The maximum acceleration of the 5, kg block i | vedclass

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(C) Let the $10\, kg$ block be $m_1$ and the $5\, kg$ block be $m_2$. The coefficient of friction between $m_1$ and $m_2$ is $\mu_1 = 0.1$,and between

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$0$

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(C) Let the $10\, kg$ block be $m_1$ and the $5\, kg$ block be $m_2$. The coefficient of friction between $m_1$ and $m_2$ is $\mu_1 = 0.1$,and between $m_2$ and the ground is $\mu_2 = 0.3$.The maximum static friction force between $m_1$ and $m_2$ is $f_{1,max} = \mu_1 m_1 g = 0.1 	imes 10 	imes 10 = 10\, N$.The maximum static friction force between $m_2$ and the ground is $f_{2,max} = \mu_2 (m_1 + m_2) g = 0.3 	imes (10 + 5) 	imes 10 = 0.3 	imes 150 = 45\, N$.For the $5\, kg$ block $(m_2)$ to move,the force applied by the $10\, kg$ block on it (which is the friction $f_1$) must exceed the maximum static friction force from the ground $(f_{2,max} = 45\, N)$.Since the maximum force that $m_1$ can exert on $m_2$ is $f_{1,max} = 10\, N$,and $10\, N Therefore,the acceleration of the $5\, kg$ block is $0\, m/s^2$.

The maximum acceleration of the $5\, kg$ block is ...... $m/s^2$.

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