A body of mass m is moving in a circle of rad | vedclass

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(B) The work done by a force is defined as the dot product of force and displacement,given by $W = \vec{F} \cdot \vec{s} = Fs \cos 	heta$,where $	he

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Zero

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(B) The work done by a force is defined as the dot product of force and displacement,given by $W = \vec{F} \cdot \vec{s} = Fs \cos 	heta$,where $	heta$ is the angle between the force and the displacement vector.In uniform circular motion,the centripetal force acts towards the center of the circle,while the instantaneous displacement (tangential to the circle) is always perpendicular to the radius.Since the force is always directed towards the center and the displacement is along the tangent,the angle $	heta$ between the force and the displacement is always $90^\circ$.Therefore,the work done is $W = Fs \cos(90^\circ) = Fs 	imes 0 = 0$.Thus,the work done by the centripetal force in moving the body over any part of the circular path,including half the circumference,is zero.

$A$ body of mass $m$ is moving in a circle of radius $r$ with a constant speed $v$. The force on the body is $\frac{mv^2}{r}$ 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?

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