A) \[\frac{5}{19}\,J\]
B) \[\frac{3}{18}\,J\]
C) \[\frac{8}{3}\,J\]
D) \[\frac{19}{5}\,J\]
Correct Answer: C
Solution :
Key Idea: If a constant force is applied on the object causing a displacement in it, then it is said that work has been done on the body to displace it. Work done by the force = Force × Displacement or \[W=F\times s....(i)\] But from Newtno?s 2nd law, we have Force = Mass × Acceleration i.e., 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, \[s=\frac{1}{3}{{t}^{2}}\] \[\therefore \frac{{{d}^{2}}s}{d{{t}^{2}}}=\frac{d}{dt}\left[ \frac{d}{dt}\left( \frac{1}{3}{{t}^{2}} \right) \right]\] \[=\frac{d}{dt}\times \left( \frac{2}{3}t \right)\] \[=\frac{2}{3}\,\frac{dt}{dt}\] \[=\frac{2}{3}\] Hence, Eq. (iii) becomes \[W=\frac{2}{3}ms=\frac{2}{3}m\times \frac{1}{3}{{t}^{2}}\] \[=\frac{2}{9}\,m{{t}^{2}}\] We have given m = 3 kg, t = 2 s \[\therefore W=\frac{2}{9}\times 3\times {{(2)}^{2}}=\frac{8}{3}\,J\]
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