Problem 142 - A Complete Resource Book in Physics for JEE Main

Question

Two blocks of masses \(5 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) are placed on a smooth horizontal surface. A horizontal force \(F=16 \mathrm{~N}\) is applied on \(5 \mathrm{~kg}\) as shown. Find normal force between the blocks.

Step-by-Step Solution

Answer

The normal force between the 5 kg and 3 kg blocks is \(6 \mathrm{~N}\).

Step 1: Find the acceleration of the blocks

To find the acceleration of the blocks, we can use Newton's second law of motion for the entire system, where the force applied on the 5 kg block causes the blocks to accelerate together (since there is no slipping between them). We can write the equation as: \[F = (m_1 + m_2)a\] Where \(F\) is the applied force, \(m_1\) and \(m_2\) are the masses of the 5 kg and 3 kg blocks respectively, and \(a\) is the acceleration of the blocks. Plugging in the values provided in the problem statement, we can solve for the acceleration: \[16 \mathrm{~N} = (5 \mathrm{~kg} + 3 \mathrm{~kg})a\]

Step 2: Find the value of acceleration

Divide both sides of the equation by the sum of the masses to get the value of acceleration: \[a = \frac{16 \mathrm{~N}}{5 \mathrm{~kg} + 3 \mathrm{~kg}} = \frac{16 \mathrm{~N}}{8 \mathrm{~kg}} = 2 \mathrm{~\frac{m}{s^2}}\] The acceleration of the blocks together is 2 m/s².

Step 3: Find the normal force between the blocks

Now we can use Newton's second law of motion for the 3 kg block in the horizontal direction. The only horizontal force acting on the 3 kg block is the normal force of the 5 kg block \(N\). The equation can be written as: \[N = m_2 a\] Plug in the values for \(m_2\) and \(a\) to find the value of the normal force \(N\): \[N = (3 \mathrm{~kg})(2 \mathrm{~\frac{m}{s^2}}) = 6 \mathrm{~N}\] The normal force between the 5 kg and 3 kg blocks is 6 N.