A block of mass m is in contact with the cart | vedclass

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(C) To prevent the block from falling,the upward frictional force $f$ must balance the downward gravitational force $mg$. $1$. The cart accelerates wi

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$\alpha \ge \frac{g}{\mu}$

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(C) To prevent the block from falling,the upward frictional force $f$ must balance the downward gravitational force $mg$. $1$. The cart accelerates with $\alpha$. In the frame of the cart,a pseudo force $F_{fic} = m\alpha$ acts on the block,pushing it against the cart. This force provides the normal reaction $N = m\alpha$. $2$. The maximum static frictional force available is $f_{max} = \mu N = \mu m\alpha$. $3$. For the block not to fall,the frictional force $f$ must be at least equal to the weight of the block: $f \ge mg$. $4$. Substituting $f = \mu m\alpha$,we get $\mu m\alpha \ge mg$. $5$. Solving for $\alpha$,we obtain $\alpha \ge \frac{g}{\mu}$.

$A$ block of mass $m$ is in contact with the cart $C$ as shown in the figure. The coefficient of static friction between the block and the cart is $\mu$. The acceleration $\alpha$ of the cart that will prevent the block from falling satisfies:

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