Question

Determine in time in which the smaller block reaches other end of bigger block in the figure.

• A. $4s$
• B. $8$
• C. $2.19s$
• D. $2.13s$

Answer 1

The correct option is C $2.19s$

Both block will not move together because friction force between block is not sufficient to accelerate 8Kg block with same acceleration

Drawing $FBD$ of $2kg$ block

$f_r=$ frictional force

$f_r=\mu Mg$

for vertical balance, $M=20N$

$fr=0.3\times 20=6N$

Now let the block moves with acceleration $a_1$

$\Rightarrow f-fr=m{a}_1$

$10-6=2\times a_1$

$\Rightarrow a_1=2m/{\sec }^2$

Drawing $FBD$ of block of $8kg$ mass

$\Rightarrow$ Let $a_2$ be its acceleration

$a_2=\frac{f_r}{m_2}=\frac{.3\times 20}{8}$

$a_2=\frac{6}{8}=\frac{3}{4}m/{\sec }^2$

Relative acceleration of both the blocks

$a=a_1-a_2=2-\frac{3}{4}=\frac{5}{4}$

distance to be covered $=3m$ initial velocity $=0$

$S=ut+\frac{1}{2}a{t}^2$

$=3=0+\frac{1}{2}\times \frac{5}{4}\times t^2$

$\Rightarrow t=2.19\sec$

Option $C$