Question

A body slides down a fixed curved track that
is one quadrant of a circle of radius R, as in
the figure. If there is no friction and the body
starts from rest, its speed at the bottom of the
track​

Answer 1

change in K.E is equal to wd by all force ....

k2-kq=wd

1/2mv²=mgRcosQ

take theta tha is given ......

Answer 2

Answer:

√2gR.

Explanation:

Since, the work is done by a non conservative force, hence the work is 0. Then the kinetic energy will be equal to the centrifugal force along the direction of the track inside.

Hence, 1/2mv^2=mv^2/R where v is the velocity of the body and R is the radius of the circle. So, on solving v will be √2gR.