Question

A horizontal force 10 N is applied to a block 4 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:

• A. Zero
• B. \(4N\)
• C. \(6N\)
• D. \(10N\)

Answer 1

The Correct Option is C

To determine the force block A exerts on block B, we will first analyze the forces on both blocks. Given a frictionless surface, the sole horizontal force acting on each block is the 10 N applied force.

The total mass of the system is calculated as follows:

\(m_A = 2 \text{ kg}\)

\(m_B = 3 \text{ kg}\)

Total mass, \(M = m_A + m_B = 2 \text{ kg} + 3 \text{ kg} = 5 \text{ kg}\)

Next, the system's acceleration is found using Newton's second law:

\(F = M \cdot a\)

\(10 \text{ N} = 5 \text{ kg} \cdot a\)

Solving for \(a\) yields:

\(a = \frac{10 \text{ N}}{5 \text{ kg}} = 2 \text{ m/s}^2\)

Now, we consider block B. The acceleration of block B is solely due to the force applied by block A.

Applying Newton's second law to block B:

\(F_{AB} = m_B \cdot a\)

\(F_{AB} = 3 \text{ kg} \cdot 2 \text{ m/s}^2 = 6 \text{ N}\)

Therefore, the force exerted by block A on block B is \(6 \, \text{N}\).