The displacement $x$ in metre of a particle of mass $m$ kg moving in one dimension under the action of a force is related to the time $t$ in second by the equation $t = \sqrt{x} + 3$, the work done by the force (in joule) in first six seconds is

Question

Dr. Abubacker · Accepted Answer

(b) Here $t = \sqrt{x} + 3$
or $x = (t - 3)^{2} = t^{2} - 6t + 9$
$v = \frac{dx}{dt} = 2t - 6$
At $t = 0\ s,v = 2 	imes 0 - 6 = - 6$
At $t = 6\ s,v = 2 	imes 6 - 6 = + 6$
Initial and final KE are same hence no work is done
$W = \frac{1}{2}m\left( v_{1}^{2} - v_{2}^{2} ight) = 0$

Q. The displacement $x$ in metre of a particle of mass $m$ kg moving in one dimension under the action of a force is related to the time $t$ in second by the equation $t = \sqrt{x} + 3$, the work done by the force (in joule) in first six seconds is

Explanation

Reviewed by Dr. Abubacker

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