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 in seconds- The work done by the force in the first two seconds is -

The correct option is A 16J v=dx/dt=t^2 W=1/2mv^2=1/2mt^4 =1/2× 2× (2)^4=16J ...

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Standard XII

Physics

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Under the act...

Question

Under the action of a force, a $2 k g$ 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 $t$ in seconds. The work done by the force in the first two seconds is :

1600 J

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160 J

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16 J

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1.6 J

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Solution

The correct option is A $16 J$
$v = \frac{d x}{d t} = t^{2}$
$W = \frac{1}{2} m v^{2} = \frac{1}{2} m t^{4}$
$= \frac{1}{2} \times 2 \times (2)^{4} = 16 J$

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

The correct option is A 16Jv=dxdt=t2W=12mv2=12mt4=12×2×(2)4=16J

Under the action of a force, a $2 k g$ 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 $t$ in seconds. The work done by the force in the first two seconds is :

The correct option is A $16 J$
$v = \frac{d x}{d t} = t^{2}$
$W = \frac{1}{2} m v^{2} = \frac{1}{2} m t^{4}$
$= \frac{1}{2} \times 2 \times (2)^{4} = 16 J$