Question

On a smooth horizontal table, 2 blocks of mass 20 kg and 10 kg are placed in contact. A horizontal force of 5 N is applied to the 20 kg block. With what force will this block press the 10 kg block?

Answer 1

According to the given data and the situation of the blocks we can draw the figure as

Since the surface is frictionless, both the blocks will move with a common

acceleration, $a=\frac{5}{20+10}=\frac{1}{6}m/s^2$

Now,
Drawing the FBD of both blocks,

Now we can write equation of force
For the 10 kg block:
$N=10a=10\times \frac{1}{6}=\frac{5}{3}newtons$

If we take 20 kg block then

$F-N=20\times \frac{1}{6}$

$Or,5-N=\frac{10}{3}$

$Or,N=\frac{5}{3}N$

Hence the force exerted by blocks at each other is $\frac{5}{3}N$