Four blocks A, B, C and D of masses 6 ,kg, 3 | vedclass

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(C) Given: $m_A=6 \,kg, m_B=3 \,kg, m_C=6 \,kg, m_D=1 \,kg$ and coefficient of friction $\mu=0.2$.The total mass of the system is $M = m_A + m_B + m_C

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) Given: $m_A=6 \,kg, m_B=3 \,kg, m_C=6 \,kg, m_D=1 \,kg$ and coefficient of friction $\mu=0.2$.The total mass of the system is $M = m_A + m_B + m_C + m_D = 6 + 3 + 6 + 1 = 16 \,kg$.The driving force is the weight of block $C$ $(m_C g)$, and the opposing forces are the weight of block $D$ $(m_D g)$ and the kinetic friction force $f_k = \mu N = \mu(m_A + m_B)g$.The acceleration $a$ of the system is given by:$a = \frac{m_C g - m_D g - \mu(m_A + m_B)g}{m_A + m_B + m_C + m_D}$$a = \frac{6 	imes 10 - 1 	imes 10 - 0.2(6 + 3) 	imes 10}{16} = \frac{60 - 10 - 18}{16} = \frac{32}{16} = 2 \,ms^{-2}$.Now, consider the free-body diagram of block $D$. The tension $T_Q$ in string $Q$ acts upwards, and the weight $m_D g$ acts downwards. Since the system accelerates in the direction of $C$, block $D$ moves upwards with acceleration $a$:$T_Q - m_D g = m_D a$$T_Q = m_D(a + g) = 1 	imes (2 + 10) = 12 \,N$.Thus, the tension in string $Q$ is $12 \,N$.

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