Two blocks of masses 10kg and 20kg are connected by a massless spring and are placed on a smooth

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?

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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 $\frac{5}{3}N$