The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation $$t=x+3$$, where x is in meters and t is in seconds. The work done (in$$ $$J) by the force in the first $$6$$ seconds is

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The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation $t = \sqrt{x} + 3$ , where x is in meters and t is in seconds. The work done (in J) by the force in the first 6 seconds is

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Solution

$x = {(t - 3)}^{2} \Rightarrow v = \frac{dx}{dt} = 2 (t - 3)$
at $t = 0$ ; $v_{1} = - 6 m / s$ and at $t = 6 \sec, v_{2} = 6 m / s$
so, change in kinetic energy $= W = \frac{1}{2} {mv}_{2}^{2} - \frac{1}{2} {mv}_{1}^{2} = 0$

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