# Question a5256

Nov 27, 2016

Let

$a \to \text{downward acceleration of M }$

$T \to \text{tension on rope }$

$g \to \text{ acceleration due to gravity } = 9.8 m {s}^{-} 2$

Considering the forces on 9 kg block we can write

$T - \mu 12 g = 9 a \ldots \ldots \ldots . . \left(1\right)$

Considering the forces on $M = 5 k g$ block we can write

$5 g - T = 5 a \ldots \ldots \ldots . . \left(2\right)$

Adding (1) and (2) we get

$\left(5 g - \mu 12 g\right) = 14 a$

$a = \frac{5 g - \mu 12 g}{14} = \frac{\left(5 - 0.3 \times 12\right) 9.8}{14} = 0.98 m {s}^{-} 2$

Apr 3, 2017

$a = 1.61 \text{ } \frac{m}{s} ^ 2$

#### Explanation:

${m}_{1} = 12 \text{ } k g$

${m}_{2} = 9 \text{ } k g$

$M = 5 \text{ } k g$

${\mu}_{k} = 0.30$

$\text{Total act on OB surface is } F = {m}_{1} g + {m}_{2} g$

$\text{Reaction Force of OB surface is } {F}_{a} = - F = - \left({m}_{1} g + {m}_{2} g\right)$

$\text{Effect of m1 mass on G H surface is } {m}_{1} g - F = {m}_{1} g - \left({m}_{1} g + {m}_{2} g\right)$

$N = - {m}_{2} g$

${f}_{s} = {\mu}_{k} \cdot N \text{ , "f_s=0.3*9*9.81=26.487" } N$

a=(Mg-f_s)/(M+m_2#

$a = \frac{5 \cdot 9.81 - 26.487}{5 + 9}$

$a = \frac{49.05 - 26.487}{14}$

$a = 1.61 \text{ } \frac{m}{s} ^ 2$