Two blocks (A) of 2,kg and (B) of 5,kg rest o | vedclass

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(A) For block $A$ to move with block $B$ without any relative motion,the maximum acceleration $a_{max}$ of the system is provided by the friction forc

Vedclass · Accepted Answer

$42\,N$

Vedclass · Answer

(A) For block $A$ to move with block $B$ without any relative motion,the maximum acceleration $a_{max}$ of the system is provided by the friction force between $A$ and $B$.The limiting friction force on block $A$ is $f_{max} = \mu N_A = \mu m_A g$.Using Newton's second law for block $A$,$f_{max} = m_A a_{max}$.Therefore,$m_A a_{max} = \mu m_A g$,which gives $a_{max} = \mu g$.Given $\mu = 0.60$ and $g = 10\,m/s^2$,we have $a_{max} = 0.60 	imes 10 = 6\,m/s^2$.Now,considering both blocks $A$ and $B$ as a single system of mass $(m_A + m_B)$,the maximum force $F_{max}$ that can be applied to $B$ is $F_{max} = (m_A + m_B) a_{max}$.Substituting the values,$F_{max} = (2 + 5) 	imes 6 = 7 	imes 6 = 42\,N$.

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