Question

The blocks A and B weighing 100 N and 250 N respectively are placed one over the other as shown in the figure. Block B rests on a smooth surface. The coefficient of static friction between A and B is 0.4. When \( F = 250 \, {N} \), the acceleration of the upper block is (Take acceleration due to gravity, \( g = 10 \, {m/s}^2 \)):

• A. \( 8.4 \, {ms}^{-2} \)
• B. \( 25 \, {ms}^{-2} \)
• C. \( 6 \, {ms}^{-2} \)
• D. \( 21 \, {ms}^{-2} \)

Hint:

In problems involving friction, always calculate the frictional force first. Use the net force acting on the system to find the acceleration.

Answer 1

The Correct Option is C

We apply Newton's second law to the system of two blocks.

- The mass of block \( A \) is:
\[ m_A = \frac{100}{g} = \frac{100}{10} = 10 \, \text{kg} \] - The mass of block \( B \) is:
\[ m_B = \frac{250}{g} = \frac{250}{10} = 25 \, \text{kg} \] - The force applied is:
\[ F = 250 \, \text{N} \] - The coefficient of friction is:
\[ \mu = 0.4 \] - The frictional force between the blocks is:
\[ f_{\text{friction}} = \mu \cdot N = 0.4 \times 10 \times 10 = 40 \, \text{N} \] The total force on the system is:
\[ F_{\text{total}} = F - f_{\text{friction}} = 250 - 40 = 210 \, \text{N} \] The total mass of the system is:
\[ m_{\text{total}} = 10 + 25 = 35 \, \text{kg} \] Thus, the acceleration is:
\[ a = \frac{F_{\text{total}}}{m_{\text{total}}} = \frac{210}{35} = 6 \, \text{m/s}^2 \]