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Tension in a String

The accelerat...

Question

The acceleration of blocks $m_{1}, m_{2}, m_{3}$ blocks as shown in figure given below will be:

\frac{F}{m_{2}}

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\frac{F}{m_{1} + m_{2}}

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\frac{F}{m_{1} + m_{2} + m_{3}}

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\frac{F}{m_{1} + 2 m_{2} + m_{3}}

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Solution

The correct option is C $\frac{F}{m_{1} + m_{2} + m_{3}}$

Considering the blocks $m_{1}, m_{2}, m_{3}$ as a system all tension forces will become internal force, only external force acting will be $F$ .

Applying newton's $2^{n d}$ law in the direction of acceleration( $a$ );

$F_{E x t e r n a l} = m_{s y s t e m} \times a_{s y s t e m} F = (m_{1} + m_{2} + m_{3}) a ∴ a = \frac{F}{m_{1} + m_{2} + m_{3}}$

Hence option C is the correct answer

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Similar questions

m_{1}, m_{2}, m_{3}

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

(g = 10 m / s^{2})

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

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