• question_answer
                    A force acts on a 3.0 g particle in such a way that the position of the particle as a function of time is given by \[x=3t-4{{t}^{2}}+{{t}^{3}}\], where \[x\] is in metre and \[t\] in second. The work done during the first 4 s is:

    A)                                                                                                                                                                                 570 mJ                 

    B)                 450 mJ

    C)                                            490 mJ

    D)                           528 mJ

    Correct Answer: D

    Solution :

                    Key Idea: Work done during the first 4s is equal to gain in kinetic energy.                 We have given,                 \[x=3t-4{{t}^{2}}+{{t}^{3}}\]                 So, velocity                 \[v=\frac{dx}{dt}=3-8\,t+3\,{{t}^{2}}\]                 At t = 0,    \[{{v}_{1}}=3-0+0=3\,m/s\]                 At t = 4 s, \[{{v}_{2}}=3-8\times 4+3\times {{4}^{2}}\]                 \[=3-32+48=19\,m/s\]                 Now work done during t = 0 and t = 4s                 = gain in kinetic energy                 \[=\frac{1}{2}mv_{2}^{2}-\frac{1}{2}mv_{1}^{2}=\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})\]                 \[=\frac{1}{2}\times 3\times {{10}^{-3}}\,[{{(19)}^{2}}-{{(3)}^{2}}]\]                 \[[\text{Using}\,\,{{a}^{2}}-{{b}^{2}}=(a+b)\,(a-b)]\]                 \[=1.5\times {{10}^{-3}}\times [\,(19+3)\,(19-3)]\]                 \[=1.5\times {{10}^{-3}}\times 22\times 16\]                 \[=528\times {{10}^{-3}}\,J\]                 \[=528\,mJ\]

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