Two blocks A and B of masses m_A = 1,kg and m | vedclass

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(B) For block $A$ not to slide over $B$,the maximum acceleration $a$ of the system is limited by the friction between $A$ and $B$.The maximum friction

Vedclass · Accepted Answer

$16$

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(B) For block $A$ not to slide over $B$,the maximum acceleration $a$ of the system is limited by the friction between $A$ and $B$.The maximum frictional force on $A$ is $f_{max} = \mu_1 m_A g = 0.2 	imes 1 	imes 10 = 2\,N$.This friction provides the maximum acceleration to block $A$: $a_{max} = \frac{f_{max}}{m_A} = \frac{2}{1} = 2\,m/s^2$.Now,consider the whole system $(A+B)$ moving with acceleration $a_{max}$. The external force $F$ must overcome the friction between $B$ and the table surface $(f_{table})$ and provide the required acceleration to both blocks.The friction between $B$ and the table is $f_{table} = \mu_2 (m_A + m_B) g = 0.2 	imes (1 + 3) 	imes 10 = 0.2 	imes 4 	imes 10 = 8\,N$.Applying Newton's second law to the system $(A+B)$: $F - f_{table} = (m_A + m_B) a_{max}$.$F - 8 = (1 + 3) 	imes 2$.$F - 8 = 4 	imes 2 = 8$.$F = 8 + 8 = 16\,N$.

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