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/s^2`
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 given and find the force acting on the particle, then determine its acceleration.

### Step 1: Identify the Potential Energy Function
The potential energy \( U \) of the particle is given by:
\[ U = -6x - 8y \quad \text{(in Joules)} \]

### Step 2: Calculate the Force Acting on the Particle
The force \( \mathbf{F} \) acting on the particle can be found using the negative gradient of the potential energy:
\[ \mathbf{F} = -\nabla U \]
Calculating the partial derivatives:
\[ F_x = -\frac{\partial U}{\partial x} = -(-6) = 6 \, \text{N} \]
\[ F_y = -\frac{\partial U}{\partial y} = -(-8) = 8 \, \text{N} \]
Thus, the force vector is:
\[ \mathbf{F} = 6 \hat{i} + 8 \hat{j} \, \text{N} \]

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

### Step 4: Calculate the Acceleration of the Particle
Using Newton's second law, the acceleration \( \mathbf{a} \) can be calculated as:
\[ \mathbf{a} = \frac{\mathbf{F}}{m} \]
Where \( m = 2 \, \text{kg} \):
\[ \mathbf{a} = \frac{10 \, \text{N}}{2 \, \text{kg}} = 5 \, \text{m/s}^2 \]

### Step 5: Determine the Direction of Motion
Since the force has components in both the x and y directions, the particle will accelerate in both directions. However, we need to analyze the initial position of the particle, which is at (6, 4) m.

### Conclusion
The particle experiences an acceleration of \( 5 \, \text{m/s}^2 \) in the direction of the force, which is directed towards the positive x and y axes. The particle will move towards the origin along the line defined by the force vector.