• # question_answer A body of mass 3 kg is under a force, which causes a displacement in it is given by $S=\frac{{{t}^{3}}}{3}$ (in m). Find the work done by the force in first 2 seconds       [BHU 1998] A)             2 J       B)             3.8 J C)             5.2 J    D)             24 J

$S=\frac{{{t}^{3}}}{3}$\ $dS={{t}^{2}}dt$             $a=\frac{{{d}^{2}}S}{d{{t}^{2}}}=\frac{{{d}^{2}}}{d{{t}^{2}}}\left[ \frac{{{t}^{3}}}{3} \right]=2t\ m/{{s}^{2}}$             Now work done by the force $W=\int\limits_{0}^{2}{F.dS}=\int\limits_{0}^{2}{ma.dS}$             $\int\limits_{0}^{2}{3\times 2t\times {{t}^{2}}dt}$$=\int\limits_{0}^{2}{6{{t}^{3}}dt}$= $\frac{3}{2}\left[ {{t}^{4}} \right]_{0}^{2}$= 24 J