A rigid body of mass m is moving on circular path of radius r & with a constant speed v, net force on body is F, then during half rotation work done by F is,

F(2r)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F(r

\sqrt{2}

)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(π r)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

zero

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D zero
Since speed = constant, K.E.=constant

so, $W_{F} = 0$

Suggest Corrections

Similar questions

Q. A rigid body of mass m is moving on circular path of radius r & with a constant speed v, net force on body is F, then during half rotation work done by F is,

Q. A body of mass m is moving in a circle of radius r with a constant speed

v

. If a force

\frac{m v^{2}}{r}

is acting on the body towards the centre, then what will be the work done by this force in moving the body over half the circumference of the circle?

Q. A block of mass m is rotating in a circular path of radius r at constant speed v . Calculate the work done by the centripetal force during one half of a revolution?

Q. A body of mass m is moving with a constant speed v on a circular path of radius r. The centripetal force acting on the body is

A body of mass m is moving in a circle of radius r with a constant speed v. The force on the body is $\frac{m v^{2}}{r}$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle

Join BYJU'S Learning Program

A rigid body of mass m is moving on circular path of radius r & with a constant speed v, net force on body is F, then during half rotation work done by F is,

The correct option is D zeroSince speed = constant, K.E.=constant so, WF=0

The correct option is D zero
Since speed = constant, K.E.=constant

so, $W_{F} = 0$