JEE Main & Advanced Physics Work, Energy, Power & Collision Question Bank Work Done by Variable Force

  • 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

    Correct Answer: D

    Solution :

                    \[S=\frac{{{t}^{3}}}{3}\]\ \[dS={{t}^{2}}dt\]             \[a=\frac{{{d}^{2}}S}{d{{t}^{2}}}=\frac{{{d}^{2}}}{d{{t}^{2}}}\left[ \frac{{{t}^{3}}}{3} \right]=2t\ m/{{s}^{2}}\]             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


You need to login to perform this action.
You will be redirected in 3 sec spinner