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 find the force exerted by block A on block B, we'll start by analyzing the forces acting on both blocks. Since the surface is frictionless, the only horizontal force acting on block A and block B is the applied force of 10 N.

First, calculate the total mass of the system:

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

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

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

Next, find the acceleration of the system using Newton's second law:

\(F = M \cdot a\)

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

Solving for \(a\) gives:

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

Now, focus on block B. The only force responsible for the acceleration of block B is the force exerted by block A on it.

Use Newton's second law for block B:

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

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

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

Answer 2

Step 1: Calculate the acceleration of the system. 

- The total mass of the system is: M = mA + mB = 2kg + 3kg = 5kg. - Using Newton’s second law: \[ F = M \cdot a \Rightarrow a = \frac{F}{M} = \frac{10}{5} = 2 \text{ m/s}^2 \] Step 2: Calculate the force exerted by block A on block B. - The force on block B due to block A is: \[ F_B = m_B \cdot a = 3 \times 2 = 6N \] Step 3: Conclude. - The force exerted by block A on block B is 6N.