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Question

(A) The power delivered by a force is given by $P = \vec{F} \cdot \vec{v}$.Since the speed of the particle is increasing,the work done by the net forc

Vedclass · Accepted Answer

$\vec{F} \cdot \vec{v} > 0$

Vedclass · Answer

(A) The power delivered by a force is given by $P = \vec{F} \cdot \vec{v}$.Since the speed of the particle is increasing,the work done by the net force on the particle must be positive.In circular motion,the net force $\vec{F}$ can be resolved into two components: the centripetal force $\vec{F}_c$ (directed towards the center) and the tangential force $\vec{F}_t$ (directed along the tangent).The centripetal force $\vec{F}_c$ is always perpendicular to the velocity $\vec{v}$,so $\vec{F}_c \cdot \vec{v} = 0$.The tangential force $\vec{F}_t$ is responsible for the change in speed. Since the speed is increasing,$\vec{F}_t$ must be in the same direction as the velocity $\vec{v}$.Therefore,$\vec{F} \cdot \vec{v} = (\vec{F}_c + \vec{F}_t) \cdot \vec{v} = \vec{F}_c \cdot \vec{v} + \vec{F}_t \cdot \vec{v} = 0 + F_t v = F_t v > 0$.Thus,$\vec{F} \cdot \vec{v} > 0$.

$A$ particle is moving in a circular path of radius $r$ under the action of a force $F$. If at an instant the velocity of the particle is $\vec{v}$,and the speed of the particle is increasing,then:

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