A) \[\frac{m{{k}^{4}}}{{{t}^{2}}}\]
B) \[\frac{m{{k}^{4}}{{t}^{2}}}{4}\]
C) \[\frac{m{{k}^{4}}{{t}^{2}}}{8}\]
D) \[\frac{m{{k}^{4}}{{t}^{2}}}{16}\]
Correct Answer: C
Solution :
[c] Given \[v=k\sqrt{x}\] Or \[\frac{dx}{dt}=k\sqrt{x}\]or \[{{x}^{\frac{1}{2}}}dx=k\,dt\] Integrating both sides, we get \[\frac{{{x}^{\frac{1}{2}}}}{\frac{1}{2}}=kt+C;\]Assuming \[x(0)=0\] Therefore, \[C=0\] \[2\sqrt{x}=kt\Rightarrow x=\frac{{{k}^{2}}{{t}^{2}}}{4}\]or \[v=\frac{{{k}^{2}}t}{2}\] Therefore, work done, \[\Delta W=\]Increase in KE \[=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{(0)}^{2}}=\frac{1}{2}m{{\left[ \frac{{{k}^{2}}t}{2} \right]}^{2}}=\frac{1}{8}m{{k}^{4}}{{t}^{2}}\]
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