The potential energy `phi` in joule of a particle of mass `1 kg` moving in x-y plane obeys the law, `phi=3x + 4y`. Here, x and y are in metres. If the particle is at rest at `(6m, 8m)` at time 0, then the work done by conservative force on the particle from the initial position to the instant when it crosses the x-axis is . a) 25 J b) 25 J c) 50 J d) - 50 J

Options

A. `25 J`
B. `25 J`
C. `50 J`
D. `-50 J`

Explanation

To solve the problem, we need to calculate the work done by the conservative force on the particle as it moves from its initial position to the point where it crosses the x-axis. The potential energy function is given as:

\[ \phi = 3x + 4y \]

1. **Determine the initial potential energy**:
The particle starts at the position \((6m, 8m)\). We can substitute these values into the potential energy function to find the initial potential energy (\(\phi_i\)).

\[ \phi_i = 3(6) + 4(8) \]
\[ \phi_i = 18 + 32 = 50 \text{ J} \]

2. **Determine the final potential energy**:
The particle crosses the x-axis when \(y = 0\). We need to find the x-coordinate at this point. Since the potential energy is a function of x and y, we can express the final potential energy (\(\phi_f\)) as:

\[ \phi_f = 3x + 4(0) = 3x \]

We need to find the x-coordinate when the particle crosses the x-axis. The particle moves from \((6m, 8m)\) to \((x, 0)\). The work done by the conservative force is equal to the change in potential energy:

\[ W = \phi_i - \phi_f \]

3. **Calculate the work done**:
We can express the work done in terms of the initial and final potential energies:

\[ W = 50 - \phi_f \]

Since we don't have the exact value of \(x\) when it crosses the x-axis, we can assume that the particle crosses the x-axis at some point, say \(x = 6\) (it could be any point, but for this calculation, we will take the initial x-coordinate). Thus:

\[ \phi_f = 3(6) = 18 \text{ J} \]

Now substituting back into the work done equation:

\[ W = 50 - 18 = 32 \text{ J} \]

However, since we need to find the work done until it crosses the x-axis, we need to consider the potential energy when it reaches \(y=0\) at some point. If we consider the particle moving to \(x = 0\):

\[ \phi_f = 3(0) + 4(0) = 0 \text{ J} \]

Then the work done becomes:

\[ W = 50 - 0 = 50 \text{ J} \]

4. **Final answer**:
The work done by the conservative force on the particle from the initial position to the instant when it crosses the x-axis is:

\[ \boxed{50 \text{ J}} \]