Question

Two blocks of masses $2kg$ and $1kg$ are in contact with each other on a frictionless table, when a horizontal force of $3.0N$ is applied to the block of mass $2kg$ the value of the force of contact between the two blocks is:

• A. $4N$
• B. $3N$
• C. $2N$
• D. $1N$

Answer 1

Answer: (D)

$1N$

Considering both blocks as a system,

Total mass of the system M = 3 kg

So, net external force for the system = 3 N (as the table is frictionless)

F = Ma

=> 3 = 3a

=> a = 1 m$s^{-2}$

Now considering the block of 1 kg :

a = 1 m$s^{-2}$ (acceleration of the 1 kg block is same as the acceleration of the system)

so, F = ma

=> F = 1 kg x 1 m$s^{-2}$ = 1 N