Under the action of a force, a 2 kg body moves such that its position x as a function of time is given by ,where xis in meters and t is in seconds. The work done by the force in the first two seconds starting form rest is

JEE MainPhysicsModerateWork Power & Energy

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

Show answer and solution

x = \frac{t^{3}}{3}

\frac{dx}{dt} = \frac{3t^{2}}{3}

V = t^{2}

V(2) = 4

V(0) = 0

\Rightarrow \Delta K = \frac{1}{2}m\left( 4^{2} \right) - \frac{1}{2}m\left( 0^{2} \right)

\Delta K = 16\ J

From work energy theorem
Work done by force =

\Delta K = 16\ J

Source: Grass-G / P-06 Physics paper 14 October.docx

Scan any doubt from your notes and get a whiteboard explanation.