A block of 200, g mass moves with a uniform s | vedclass

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(D) The normal force $N$ provides the necessary centripetal force for circular motion.$N = F_c = m \omega^2 R$Since $\omega = \frac{2\pi}{T}$,we have

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$9.859 	imes 10^{-4}\, N$

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(D) The normal force $N$ provides the necessary centripetal force for circular motion.$N = F_c = m \omega^2 R$Since $\omega = \frac{2\pi}{T}$,we have $N = m \left( \frac{2\pi}{T} ight)^2 R = m \frac{4\pi^2}{T^2} R$.Given values: $m = 200\, g = 0.2\, kg$,$R = 20\, cm = 0.2\, m$,and $T = 40\, s$.Substituting these values into the formula:$N = 0.2 	imes \frac{4 	imes (3.14159)^2}{(40)^2} 	imes 0.2$$N = 0.2 	imes \frac{4 	imes 9.8696}{1600} 	imes 0.2$$N = 0.04 	imes \frac{39.4784}{1600}$$N = 0.04 	imes 0.024674 = 9.8696 	imes 10^{-4}\, N$.Rounding to the nearest option,we get $N \approx 9.859 	imes 10^{-4}\, N$.

$A$ block of $200\, g$ mass moves with a uniform speed in a horizontal circular groove,with vertical side walls of radius $20\, cm$. If the block takes $40\, s$ to complete one round,the normal force by the side walls of the groove is

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