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 by the force in the first $6$ second is

According to the question,

$x =( t-3)^2 \Rightarrow v = \dfrac {dx}{dt} = 2(t -3)$

at $t = 0 ; v_1 = -6 m/s$ and at $t =6 sec , v_2 = 6 m/s$

So change in kinetic energy $= W = \dfrac {1}{2} mv^2_2 - \dfrac {1}{2} mv^2_1 = 0 J$