A) \[\frac{5}{19}J\]
B) \[\frac{3}{8}J\]
C) \[\frac{8}{3}J\]
D) \[\frac{19}{5}J\]
Correct Answer: C
Solution :
Work done by the force = force \[\times \] displacement or \[W=F\times s\] .....(i) But from Newtons second law, we have Force = Mass \[\times \] Acceleration ie \[F=ma\] ?..(ii) Hence, from Eqs. (i) and (ii), we get \[W=mas\] \[=m\left( \frac{{{d}^{2}}s}{d{{t}^{2}}} \right)s\] ?..(iii) \[\left( \because \,\,a=\frac{{{d}^{2}}s}{d{{t}^{2}}} \right)\] Now, we have, \[s=\frac{1}{3}{{t}^{2}}\] \[\therefore \] \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=\frac{d}{dt}\left[ \frac{d}{dt}\left( \frac{1}{3}{{t}^{2}} \right) \right]\] \[=\frac{d}{dt}\times \left( \frac{2}{3}t \right)\] \[=\frac{2}{3}\] Hence, Eq. (iii) becomes \[W=\frac{2}{3}m\,s=\frac{2}{3}m\times \frac{1}{3}{{t}^{2}}\] \[=\frac{2}{9}m{{t}^{2}}\] Given, \[m=3kg,\,t=2s\] \[\therefore \] \[W=\frac{2}{9}\times 3\times {{(2)}^{2}}=\frac{8}{3}J\]
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