The potential energy of a force filed `vec(F)` is given by `U(x,y)=sin(x+y)`. The force acting on the particle of mass m at `(0,(pi)/4)` is

Options

A. 1
B. `sqrt(2)`
C. `1/(sqrt(2))`
D. 0

Explanation

To find the force acting on a particle of mass \( m \) at the point \( (0, \frac{\pi}{4}) \) given the potential energy function \( U(x,y) = \sin(x+y) \), we can follow these steps:

### Step 1: Understand the relationship between force and potential energy
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 function.

### Step 2: Calculate the gradient of \( U(x,y) \)
The gradient \( \nabla U \) in two dimensions is given by:
\[ \nabla U = \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y} \right) \]
We need to compute the partial derivatives of \( U(x,y) = \sin(x+y) \).

- **Partial derivative with respect to \( x \)**:
\[ \frac{\partial U}{\partial x} = \cos(x+y) \]

- **Partial derivative with respect to \( y \)**:
\[ \frac{\partial U}{\partial y} = \cos(x+y) \]

Thus, we have:
\[ \nabla U = \left( \cos(x+y), \cos(x+y) \right) \]

### Step 3: Write the force vector
Using the gradient calculated above, we can express the force vector:
\[ \vec{F} = -\nabla U = -\left( \cos(x+y), \cos(x+y) \right) = \left( -\cos(x+y), -\cos(x+y) \right) \]

### Step 4: Substitute the point \( (0, \frac{\pi}{4}) \)
Now we substitute \( x = 0 \) and \( y = \frac{\pi}{4} \) into the force vector:
\[ \vec{F} = \left( -\cos(0 + \frac{\pi}{4}), -\cos(0 + \frac{\pi}{4}) \right) \]
Calculating \( \cos(\frac{\pi}{4}) \):
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus, the force vector becomes:
\[ \vec{F} = \left( -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right) \]

### Step 5: Final result
The force acting on the particle at the point \( (0, \frac{\pi}{4}) \) is:
\[ \vec{F} = -\frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} \]