Question

Coefficient of friction between $5kg$ and $10kg$ block is $0.5$. If friction between them is $20N$, the value of force being applied on $5kg$ (in $N$) is (Assume that the floor is frictionless)

Answer 1

The limiting friction between $5kg$ and $10kg$ is $f_{l2}=50\times 0.5=25N.$
Since $20<25,$ the two blocks are moving with a common acceleration.
$a_{com}=\frac{F-F_{l1}}{5+10}(\because f_{l1}=0$, since floor is smooth )
$\Rightarrow F=15a_{com}$ ......$\left(i\right)$
FBD of $10kg$ block alone is,

$\Rightarrow 20=10a_{com}\Rightarrow a_{com}=2$
In $\left(i\right)$
$F=15\left(2\right)=30N.$