A block A of mass m_1 rests on a horizontal t | vedclass

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Question

(A) For block $B$ (mass $m_2$) moving downwards with acceleration $a$: $m_2 g - T = m_2 a$  (Equation $1$)For block $A$ (mass $m_1$) moving horizontal

Vedclass · Accepted Answer

$\frac{m_1 m_2 (1 + \mu_k) g}{m_1 + m_2}$

Vedclass · Answer

(A) For block $B$ (mass $m_2$) moving downwards with acceleration $a$: $m_2 g - T = m_2 a$  (Equation $1$)For block $A$ (mass $m_1$) moving horizontally with acceleration $a$: $T - \mu_k m_1 g = m_1 a$  (Equation $2$)Adding Equation $1$ and Equation $2$:$(m_2 g - T) + (T - \mu_k m_1 g) = m_2 a + m_1 a$$m_2 g - \mu_k m_1 g = (m_1 + m_2) a$$a = \frac{(m_2 - \mu_k m_1) g}{m_1 + m_2}$Now,substitute the value of $a$ into Equation $1$:$T = m_2 g - m_2 a = m_2 (g - a)$$T = m_2 \left( g - \frac{(m_2 - \mu_k m_1) g}{m_1 + m_2} \right)$$T = m_2 g \left( \frac{m_1 + m_2 - m_2 + \mu_k m_1}{m_1 + m_2} \right)$$T = \frac{m_1 m_2 (1 + \mu_k) g}{m_1 + m_2}$

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