The coefficient of friction between a $5kg$ and a $10kg$ is 0.5(Fig.E5.28) If the friction between them is $20N$ , what is the value of force being applied on $5kg?$ The floor is frictionless.

Free body Diagram of $5Kg.$

Force body diagram of $10kg.$

$$\begin{gathered} M=10Kg \\ m=5kg \end{gathered}$$

We know that minimum friction:

$N_1=5\times 10=50N$

According to given data:

Normal force of$5kg.$

$F=\mu N_1=25N$

Given friction force is $20N,$ which is less than maximum friction force.

Therefore x 's static case which means both of them will move together.

Acceleration of $10kg$ Block is given by;

$a=\frac{F\text{'}}{M}$ Here $F\text{'}=$given friction force

$M=10kg$

$a=\frac{20}{10}=2m/se{c}^2$

Now applying newton's second law on $5kg.$

$$\begin{gathered} F-F\text{'}=ma \\ F-20=5\times 2\text{'} \\ F=30Newton \end{gathered}$$

So, value of force on $5kg$ is $30N$.