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

Question

Dr. Abubacker · Accepted Answer

(b) $v = \frac{dx}{dt} = \frac{d}{dt}\left( \frac{t^{3}}{3} ight) = t^{2}$
When $t = 0$, then $v = 0$, when $t = 2$, then $v = 4\ m/s$
Work done in first two second = change in KE
$W = \frac{1}{2}m\left\lbrack (4)^{2} - (0)^{2} ightbrack = \frac{1}{2} 	imes 2 	imes 16 = 16\ J$

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

Explanation

Reviewed by Dr. Abubacker

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