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From the below figure, if the blocks A, B and C of mass $m$ each have acceleration $a_{1}, a_{2} and a_{3}$ and external forces $F_{1} and F_{2}$ of magnitude $2 mg$ and $mg$ , respectively, then the statement that is true on acceleration will be_

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a_{1} = a_{2} = a_{3}

a_{1} > a_{3} > a_{2}

a_{1} > a_{2} > a_{3}

a_{1} > a_{2}, a_{2} = a_{3}

answer is C.

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Detailed Solution

If the blocks A, B and C of mass m each have acceleration

a_{1}, a_{2} and a_{3}

and external forces

F_{1} and F_{2}

of magnitude

2 mg

and

mg

respectively, then the statement that is true on acceleration will be

a_{1} > a_{2} > a_{3} .

Given:
Mass of each block

A, B and C is m

Acceleration of

A, B and C is a_{1}, a_{2} and a_{3}

respectively.

F_{1} = 2 mg

and

F_{2} = mg

Now, using

F = ma

First, on block A we will apply Newton's law.

F_{net} = m a_{1}

\Rightarrow 2 mg - mg = m a_{1}

{\Rightarrow a}_{1} = g

Now, on block B we will apply Newton's law.

F_{net} = m a_{2}

\Rightarrow 2 mg - mg = (2 m + m) a_{2}

\Rightarrow mg = 3 m a_{2}

\Rightarrow a_{2} = \frac{g}{3}

Again, on block C we will apply Newton's law.

F_{net} = m a_{3}

\Rightarrow mg - mg = (m + m) a_{3}

\Rightarrow 0 = 2 m a_{3}

{\Rightarrow a}_{3} = 0

Therefore

g > \frac{g}{2} > 0 \Rightarrow a_{1} > a_{2} > a_{3}

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