A block of mass 10; kg is in contact with the | vedclass

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(C) For the block to remain stationary,the upward frictional force must balance the downward gravitational force.$f_s = mg$The normal force $N$ provid

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$10$

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(C) For the block to remain stationary,the upward frictional force must balance the downward gravitational force.$f_s = mg$The normal force $N$ provided by the wall acts as the centripetal force for the block's circular motion:$N = mr\omega^2$The maximum static frictional force is given by:$f_{L} = \mu N = \mu mr\omega^2$For the block to stay in equilibrium,the frictional force must be at least equal to the weight:$f_{L} \geq mg$$\mu mr\omega^2 \geq mg$Solving for $\omega$:$\omega^2 \geq \frac{g}{\mu r}$$\omega \geq \sqrt{\frac{g}{\mu r}}$Substituting the given values ($g = 10\; m/s^2$,$\mu = 0.1$,$r = 1\; m$):$\omega_{\min} = \sqrt{\frac{10}{0.1 	imes 1}} = \sqrt{\frac{10}{0.1}} = \sqrt{100} = 10\; rad/s$

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