Question

A block of mass \(m\) is placed in equilibrium on a moving horizontal plank. The maximum horizontal acceleration of the plank for \(\mu = 0.2\) is (Acceleration due to gravity = \(10 \, {ms}^{-2}\)).

• A. 2 ms\(^{-2}\)
• B. 3 ms\(^{-2}\)
• C. 4 ms\(^{-2}\)
• D. 5 ms\(^{-2}\)

Hint:

Remember that the maximum acceleration before slipping occurs is directly proportional to the coefficient of friction and the acceleration due to gravity.

Answer 1

The Correct Option is A

The frictional force \( f \) between the block and the plank is given by:

\[ f = \mu mg \]

where:

• \( \mu = 0.2 \) is the coefficient of friction,
• \( m \) is the mass of the block,
• \( g = 10 \, \text{m/s}^2 \) is the acceleration due to gravity.

For the block to remain in equilibrium on the plank, the frictional force must equal the force required to accelerate the block with the plank. Let the maximum horizontal acceleration of the plank be \( a \).

The force required to accelerate the block is:

\[ F = ma \]

The frictional force provides the maximum force that can be applied without the block slipping. Hence, the maximum acceleration \( a_{\text{max}} \) of the plank is:

\[ ma_{\text{max}} = \mu mg \]

Simplifying:

\[ a_{\text{max}} = \mu g = 0.2 \times 10 = 2 \, \text{m/s}^2 \]

Thus, the correct answer is Option (1), with the maximum horizontal acceleration being \( 2 \, \text{m/s}^2 \).