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 kinetic energy of the object iss

Answer 1

Please find the answer below:

Keeping in view that; 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 kinetic energy of the object is FR/2

Here,

F=Force.

R=Radius.

Explanation:

Force F is providing the centripetal force keeping the object moving in a circular path.

F = mv²/R

FR = mv²

Hence,

Kinetic Energy(KE) = ½mv² = ½FR

Answer 2

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

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

$\boxed{KE = \frac{1}{2} FR }$

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