• # 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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