A 2 , kg block is lying on a smooth table and | vedclass

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Question

(A) Let $m_1 = 2 \, kg$ be the mass on the table and $m_2 = 1 \, kg$ be the hanging mass. Since the table is smooth,there is no friction. The equation

Vedclass · Accepted Answer

$3.27 \, m/s^2, 6.54 \, N$

Vedclass · Answer

(A) Let $m_1 = 2 \, kg$ be the mass on the table and $m_2 = 1 \, kg$ be the hanging mass. Since the table is smooth,there is no friction. The equation of motion for the block on the table is $T = m_1 a$. The equation of motion for the hanging mass is $m_2 g - T = m_2 a$. Adding these two equations: $m_2 g = (m_1 + m_2) a$. Thus,the acceleration $a = \frac{m_2}{m_1 + m_2} 	imes g = \frac{1}{2 + 1} 	imes 9.8 = \frac{9.8}{3} \approx 3.27 \, m/s^2$. Now,substituting $a$ into the first equation: $T = m_1 a = 2 	imes 3.27 = 6.54 \, N$.

$A$ $2 \, kg$ block is lying on a smooth table and is connected to a body of mass $1 \, kg$ by a string passing over a pulley. The $1 \, kg$ mass is hanging vertically. Find the acceleration of the block and the tension in the string.

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