Question

Three masses of $1kg$, $5kg$ and $3kg$ are connected to each other with threads and are placed to table as shown in the figure. What is the acceleration with which the system is moving? (Take $g=10m/s^{-2}$)

• A. Zero
• B. $1.1m{s}^{-2}$
• C. $2.2m{s}^{-2}$
• D. $3.3m{s}^{-2}$

Answer 1

The correct option is C $2.2m{s}^{-2}$

$\text{Step 1: Identify all forces on each block and predict the direction of motion}$

The surface is smooth so the $1kg$ block will move upward, $5kg$ block will move rightward and $3kg$ block will move downward.

From figure, Since all blocks are connected with string, so magnitude of acceleration of all the blocks will be same, say $a$.

$\text{Step 2: Draw F.B.D. of each block individually and apply Newton's second law}$

$\text{For block 1:}$ Applying Newton's second law on on block $1$ (Upward direction is positive)

$\Sigma F=m_{1}a$

$\Rightarrow T_1-m_{1}g=m_{1}a$ $....\left(1\right)$

$\text{For block 2:}$ Applying Newton's second law on on block $2$ (Rightward direction is positive)

$\Sigma F=m_{2}a$

$\Rightarrow T_2-T_1=m_{2}a$ $....\left(2\right)$

$\text{For block 3:}$ Applying Newton's second law on on block $3$ (Downward direction is positive)

$\Sigma F=m_{3}a$

$\Rightarrow m_{3}g-T_2=m_{3}a$ $....\left(3\right)$

$\text{Step 3: Solving above equations to get required value}$

Given: $m_1=1kg,m_2=5kg,m_3=3kg$

Adding equation $\left(1\right),\left(2\right)$ and $\left(3\right)$ we get

$m_{3}g-m_{1}g=m_{1}a+m_{2}a+m_{3}a$

$\Rightarrow 3g-g=a+5a+3a$

$\Rightarrow 2g=9a$

$\Rightarrow a=\frac{2\times 10m/s^2}{9}=2.22m/s^2$

Hence the acceleration of the system is $2.22m/s^2$

Hence, Option C is correct