An inclined plane of length 5.60 m making an | vedclass

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(B) Given: $	heta = 45^{\circ}$,$d = 5.60 \ m$,$E = 100 \ V \ m^{-1}$,$m = 1 \ kg$,$\mu = 0.1$,$q = 10^{-2} \ C$,$v_0 = 0$.From the free body diagram

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$1.41 \ s$

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(B) Given: $	heta = 45^{\circ}$,$d = 5.60 \ m$,$E = 100 \ V \ m^{-1}$,$m = 1 \ kg$,$\mu = 0.1$,$q = 10^{-2} \ C$,$v_0 = 0$.From the free body diagram,the normal force $N$ is:$N = mg \cos 45^{\circ} + qE \sin 45^{\circ}$$N = (1 	imes 10 	imes \frac{1}{\sqrt{2}}) + (10^{-2} 	imes 100 	imes \frac{1}{\sqrt{2}})$$N = \frac{10}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{11}{\sqrt{2}} \approx 7.778 \ N$The net force $F$ acting down the incline is:$F = mg \sin 45^{\circ} - qE \cos 45^{\circ} - \mu N$$F = (1 	imes 10 	imes \frac{1}{\sqrt{2}}) - (10^{-2} 	imes 100 	imes \frac{1}{\sqrt{2}}) - (0.1 	imes 7.778)$$F = \frac{10}{\sqrt{2}} - \frac{1}{\sqrt{2}} - 0.7778 = \frac{9}{\sqrt{2}} - 0.7778 \approx 6.364 - 0.778 = 5.586 \ N$Acceleration $a = \frac{F}{m} = \frac{5.586}{1} = 5.586 \ m \ s^{-2}$.Using $d = v_0 t + \frac{1}{2} a t^2$:$5.60 = 0 + \frac{1}{2} 	imes 5.586 	imes t^2$$t^2 = \frac{2 	imes 5.60}{5.586} \approx 2$$t = \sqrt{2} \approx 1.41 \ s$.

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