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

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(B) The work done $W$ by a force $\vec{F}$ is given by the dot product $W = \int \vec{F} \cdot d\vec{s}$.In uniform circular motion,the centripetal fo

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Zero

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(B) The work done $W$ by a force $\vec{F}$ is given by the dot product $W = \int \vec{F} \cdot d\vec{s}$.In uniform circular motion,the centripetal force $\vec{F}$ is always directed towards the centre of the circle.The displacement vector $d\vec{s}$ at any point on the circular path is always tangent to the circle.Since the radius (and thus the force) is always perpendicular to the tangent (displacement),the angle $	heta$ between the force $\vec{F}$ and the displacement $d\vec{s}$ is always $90^\circ$.Therefore,the work done $W = \int F \cdot ds \cdot \cos(90^\circ) = \int F \cdot ds \cdot 0 = 0$.Thus,the work done by the centripetal force in moving the body over any distance along the circular path 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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