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$ seconds is.....$J$
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$ seconds is.....$J$
$9$
$6$
$0$
$3 $
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$ seconds is.....$J$
$x = {(t - 3)^2}$==> $v = \frac{{dx}}{{dt}} = 2(t - 3)$
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$
Other Language