A horizontal electric field E = (mg)/q exists | vedclass

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Question

(B) According to the work-energy theorem,the total work done by all forces is equal to the change in kinetic energy:$W_e + W_g = \Delta K = \frac{1}{2

Vedclass · Accepted Answer

$\sqrt{\frac{2g}{l}}$

Vedclass · Answer

(B) According to the work-energy theorem,the total work done by all forces is equal to the change in kinetic energy:$W_e + W_g = \Delta K = \frac{1}{2}mv^2 - 0$The work done by the electrostatic force $F_e = qE$ acting horizontally is:$W_e = F_e \cdot d_x = (qE)(l \sin 	heta) = q(\frac{mg}{q})l \sin 	heta = mgl \sin 	heta$The work done by gravity is:$W_g = mg \cdot d_y = mg(l - l \cos 	heta)$Substituting these into the work-energy theorem:$mgl \sin 	heta + mgl(1 - \cos 	heta) = \frac{1}{2}mv^2$Given $	heta = 45^{\circ}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$:$mgl(\frac{1}{\sqrt{2}}) + mgl(1 - \frac{1}{\sqrt{2}}) = \frac{1}{2}mv^2$$mgl(\frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}}) = \frac{1}{2}mv^2$$mgl = \frac{1}{2}mv^2$$v^2 = 2gl$Since $v = \omega l$,we have:$(\omega l)^2 = 2gl$$\omega^2 l^2 = 2gl$$\omega = \sqrt{\frac{2g}{l}}$

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