The displacement x of a particle of mass m kg moving in one dimension, under the action of a 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 by the force in the first six seconds, in joules, is

t=x+3⇒x=t2-6t+9 v=dxdt=2t-6 At t=0,v1=-6 m/s ⇒ E1=12m(6)2=18mJ At t=6 s,v2=6 m/s ⇒E2=12m(6)2=18mJ Work done =E2-E1=0

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The displacement x of a particle of mass m kg moving in one dimension, under the action of a 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 by the force in the first six seconds, in joules, is

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Detailed Solution

$t = \sqrt{x} + 3 \Rightarrow x = t^{2} - 6 t + 9 v = \frac{dx}{dt} = 2 t - 6 At t = 0, v_{1} = - 6 m / s \Rightarrow E_{1} = \frac{1}{2} m (6)^{2} = 18 mJ At t = 6 s, v_{2} = 6 m / s \Rightarrow E_{2} = \frac{1}{2} m (6)^{2} = 18 mJ$

$Work done = E_{2} - E_{1} = 0$

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