Question

Find the acceleration of the two blocks of mass $4\text{kg}$ & $5\text{kg},$ if a force of $40\text{N}$ is applied on $4\text{kg}$ block. Friction coefficient between the respective surfaces are shown in figure. (take $g=10{\text{m/s}}^2$).

• A. $a_2=a_1=5{\text{m/s}}^2$
• B. $a_1=a_2=\frac{13}{9}{\text{m/s}}^2$
• C. $a_1=0,a_2=5{\text{m/s}}^2$
• D. $a_2=0,a_1=\frac{13}{9}{\text{m/s}}^2$

Answer 1

The correct option is C $a_1=0,a_2=5{\text{m/s}}^2$

First assume that there is no sliding between the two blocks and both are moving together with common acceleration $a$.



Hence, here friction between $4\&5\text{kg}$ block is internal friction.
$F-f=ma$
$40-{\mu }_1\times (4+5)\times 10=(4+5)a$
$40-0.3\left(9\right)\times 10=9\times a$
$a=\frac{13}{9}{\text{m/s}}^2$

FBD for block of $4\text{kg}$


$F-f_2=ma$
$40-f_2=4\times \frac{13}{9}$
$f_2=34.22\text{N}$
Here the maximum possible value of $f_{2L}$ is
$f_{2L}={\mu }_2{m}_{2}g=0.5\times 4\times 10=20\text{N}$

But $f_2$ is more than $f_{2L},$ it implies that the block of $4\text{kg}$ can never be at rest relative to $5\text{kg}$ block.
Hence, both will move seperately.

FBD for block $4\text{kg}$


$F-f_{2L}=m_2{a}_2$
$40-20=4\times a_2$
$a_2=5{\text{m/s}}^2$

FBD of block $5\text{kg}$


Here maximum possible value of $f_1$ is
$f_1={\mu }_1(m_1+m_2)g=0.3\times (4+5)\times 10$
$f_1=27\text{N}$
It is more than $20\text{N}$ so block $5\text{kg}$ will not move.
$a_1=0{\text{m/s}}^2$