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:

We are given a block of mass $$m = 200 \text{ g} = 0.2 \text{ kg}$$ moving with uniform speed in a horizontal circular groove of radius $$r = 20 \text{ cm} = 0.2 \text{ m}$$, completing one round in $$T = 40 \text{ s}$$.

Since the block moves in a horizontal circular path, the normal force from the side walls provides the centripetal force. The angular velocity is $$\omega = \frac{2\pi}{T} = \frac{2\pi}{40} = \frac{\pi}{20} \text{ rad/s}$$.

The centripetal force (which equals the normal force) is $$N = m\omega^2 r = 0.2 \times \left(\frac{\pi}{20}\right)^2 \times 0.2$$.

Computing step by step: $$\omega^2 = \frac{\pi^2}{400}$$, so $$N = 0.2 \times \frac{\pi^2}{400} \times 0.2 = \frac{0.04 \pi^2}{400} = \frac{\pi^2}{10000}$$.

Using $$\pi^2 \approx 9.8596$$, we get $$N = \frac{9.8596}{10000} = 9.859 \times 10^{-4} \text{ N}$$.