A) \[mkr{{t}_{0}}\]
B) \[\frac{mkr{{t}_{0}}^{2}}{2}\]
C) \[\frac{mkr{{t}_{0}}^{{}}}{2}\]
D) \[\frac{mkr{{t}_{0}}^{{}}}{4}\]
Correct Answer: B
Solution :
[b] Given \[{{a}_{n}}=k{{t}^{2}}\] Or \[\frac{{{v}^{2}}}{r}=k{{t}^{2}}\] or \[{{v}^{2}}=kr{{t}^{2}}\] Therefore, average power delivered =\[\frac{Total\,work\,done}{Total\,time\,elaped}\] \[=\frac{increase\,in\,KE}{Total\,time\,elapsed}\] Or \[ =\frac{\frac{1}{2}m({{v}^{2}}-{{0}^{2}})}{{{t}_{0}}}=\frac{m}{2}\frac{kr{{t}_{0}}^{2}}{{{t}_{0}}}=\frac{mkr{{t}_{0}}^{2}}{2}\] Alternative: \[v=\sqrt{krt}\Rightarrow {{a}_{t}}=\sqrt{kr}\] \[P={{F}_{t}}v=m{{a}_{t}}v=mkrt\] \[ =\frac{\int\limits_{0}^{{{t}_{0}}}{Pdt}}{{{t}_{0}}}=\frac{1}{2}mk\,r{{t}_{0}}^{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
