If $m_{1} = 10 k g, m_{2} = 4 k g$ and $m_{3} = 2 k g$ , the acceleration of the system is______

\frac{5 g}{2}

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\frac{5 g}{3}

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\frac{5 g}{8}

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\frac{5 g}{14}

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Solution

The correct option is C $\frac{5 g}{8}$
Given,

Force on block $m_{1}$ , $F_{1} = m_{1} g = 10 g N$
Force on block $m_{1}$ , $F_{2} = m_{2} g sin 30^{o} = 4 g \times 0.5 = 2 g N$

Force on block $m_{3}$ , $F_{3} = m_{3} g = 2 g N$

$F_{n e t} = F_{1} + F_{2} - F_{3}$

$\Rightarrow (m_{1} + m_{2} + m_{3}) a = 10 g + 2 g - 2 g$

$\Rightarrow a = \frac{10 g}{10 + 4 + 2} = \frac{5 g}{8} m s^{- 2}$

Acceleration of system $\frac{5 g}{8} m s^{- 2}$

Suggest Corrections

Similar questions

m_{1} = 10 k g, m_{2} = 4 k g, m_{3} = 2 k g

10 k g + 5 g =

g

m_{1} = 10 k g

m_{2} = 4 k g

m_{3} = 2 k g

5 g

100^{0} C

5 g

0^{0} C

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