Question

An object of mass m moves with constant speed in a circular path of radius R under the action of a force of constant magnitude F. The K.E of object is

Answer 1

When an objects moves in a circular path with constant velocity, the centrifugal force is equal to the centripetal force. That is

$F=\frac{mv^2}{R} \\ mv^2=FR\\ \frac{1}{2}mv^2=\frac{1}{2}FR\\ KE=\frac{1}{2}FR$

Since the KE of a moving object with velocity v is $\frac{1}{2}mv^2$.

Answer 2

$\underline{\large\bf{\mathfrak{Hello!}}}$

$According \: to \: the \: question; \: An \\ object \: of \: mass \: m \: moves \: with \\ constant \: speed \: in \: a \: circular \\ path \: of \: radius \: R \: under \: the \\ action \: of \: a \: force \: of \: constant \\ magnitude \: F. \\ \\ As \: we \: know; \\ Kinetic \: energy = \frac{1}{2} m {v}^{2} \\ (Force = mass \times acceleration) \\(F = m \times a )\\ (m = \frac{F}{a} ) \\ \\ Therefore, \: \\ K.E. = \frac{1}{2} \frac{F {v}^{2} }{a} \\ (As \: we \: know; \: Centripetal \: acceleration = \frac{ {v}^{2} }{r} ) \\ \: \: \: \: \: \: \: \: \: = \frac{1}{2} \frac{F {v}^{2} }{ \frac{ {v}^{2} }{FR} } \\ \: \: \: \: \: \: \: \: \: = \frac{1}{2} FR$
$\boxed{KE = \frac{1}{2} FR }$

$\bf{\mathfrak{Hope \: this \: helps...:)}}$