The potential energy U for a force field `vec (F)` is such that `U=-`kxy where K is a constant . Then

Options

A. `vec( F)=ky hat(i)+kx hat(j)`
B. `kx hat(i)+ky hat(j)`
C. the force `vec(F)` is a conservative force
D. the force `vec(F)` is a non-conservative force

Answer

To solve the question regarding the potential energy \( U = -kxy \) and to find the corresponding force \( \vec{F} \), we can follow these steps:

### Step 1: Understand the relationship between potential energy and force
The force \( \vec{F} \) in a conservative field can be derived from the potential energy \( U \) using the formula:
\[ \vec{F} = -\nabla U \]
where \( \nabla U \) is the gradient of the potential energy.

### Step 2: Calculate the gradient of \( U \)
Given:
\[ U = -kxy \]
We need to compute the partial derivatives of \( U \) with respect to \( x \) and \( y \).

1. **Partial derivative with respect to \( x \)**:
\[ \frac{\partial U}{\partial x} = -k \cdot y \]

2. **Partial derivative with respect to \( y \)**:
\[ \frac{\partial U}{\partial y} = -k \cdot x \]

### Step 3: Write the force components
Now, substituting these derivatives into the formula for force:
\[ \vec{F} = -\left( \frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j} \right) \]
This gives us:
\[ \vec{F} = -\left( -ky \hat{i} - kx \hat{j} \right) = ky \hat{i} + kx \hat{j} \]

### Step 4: Final expression for the force
Thus, the force \( \vec{F} \) can be expressed as:
\[ \vec{F} = ky \hat{i} + kx \hat{j} \]

### Step 5: Check if the force is conservative
To check if the force is conservative, we can compute the curl of \( \vec{F} \):
\[ \nabla \times \vec{F} = 0 \]
If the curl is zero, the force is conservative.

1. **Set up the curl**:
\[ \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ ky & kx & 0 \end{vmatrix} \]

2. **Calculate the determinant**:
\[ \nabla \times \vec{F} = \hat{i} \left( \frac{\partial (0)}{\partial y} - \frac{\partial (kx)}{\partial z} \right) - \hat{j} \left( \frac{\partial (0)}{\partial x} - \frac{\partial (ky)}{\partial z} \right) + \hat{k} \left( \frac{\partial (kx)}{\partial x} - \frac{\partial (ky)}{\partial y} \right) \]
This simplifies to:
\[ \nabla \times \vec{F} = \hat{k} (k - k) = 0 \]

Since the curl is zero, the force \( \vec{F} \) is conservative.

### Conclusion
The force \( \vec{F} \) corresponding to the potential energy \( U = -kxy \) is:
\[ \vec{F} = ky \hat{i} + kx \hat{j} \]
And it is a conservative force.

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