WB JEE Medical WB JEE Medical Solved Paper-2008

  • question_answer
    A body of mass 3 kg acted upon by a constant force is displaced by S metre, given by  relation \[S=\frac{1}{3}{{t}^{2}},\]where t is in second. Work done by the  force in 2 is

    A)  \[\frac{8}{3}J\]                                

    B)  \[\frac{19}{5}J\]

    C)  \[\frac{5}{19}J\]                              

    D)  \[\frac{3}{8}\,J\]

    Correct Answer: A

    Solution :

                     Given that, \[S=\frac{1}{3}{{t}^{2}}\] \[v=\frac{dS}{dt}=\frac{2}{3};a=\frac{{{d}^{2}}S}{d{{t}^{2}}}=\frac{2}{3}\] \[F=ma=3\times \frac{2}{3}=2N;\]Work \[=2\times \frac{1}{3}{{t}^{2}}\] At t = 2 Work \[=2\times \frac{1}{3}\times 2\times 2=\frac{8}{3}J\]


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