Email
Chat with tutors
logo

Ask Questions, Get Answers

X
 
Questions  >>  Archives  >>  JEEMAIN-2015
Answer
Comment
Share
Q)

A particle is moving in a circle of radius r under the action of a force $F= \alpha r^2$ which is directed towards center of the circle. Total mechanical energy (kinetic energy + potential energy ) of the particle is (take potential energy =0 for r=0):

$\begin{array}{1 1} \alpha r^3 \\ \frac{1}{2} \alpha r^3 \\ \frac{4}{3} \alpha r^3 \\ \frac{5}{6} \alpha r^3\end{array} $

1 Answer

Home Ask Tuition Questions
Your payment for is successful.
Continue
...