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Pulley

Two blocks of...

Question

Two blocks of masses $m_{1}$ and $m_{2}$ are connected as shown in the figure. The acceleration of the block $m_{2}$ is

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

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

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\frac{4 m_{2} g - m_{1} g}{m_{1} + m_{2}}

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\frac{m_{2} g}{m_{1} + 4 m_{2}}

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Solution

The correct option is A $\frac{m_{2} g}{m_{1} + m_{2}}$
Let acceleration of $m_{1}$ be $a_{1}$ and $m_{2}$ be $a_{2}$ .
$a_{2} = \frac{F_{n e t}}{m_{2}}$
Considering force body diagram of $m_{2}$
$a_{2} = \frac{m_{2} g - T}{m_{2}} a_{2} = g - \frac{T}{m_{2}}$

Now considering the pulley connected to mass $m_{2}$
Since pulley is massless, so net force on iy would be zero.
Hence by force balance
$T^{'} = 2 T$
where $T^{'}$ is the tension in the string joining two pulleys
Now taking the pulley above the previous pulley we get
$T^{'} = 2 T "$
where $T "$ is the tension in strings which are joining the uppermost pulley with middle one.
From this we get
$T "= T$
Taking $m_{1}$ into consideration and balancing force
$T = m_{1} a_{1}$
Now putting this into force balance equation of $m_{2}$
$a_{2} = g - \frac{m_{1} a_{1}}{m_{2}}$
Now as the middle pulley is the exact mirror symmetry of the bottom pulley we can say
$a_{1} = a_{2} = a$
$\Rightarrow a = g - \frac{m_{1} a}{m_{2}} \Rightarrow a = \frac{m_{2} g}{m_{1} + m_{2}}$

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Two blocks of masses m1 and m2 are connected as shown in the figure. The acceleration of the block m2 is

Two blocks of masses $m_{1}$ and $m_{2}$ are connected as shown in the figure. The acceleration of the block $m_{2}$ is