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

• A. $\frac{m_{2}g}{m_1+m_2}$
• B. $\frac{m_{1}g}{m_1+m_2}$
• C. $\frac{4m_{2}g-m_{1}g}{m_1+m_2}$
• D. $\frac{m_{2}g}{m_1+4m_2}$

Answer 1

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 $a_2$.
$a_2=\frac{F_{net}}{m_2}$
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
$T^{\prime }=2T$
where $T^{\prime }$ is the tension in the string joining two pulleys
Now taking the pulley above the previous pulley we get
$T^{\prime }=2T"$
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
$a_1=a_2=a$
$\Rightarrow a=g-\frac{m_{1}a}{m_2}\Rightarrow a=\frac{m_{2}g}{m_1+m_2}$