Question

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Mass mA rests on a smooth horizontal surface, mB hangs vertically.

a)If mA=12.0 kg and mB=3.0 kg in the figure (Figure 1), determine the magnitude of the acceleration of each block.?

b)If initially mA is at rest 1.300 m from the edge of the table, how long does it take to reach the edge of the table if the system is allowed to move freely?

c)If mB=1.0kg, how large must mA be if the acceleration of the system is to be kept at 1100g?

Answer 1

$\text{Forces acting on the block A:}$

$\text{In the horizontal direction:}$

$T -\text{thread tension}$

$\text{In the vertical direction:}$

$F_{Ag}- \text{force of gravity}$

$N - \text{table reaction force}$

$F_{Ag}=N,\text{ then}$

$T = m_Aa\ (1)$

$\text{Forces acting on the block B:}$

$T -\text{thread tension}$

$F_{bg}- \text{force of gravity}$

$F_{Bg}-T =m_Ba$

$F_{Bg}= m_Bg$

$T= m_Bg-m_Ba\ (2)$

$\text{From the equalities }(1) \text{ and }(2):$

$a = \frac{m_Bg}{m_B+m_A}(3)$

$a) m_A = 12kg;m_B=3kg$

$g =9.8\frac{m}{s^2}$

$a_A = a_B = a$

$\text{From formule }(3)$

$a = \frac{m_Bg}{m_B+m_A}=\frac{3*9.8}{12+3}=1.96\frac{m}{s^2}$

$\text{Answer: }a_A = 1.96\frac{m}{s^2};a_B = 1.96\frac{m}{s^2};$

$b)v_0 =0 ; a = 1.96\frac{m}{s^2}$

$S = 1.3m$

$S = v_0t+\frac{at^2}{2}$

$t = \sqrt{\frac{2S}{a}}=\sqrt{\frac{2*1.3}{1.96}}=1.15 \ s$

$\text{Answer: }1.15s$

$c) m_B= 1$

$a =1100g$

$\text{From }a = \frac{m_Bg}{m_B+m_A}$

$\frac{m_B}{m_B+m_A}<1$

$\text{Hence always }a<g$

$\text{Answer: The conditions of the question are not correct}$