Question

A horizontal force of 10 N is applied to a block A as shown in the figure. The mass of blocks A and B are 2 kg and 3 kg, respectively. The blocks slide over a surface where friction coefficient is 0.2. The contact force exerted by block A on block B is

• A. Zero
• B. 6 N
• C. 4 N
• D. 10 N

Answer 1

Answer: (B)

6 N

Answer 2

The contact force exerted by block A on block B can be found by analyzing the forces acting on each block.

1. Calculate the friction forces:

   • Friction on A: $f_A = \mu m_A g = 0.2 \times 2 \times 10 = 4\ \text{N}$
   • Friction on B: $f_B = \mu m_B g = 0.2 \times 3 \times 10 = 6\ \text{N}$

2. Analyze forces on Block B:

   • The only horizontal forces on Block B are the contact force $F_{AB}$ (exerted by A on B) to the right and friction $f_B$ to the left.
   • Equation: $F_{AB} - f_B = m_B a$

3. Analyze the system as a whole:

   • Total mass $= 2 + 3 = 5\ \text{kg}$
   • Total friction $= 4 + 6 = 10\ \text{N}$
   • Net force: $10\ \text{N (applied)} - 10\ \text{N (friction)} = 0$
   • Therefore, the acceleration $a = 0$

4. Determine the contact force:

   • Substitute $a = 0$ into the equation for Block B: $F_{AB} - 6 = 3 \times 0$ $F_{AB} = 6\ \text{N}$

Therefore, the contact force exerted by block A on block B is 6 N.