The potential energy function of a particle due to some gravitational field is given by `U=6x+4y`. The mass of the particle is `1kg` and no other force is acting on the particle. The particle was initially at rest at a point `(6,8)`. Then the time (in sec). after which this particle is going to cross the `'x'` axis would be:

Explanation

To solve the problem, we need to determine the time it takes for a particle to cross the x-axis given its potential energy function and initial conditions.

### Step-by-Step Solution:

1. **Identify the Potential Energy Function:**
The potential energy function is given as:
\[ U = 6x + 4y \]

2. **Calculate the Force Acting on the Particle:**
The force acting on the particle can be derived from the potential energy function using the formula:
\[ \vec{F} = -\nabla U \]
where \(\nabla U\) is the gradient of the potential energy function.

The components of the force are:
\[ F_x = -\frac{\partial U}{\partial x} = -6 \]
\[ F_y = -\frac{\partial U}{\partial y} = -4 \]

Thus, the force vector is:
\[ \vec{F} = -6 \hat{i} - 4 \hat{j} \]

3. **Determine the Acceleration of the Particle:**
Using Newton's second law, \(F = ma\), we can find the acceleration:
\[ a_x = \frac{F_x}{m} = \frac{-6}{1} = -6 \, \text{m/s}^2 \]
\[ a_y = \frac{F_y}{m} = \frac{-4}{1} = -4 \, \text{m/s}^2 \]

4. **Initial Conditions:**
The particle starts at rest at the point \((6, 8)\), which means:
\[ u_x(0) = 0 \quad \text{and} \quad u_y(0) = 0 \]

5. **Use Kinematic Equations to Find Time to Cross the X-axis:**
The particle will cross the x-axis when \(y = 0\). We can use the kinematic equation for motion in the y-direction:
\[ y = y_0 + u_y(0)t + \frac{1}{2} a_y t^2 \]
Substituting the known values:
\[ 0 = 8 + 0 \cdot t + \frac{1}{2}(-4)t^2 \]
Simplifying gives:
\[ 0 = 8 - 2t^2 \]
Rearranging:
\[ 2t^2 = 8 \quad \Rightarrow \quad t^2 = 4 \]
Taking the square root:
\[ t = \sqrt{4} = 2 \, \text{seconds} \]

### Final Answer:
The particle will cross the x-axis after **2 seconds**.