Maximum Trolley Acceleration Calculation
The problem asks for the maximum horizontal acceleration ($a_{max}$) a trolley can achieve while a box placed on it remains stationary relative to the trolley. This scenario is governed by the principles of static friction.
Physics Principles Involved
For the box to remain stationary on the accelerating trolley, the static friction force ($f_s$) acting on the box must provide the necessary centripetal force (or in this case, the force causing linear acceleration).
- According to Newton's second law, the net force on the box horizontally is $F_{net} = m \times a$.
- The static friction force provides this net force: $f_s = m \times a$.
- The static friction force has a maximum value, $f_{s,max}$, given by $f_{s,max} = \mu_s \times N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force.
- Since the trolley is moving horizontally and the box is on it, the normal force ($N$) exerted by the trolley on the box equals the weight of the box ($mg$), assuming no vertical acceleration. Thus, $N = mg$.
- Substituting $N = mg$ into the equation for maximum static friction gives: $f_{s,max} = \mu_s \times mg$.
Determining Maximum Acceleration
To find the maximum acceleration ($a_{max}$) the trolley can have while keeping the box stationary, we set the required force ($m \times a_{max}$) equal to the maximum available static friction force ($f_{s,max}$):
$m \times a_{max} = f_{s,max}$
$m \times a_{max} = \mu_s \times mg$
The mass of the box ($m$) cancels out from both sides:
$a_{max} = \mu_s \times g$
Calculation
Given:
- Coefficient of static friction, $\mu_s = 0.12$
- Acceleration due to gravity, $g = 10 \text{ m/s}^2$
Substitute the values into the formula:
$a_{max} = 0.12 \times 10 \text{ m/s}^2$
$a_{max} = 1.2 \text{ m/s}^2$
Therefore, the maximum acceleration with which the trolley can be moved horizontally while keeping the box stationary is $1.2 \text{ m/s}^2$. This corresponds to Option A.
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