Question

A box of mass 15 kg is kept on the floor of a stationary trolley. The coefficient of static friction between the box and the trolley is 0.12. Keeping the box in stationary state over the trolley, the maximum acceleration with which the trolley can be moved horizontally in m s⁻² is: ____.

• A. 1.2
• B. 1.8
• C. 1.5
• D. 2.1

Hint:

Notice that the mass of the box (15 kg) is irrelevant to the final answer. In friction-limited acceleration problems, the mass always cancels out!

Answer 1

The Correct Option is A

Step 1: Understanding the Topic:
This problem involves "Laws of Motion," specifically focusing on friction in non-inertial frames. When a trolley accelerates, an object on it experiences a "pseudo force." Friction acts to prevent the object from sliding due to this force.
Step 2: Key Formulas and Approach:

Pseudo Force ($F_p$) = $m \times a$ (acting opposite to acceleration).
Limiting Static Friction ($f_{max}$) = $\mu_s \times N = \mu_s \times m \times g$.
For the box to remain stationary relative to the trolley: $F_p \leq f_{max}$.

Step 3: Detailed Explanation:

Analyze forces: As the trolley accelerates forward with acceleration '$a$', the box (mass $m$) feels a force $ma$ pushing it backward relative to the trolley. To stop this motion, the static friction $f_s$ acts forward.
Find the limit: The box starts to slip when the required force to keep it stationary exceeds the maximum possible friction the surface can provide. \[ m \cdot a = \mu_s \cdot m \cdot g \]
Simplify: Notice that the mass '$m$' appears on both sides of the equation. Dividing both sides by $m$: \[ a = \mu_s \cdot g \]
Calculate: Substitute the given values ($\mu_s = 0.12$ and $g = 10 \text{ m/s}^2$): \[ a = 0.12 \times 10 = 1.2 \text{ m/s}^2 \]
Even though the mass was given as 15 kg, it does not affect the maximum possible acceleration.
Step 4: Final Answer:
The maximum acceleration is 1.2 m/s².