Block A of mass 30, kg is resting on a fricti | vedclass

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Question

(A) $1$. First,calculate the maximum static friction force between the blocks: $f_{s,max} = \mu_s N = \mu_s m_B g = 0.4 	imes 5 	imes 10 = 20\, N$.$

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(A) $1$. First,calculate the maximum static friction force between the blocks: $f_{s,max} = \mu_s N = \mu_s m_B g = 0.4 	imes 5 	imes 10 = 20\, N$.$2$. If both blocks move together with acceleration $a$,then $F = (m_A + m_B) a$. Thus,$175 = (30 + 5) a \Rightarrow a = 175 / 35 = 5\, m/s^2$.$3$. The force required to accelerate block $A$ at $5\, m/s^2$ is $F_A = m_A a = 30 	imes 5 = 150\, N$. Since $150\, N > f_{s,max} (20\, N)$,the blocks will slip.$4$. Since they slip,the kinetic friction force $f_k = \mu_k m_B g = 0.3 	imes 5 	imes 10 = 15\, N$ acts on block $A$.$5$. The acceleration of block $A$ is $a_A = f_k / m_A = 15 / 30 = 0.5\, m/s^2$.

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