position x as a function of time is given by x= t33 , where x is in metre and t is in second. The work done by the force in the first two seconds is

displacement x=t33,differentiate with respect to time to get Velocity=dxdt=t2Work done=change in KEW=12mv2substitute v=t2work done=12m(t2)2work done=12×2×34work done=81J

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under the action of a force, a 2 kg body moves such that its position x as a function of time is given by x= $\frac{t^{3}}{3}$ , where x is in metre and t is in second. The work done by the force in the first two seconds is

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160 J

81 J

1600 J

8.1 J

answer is C.

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Detailed Solution

$\begin{matrix} d i s p l a c e m e n t x = \frac{t^{3}}{3}, d i f f e r e n t i a t e w i t h r e s p e c t t o t i m e t o g e t V e l o c i t y = \frac{d x}{d t} = t^{2} \\ W o r k d o n e = c h a n g e i n K E \\ W = \frac{1}{2} m v^{2} \\ s u b s t i t u t e v = t^{2} \\ w o r k d o n e = \frac{1}{2} m {(t^{2})}^{2} \\ w o r k d o n e = \frac{1}{2} \times 2 \times 3^{4} \\ w o r k d o n e = 81 J \end{matrix}$