A) \[2\pi m{{k}^{2}}{{r}^{2}}t\]
B) \[m{{k}^{2}}{{r}^{2}}t\]
C) \[\frac{\left( m{{k}^{4}}{{r}^{2}}{{t}^{5}} \right)}{3}\]
D) zero
Correct Answer: B
Solution :
[b] The centripetal acceleration \[{{a}_{c}}={{k}^{2}}\,r\,{{t}^{2}}or\,\frac{{{v}^{2}}}{r}={{k}^{2}}r{{t}^{2}}\,\,\,\,\therefore v=krt\] So, tangential acceleration, \[{{\operatorname{a}}_{t}} = \frac{dv}{dt} = kr\] Work is done by tangential force. Power \[= {{F}_{i}}.v.cos 0{}^\circ = \left( m{{a}_{t}} \right)\left( krt \right) = \left( mkr \right)\left( krt \right)\] \[=m{{k}^{2}}{{r}^{2}}t\]
You need to login to perform this action.
You will be redirected in
3 sec
