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 problem, we need to find the force vector \( \vec{F} \) given the potential energy \( U = -kxy \). Here are the steps to derive the force from the potential energy:

### Step 1: Understand the relationship between force and potential energy
The force \( \vec{F} \) is related to the potential energy \( U \) by 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. **Calculate \( \frac{\partial U}{\partial x} \)**:
\[ \frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(-kxy) = -ky \]

2. **Calculate \( \frac{\partial U}{\partial y} \)**:
\[ \frac{\partial U}{\partial y} = \frac{\partial}{\partial y}(-kxy) = -kx \]

### Step 3: Write the force vector \( \vec{F} \)
Using the results from the gradient:
\[ \vec{F} = -\left(\frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j}\right) \]
Substituting the partial derivatives:
\[ \vec{F} = -\left(-ky \hat{i} - kx \hat{j}\right) = ky \hat{i} + kx \hat{j} \]

### Step 4: Conclude the expression for the force
Thus, the force vector \( \vec{F} \) is:
\[ \vec{F} = k(y \hat{i} + x \hat{j}) \]

### Step 5: Determine if the force is conservative
To check if the force is conservative, we need to compute the curl of the force \( \vec{F} \):
\[ \nabla \times \vec{F} = 0 \]
This can be shown by calculating the determinant of the matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of \( \vec{F} \):
\[ \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & 0 \\ ky & kx & 0 \end{vmatrix} \]
Calculating this determinant gives:
\[ \frac{\partial}{\partial x}(kx) - \frac{\partial}{\partial y}(ky) = k - k = 0 \]
Since the curl is zero, the force is conservative.

### Final Answer
Thus, the force \( \vec{F} \) is:
\[ \vec{F} = k(y \hat{i} + x \hat{j}) \]
And the force is conservative.

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