A horizontal force 10 N is applied to a block A as shown in figure. The mass of blocks A and B are 2 kg and 3 kg, respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is :

Using Newton’s second law for the whole system:

$a\; =\; \backslash frac\{F\_\{\backslash text\{ext\}\}\}\{m\_\{\backslash text\{total\}\}\}$

where F_ext = 10 N and m_total = m_A + m_B = 2 kg + 3 kg = 5 kg.

Calculate the acceleration:

$a\; =\; \backslash frac\{10\backslash \; \backslash text\{N\}\}\{5\backslash \; \backslash text\{kg\}\}\; =\; 2\backslash \; \backslash text\{m/s\}^2$

Now consider block B alone. The only horizontal force acting on block B is the contact force exerted by block A, which we denote as F_AB. Applying Newton’s second law to block B:

$F\_\{AB\}\; =\; m\_\{B\}\backslash ,\; a$

Substituting m_B = 3 kg and a = 2 m/s²:

$F\_\{AB\}\; =\; 3\backslash \; \backslash text\{kg\}\; \backslash times\; 2\backslash \; \backslash text\{m/s\}^2\; =\; 6\backslash \; \backslash text\{N\}$

Therefore, the force that block A exerts on block B is **6 N**.