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 will follow these steps:

### Step 1: Calculate the Gradient of the Potential Energy
The force \( \vec{F} \) in a conservative field is related to the potential energy \( U \) by the equation:
\[ \vec{F} = -\nabla U \]
where \( \nabla U \) is the gradient of \( U \).

### Step 2: Compute the Partial Derivatives
We need to compute the partial derivatives of \( U(x, y) \):
1. \( \frac{\partial U}{\partial x} \)
2. \( \frac{\partial U}{\partial y} \)

Given \( U(x, y) = \sin(x + y) \):
- For \( \frac{\partial U}{\partial x} \):
\[ \frac{\partial U}{\partial x} = \cos(x + y) \]
- For \( \frac{\partial U}{\partial y} \):
\[ \frac{\partial U}{\partial y} = \cos(x + y) \]

### Step 3: Evaluate the Partial Derivatives at the Point \( (0, \frac{\pi}{4}) \)
Now we substitute \( x = 0 \) and \( y = \frac{\pi}{4} \):
- Calculate \( x + y \):
\[ x + y = 0 + \frac{\pi}{4} = \frac{\pi}{4} \]
- Now evaluate the partial derivatives:
\[ \frac{\partial U}{\partial x} \bigg|_{(0, \frac{\pi}{4})} = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
\[ \frac{\partial U}{\partial y} \bigg|_{(0, \frac{\pi}{4})} = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Step 4: Calculate the Force Vector
Using the gradient values, we can find the force vector:
\[ \vec{F} = -\left(\frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j}\right) \]
Substituting the values we found:
\[ \vec{F} = -\left(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}\right) = -\frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} \]

### Step 5: Magnitude of the Force
Now, we can find the magnitude of the force vector:
\[ |\vec{F}| = \sqrt{\left(-\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1 \]

### Final Result
Thus, the force acting on the particle at the point \( (0, \frac{\pi}{4}) \) is:
\[ \vec{F} = -\frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} \]
and its magnitude is \( 1 \, \text{N} \).