The potential energy of a particle of mass 2 kg moving in a plane is given by `U = (-6x -8y)J`. The position coordinates x and y are measured in meter. If the particle is initially at rest at position (6, 4)m, then

Options

A. Its acceleration is of magnitude 5 m/s2
B. It crosses the y-axis at (x = 0, y = 6 m)
C. Its speed when it crosses the y-axis is 10 m/s
D. Both (1) & (3)

Answer

To solve the problem, we need to analyze the potential energy function provided and derive the necessary quantities step by step.

### Step 1: Identify the potential energy function
The potential energy \( U \) of the particle is given by:
\[ U = -6x - 8y \quad \text{(in Joules)} \]
where \( x \) and \( y \) are the position coordinates in meters.

### Step 2: Calculate the force acting on the particle
The force \( \vec{F} \) acting on the particle can be derived from the potential energy using the relation:
\[ \vec{F} = -\nabla U \]
This means we need to compute the partial derivatives of \( U \) with respect to \( x \) and \( y \).

Calculating the partial derivatives:
\[ F_x = -\frac{\partial U}{\partial x} = -(-6) = 6 \quad \text{(in Newtons)} \]
\[ F_y = -\frac{\partial U}{\partial y} = -(-8) = 8 \quad \text{(in Newtons)} \]

Thus, the force vector is:
\[ \vec{F} = 6 \hat{i} + 8 \hat{j} \quad \text{(in Newtons)} \]

### Step 3: Calculate the magnitude of the force
The magnitude of the force \( F \) can be calculated using the Pythagorean theorem:
\[ F = \sqrt{F_x^2 + F_y^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \quad \text{(in Newtons)} \]

### Step 4: Calculate the acceleration of the particle
Using Newton's second law, the acceleration \( \vec{a} \) can be calculated as:
\[ \vec{a} = \frac{\vec{F}}{m} \]
where \( m = 2 \, \text{kg} \).

Thus, the acceleration is:
\[ \vec{a} = \frac{10}{2} = 5 \quad \text{(in m/s}^2\text{)} \]

### Step 5: Analyze the motion of the particle
Since both components of the force are positive, the particle will accelerate in both the \( x \) and \( y \) directions. Therefore, it will not cross the x-axis or y-axis as both coordinates will continue to increase.

### Conclusion
From the analysis, we find that:
- The acceleration of the particle is \( 5 \, \text{m/s}^2 \).
- The particle will not cross the y-axis or x-axis.

### Final Result
The only correct option is:
- Option A: The acceleration is \( 5 \, \text{m/s}^2 \).