A bead of mass frac{1}{2} , kg starts from re | vedclass

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Question

(A) According to the work-energy theorem,the total work done on the bead is equal to the change in its kinetic energy.$W_{F} + W_{mg} + W_{N} = \Delta

Vedclass · Accepted Answer

$14.14$

Vedclass · Answer

(A) According to the work-energy theorem,the total work done on the bead is equal to the change in its kinetic energy.$W_{F} + W_{mg} + W_{N} = \Delta K$Here,$W_{F}$ is the work done by the constant horizontal force $F$,$W_{mg}$ is the work done by gravity,and $W_{N}$ is the work done by the normal force.Since the ring is smooth,the normal force $N$ is always perpendicular to the displacement,so $W_{N} = 0$.The horizontal force $F$ acts in the direction of the horizontal displacement,which is $R = 5 \, m$. Thus,$W_{F} = F 	imes R = 5 \, N 	imes 5 \, m = 25 \, J$.The bead moves from $A$ to $B$,which is a vertical downward displacement of $R = 5 \, m$. Thus,$W_{mg} = mgR = (\frac{1}{2} \, kg) 	imes (10 \, m/s^2) 	imes (5 \, m) = 25 \, J$.The change in kinetic energy is $\Delta K = \frac{1}{2}mv^2 - 0 = \frac{1}{2} 	imes (\frac{1}{2} \, kg) 	imes v^2 = \frac{1}{4}v^2$.Equating the work done to the change in kinetic energy:$25 + 25 = \frac{1}{4}v^2$$50 = \frac{1}{4}v^2$$v^2 = 200$$v = \sqrt{200} = 10\sqrt{2} \approx 14.14 \, m/s$.

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