As shown in the figure, block m and wedge M m | vedclass

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(C) For the block $m$ to move with the wedge $M$, the vertical forces must balance. The pseudo force acting on $m$ is $F_p = ma$ in the backward direc

Vedclass · Accepted Answer

As acceleration of the system is slightly increased, friction force between $m$ and $M$ remains constant.

Vedclass · Answer

(C) For the block $m$ to move with the wedge $M$, the vertical forces must balance. The pseudo force acting on $m$ is $F_p = ma$ in the backward direction. This pseudo force acts as the normal force $N$ between $m$ and $M$, so $N = ma = 1 	imes 20 = 20\, N$.The maximum static friction force is $f_{max} = \mu N = 0.6 	imes 20 = 12\, N$.The weight of the block is $mg = 1 	imes 10 = 10\, N$.Since $f_{max} > mg$, the block $m$ will not slide down. The static friction force $f$ balances the weight of the block, so $f = mg = 10\, N$.As long as the block does not slide, the friction force $f$ remains equal to $mg$, which is constant, regardless of small changes in acceleration $a$, provided $f_{max} \ge mg$ (i.e., $\mu ma \ge mg \implies a \ge g/\mu = 10/0.6 = 16.67\, m/s^2$).Since the current acceleration $20\, m/s^2 > 16.67\, m/s^2$, the friction force is constant at $10\, N$.

As shown in the figure, block $m$ and wedge $M$ move together with a horizontal acceleration of $20\, m/s^2$. Given $m = 1\, kg$, $\mu = 0.6$ (between $m$ and $M$) and $g = 10\, m/s^2$. Choose the $\text{CORRECT}$ alternative:

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