Consider the situations of the previous proble. Let the water push the left wass by a force `F_1` and the right wall byi as force `F_2`.

Options

A. `F_1=F_2`
B. `F_1gtF_2`
C. `F_1ltF_2`
D. the information in insufficient to know the relation between `F_1 and F_2`

Answer

To solve the problem, we need to analyze the forces acting on the water in the box that is accelerating to the right. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Situation
The box is filled with water and is accelerating to the right with an acceleration \( a \). As a result, the water inside the box will experience a pseudo force acting to the left due to the acceleration of the box.

### Step 2: Identify the Forces
- The force \( F_1 \) is the force exerted by the water on the left wall of the box.
- The force \( F_2 \) is the force exerted by the water on the right wall of the box.
- The pseudo force acting on the water due to the acceleration of the box is given by \( F_{pseudo} = m \cdot a \), where \( m \) is the mass of the water.

### Step 3: Apply Newton's Second Law
Since the water is not moving relative to the box, we can apply Newton's second law to the water:
- The net force acting on the water in the horizontal direction must be zero.
- Therefore, we can write the equation:
\[ F_2 - F_1 - F_{pseudo} = 0 \]
Rearranging gives:
\[ F_2 = F_1 + F_{pseudo} \]

### Step 4: Substitute the Pseudo Force
Substituting the expression for the pseudo force:
\[ F_2 = F_1 + m \cdot a \]

### Step 5: Analyze the Relationship
From the equation \( F_2 = F_1 + m \cdot a \), we can see that:
- \( F_2 \) is greater than \( F_1 \) because \( m \cdot a \) is a positive quantity.
- Therefore, we conclude that:
\[ F_1 < F_2 \]

### Conclusion
The relationship between the forces is:
\[ F_1 < F_2 \]