In the shown arrangement,the mass of A = 1,kg | vedclass

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Question

(A) $1$. Assume both blocks $A$ and $B$ move together with a common acceleration $a$. The total mass of the system is $M = m_A + m_B = 1 + 2 = 3\,kg$.

Vedclass · Accepted Answer

$\frac{5}{3}\,N$

Vedclass · Answer

(A) $1$. Assume both blocks $A$ and $B$ move together with a common acceleration $a$. The total mass of the system is $M = m_A + m_B = 1 + 2 = 3\,kg$.$2$. The external force applied is $F = 5\,N$. The acceleration of the system is $a = \frac{F}{M} = \frac{5}{3}\,m/s^2$.$3$. For block $A$ to move with this acceleration,the required force is $f = m_A \cdot a = 1 \cdot \frac{5}{3} = \frac{5}{3}\,N$.$4$. The maximum static friction force between $A$ and $B$ is $f_{max} = \mu \cdot m_A \cdot g = 0.2 \cdot 1 \cdot 10 = 2\,N$.$5$. Since the required force $f = \frac{5}{3}\,N \approx 1.67\,N$ is less than $f_{max} = 2\,N$,the blocks move together.$6$. Therefore,the frictional force exerted by $A$ on $B$ is equal to the force required to accelerate $A$,which is $\frac{5}{3}\,N$.

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