Question

Two blocks of mass 2 kg and 4 kg are kept in contact with each other on a smooth horizontal surface. A horizontal force of 12 N is applied on the first block due to which they move with certain constant acceleration. Calculate the force between the blocks.

• A. 2 N
• B. 4 N
• C. 8 N
• D. 16 N

Answer 1

The correct option is C

8 N

Let a be the common acceleration of blocks.
R = Force between the two blocks
From FBD of block 1
F - R = $m_1$a . . . (i)
From FBD of block 2,
R = $m_2$ a . . . (ii)
$\therefore$ From (i) & (ii), $F=(m_1+m_2)a$
$\therefore a=\frac{F}{m_1+m_2}=\frac{12}{6}=2m/se{c}^2$
$\therefore$ Force between blocks $a=(4\times 2)N=8N$