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 \[\times \] Displacement |
| or \[W=F\times s\] (i) |
| But from Newtons 2nd law, we have |
| Force = Mass \[\times \] 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 |
| \[m=3\,kg,\,\,t=\,2s\] |
| \[\therefore \] \[W=\frac{2}{9}\times 3\times {{(2)}^{2}}=\frac{8}{3}\,J\] |
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