Under the action of force a 2 kg body moves such that its position x as a function of time ‘t’ as given x=t33,x is in metre and t in sec calculate work done by force in first 2 second

x=t33 v=dxdt=t2At t=0, vi=0, at t=2, vf=4 m/s ω=12m(vf2−vi2) =12×2×(42−02)=16J

Under the action of force a 2 kg body moves such that its position x as a function of time ‘t’ as given x=t33,x is in metre and t in sec calculate work done by force in first 2 second

x=t33 v=dxdt=t2At t=0, vi=0, at t=2, vf=4 m/s ω=12m(vf2−vi2) =12×2×(42−02)=16J

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Under the action of force a 2 kg body moves such that its position $x$ as a function of time ‘t’ as given $x = \frac{t^{3}}{3}, x$ is in metre and t in sec calculate work done by force in first 2 second

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10 J

0 J

16 J

20 J

answer is C.

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Detailed Solution

$x = \frac{t^{3}}{3}$
$v = \frac{d x}{d t} = t^{2}$
At t=0, v_i=0, at t=2, v_f=4 m/s

$ω = \frac{1}{2} m (v_{f}^{2} - {v_{i}}^{2})$
$= \frac{1}{2} \times 2 \times (4^{2} - 0^{2})$
$= 16 J$

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Under the action of force a 2 kg body moves such that its position x as a function of time ‘t’ as given x=t33,x is in metre and t in sec calculate work done by force in first 2 second

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Under the action of force a 2 kg body moves such that its position $x$ as a function of time ‘t’ as given $x = \frac{t^{3}}{3}, x$ is in metre and t in sec calculate work done by force in first 2 second