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WB JEE Medical WB JEE Medical Solved Paper-2011

  • question_answer
    A body of mass 6 kg is acted upon by a force which causes a displacement in it given by \[x=\frac{{{t}^{2}}}{4}\] metre, where t is the time in second. 4 The work done by the force in 2 s is

    A)  12J                                        

    B)  9J

    C)  6J                                          

    D)  3J

    Correct Answer: D

    Solution :

                     From the question, we see that the mass of the body is 6 kg and the displacement is t2 given by \[x=\frac{{{t}^{2}}}{4}.\]So, the velocity would be equal to \[v=\frac{dx}{dt}=\frac{t}{2}.\]Thus, v(0) = 0 and \[v(2)=\frac{2}{2}=1.\] Hence the initial kinetic energy would be equal to \[K{{E}_{ini}}=\frac{1}{2}m{{[v(0)]}^{2}}=0\]and similarly the final kinetic energy would be equal to \[K{{E}_{fin}}=\frac{1}{2}m{{[v(2)]}^{2}}=\frac{1}{2}\times 6\times {{1}^{2}}=3.\]So, we see that the work done by the force is equal to \[W=K{{E}_{fin}}-K{{E}_{ini}}=3-0=3J.\]


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