Question

Find the accelerations of the three blocks shown in the figure, if a horizontal force of $10\text{N}$ is applied on $2\text{kg}$ block.
(Take $g=10m/s^2$)

• A. $a_A=a_C=0.4{\text{m/s}}^2,a_B=3{\text{m/s}}^2$
• B. $a_A=0.4{\text{m/s}}^2,a_B=a_C=3{\text{m/s}}^2$
• C. $a_A=3{\text{m/s}}^2,a_B=a_C=0.4{\text{m/s}}^2$
• D. $a_A=a_B=3{\text{m/s}}^2,a_C=0.4{\text{m/s}}^2$

Answer 1

The correct option is C $a_A=3{\text{m/s}}^2,a_B=a_C=0.4{\text{m/s}}^2$

For $2\text{kg}$ block:
The FBD for $2\text{kg}$ block is:
$N=2g=20\text{N}$
Maximum possible friction on $2\text{kg}$ block is
$f_{\text{3 max}}=\mu N=0.2\times 20=4\text{N}$
Writing equation of force to $2\text{kg}$ block
$\Rightarrow 10-4=2a_A$
$\Rightarrow a_A=3{\text{m/s}}^2$

Now for $3\text{kg}$ block:
The maximum available friction between the $2\text{kg}$ and $3\text{kg}$ blocks is $4\text{N}$, which will act as a driving force for $3\text{kg}$ block.
Now, maximum available friction between $3\text{kg}$ and $7\text{kg}$ block $\left(f_{\text{2 max}}\right)$ is

$N=5g=50\text{N}$
$f_{\text{2 max}}=\mu N=0.3\times 50=15\text{N}$
Since, maximum available driving force for $3\text{kg}$ < maximum available friction force for $3\text{kg}$, the $3\text{kg}$ cannot move relative to $7\text{kg}$ block.

For $7\text{kg}$ block:
maximum available friction between $7\text{kg}$ block and the floor is zero as it is smooth ($\mu =0$)
So, there is no friction from the floor.
Thus, $3\text{kg}$ and $7\text{kg}$ both blocks put together move with same acceleration $\left(a\right)$, which is given by
$a_B=a_C=a=\frac{f_{\text{3 max}}}{m_B+m_C}=\frac{4}{3+7}=0.4m/s^2$