A block of mass m is pulled along a circular | vedclass

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(C) The work done by a constant force is given by the dot product of force and displacement: $W = \vec{F} \cdot \vec{d} = F d \cos 	heta$.Since the f

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2} F R$

Vedclass · Answer

(C) The work done by a constant force is given by the dot product of force and displacement: $W = \vec{F} \cdot \vec{d} = F d \cos 	heta$.Since the force $F$ is constant and horizontal,the work done is $W = F 	imes (	ext{horizontal displacement})$.From the geometry of the circular arc,the horizontal displacement of the block as it moves from $A$ to $B$ is $d = R \sin 60^{\circ}$.Substituting the value of $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we get the horizontal displacement $d = R \left(\frac{\sqrt{3}}{2}ight)$.Therefore,the work done is $W = F 	imes \left(R \frac{\sqrt{3}}{2}ight) = \frac{\sqrt{3}}{2} F R$.Thus,the correct option is $C$.

$A$ block of mass $m$ is pulled along a circular arc by means of a constant horizontal force $F$ as shown. Work done by this force in pulling the block from $A$ to $B$ is ............

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