A pendulum of length 1 ,m and having a bob of | vedclass

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Question

(A) The power delivered by all forces acting on a body is equal to the rate of change of its kinetic energy,which is equivalent to the net force dot p

Vedclass · Accepted Answer

$13.5$

Vedclass · Answer

(A) The power delivered by all forces acting on a body is equal to the rate of change of its kinetic energy,which is equivalent to the net force dot product with velocity,$P = \vec{F}_{net} \cdot \vec{v}$.Since the tension force is always perpendicular to the velocity of the bob,it does no work and contributes zero power.The only force doing work is gravity. The component of gravity along the velocity is $mg \sin 	heta$.Thus,$P = (mg \sin 	heta) v$.Using conservation of energy,the velocity $v$ at angle $	heta$ when released from $	heta_0 = 60^{\circ}$ is given by $\frac{1}{2}mv^2 = mg(l \cos 	heta - l \cos 	heta_0)$.$v = \sqrt{2gl(\cos 	heta - \cos 	heta_0)}$.Given $l = 1 \,m$,$m = 10^{-3} \,kg$,$g = 10 \,ms^{-2}$,$	heta = 30^{\circ}$,and $	heta_0 = 60^{\circ}$.$v = \sqrt{2 	imes 10 	imes 1 	imes (\cos 30^{\circ} - \cos 60^{\circ})} = \sqrt{20 	imes (\frac{\sqrt{3}}{2} - 0.5)} = \sqrt{20 	imes (0.866 - 0.5)} = \sqrt{20 	imes 0.366} = \sqrt{7.32} \approx 2.705 \,ms^{-1}$.Now,$P = mg \sin 30^{\circ} 	imes v = 10^{-3} 	imes 10 	imes 0.5 	imes 2.705 = 0.013525 \,W = 13.525 \,mW$.Rounding to the nearest option,the power is $13.5 \,mW$.

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