A) \[\frac{1}{2}\frac{m{{v}^{2}}}{{{t}_{1}}^{2}}{{t}^{2}}\]
B) \[\frac{1}{2}{{\left( \frac{mv}{{{t}_{1}}} \right)}^{2}}{{t}^{2}}\]
C) \[\frac{m{{v}^{2}}}{{{t}_{1}}^{2}}{{t}^{2}}\]
D) \[\frac{2m{{v}^{2}}}{{{t}_{1}}^{2}}{{t}^{2}}\]
Correct Answer: A
Solution :
Velocity of particle accelerating uniformly in time \[{{t}_{1}}\], \[v=a{{t}_{1}}\] \[\Rightarrow \] \[a=\frac{v}{{{t}_{1}}}\] Velocity of particle in time t, \[v=at=\frac{vt}{{{t}_{1}}}\] According to work-energy theorem, work done = change in kinetic energy i.e., \[{{W}_{ext}}=\Delta K\] or \[{{W}_{ext}}=\frac{1}{2}mv{{}^{2}}-0\] \[=\frac{1}{2}\,m{{\left( \frac{vt}{{{t}_{1}}} \right)}^{2}}\] \[=\frac{1}{2}\,\frac{m{{v}^{2}}}{t_{1}^{2}}{{t}^{2}}\]
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