The potential energy of a particle is determined by the expression `U=alpha(x^(2)+y^(2))`, where `alpha` is a positive constant. The particle begins to move from a point with the co-ordinates (3, 3) only under the action of potential fields force. When it reaches the point
(1, 1) its kinetic energy is `4 Kalpha`. Find the value of K.

Answer

To solve the problem, we will follow these steps:

### Step 1: Write down the expression for potential energy
The potential energy \( U \) of the particle is given by:
\[ U = \alpha (x^2 + y^2) \]
where \( \alpha \) is a positive constant.

### Step 2: Calculate the initial potential energy at the point (3, 3)
Substituting the coordinates \( (3, 3) \) into the potential energy equation:
\[ U_{\text{initial}} = \alpha (3^2 + 3^2) = \alpha (9 + 9) = 18\alpha \]

### Step 3: Calculate the final potential energy at the point (1, 1)
Now, substituting the coordinates \( (1, 1) \) into the potential energy equation:
\[ U_{\text{final}} = \alpha (1^2 + 1^2) = \alpha (1 + 1) = 2\alpha \]

### Step 4: Determine the change in potential energy
The change in potential energy \( \Delta U \) as the particle moves from the initial to the final position is:
\[ \Delta U = U_{\text{initial}} - U_{\text{final}} = 18\alpha - 2\alpha = 16\alpha \]

### Step 5: Relate the change in potential energy to kinetic energy
According to the work-energy principle, the change in potential energy is equal to the change in kinetic energy. Therefore, the kinetic energy \( K.E. \) gained by the particle is:
\[ K.E. = \Delta U = 16\alpha \]

### Step 6: Set up the equation for kinetic energy
We are given that the kinetic energy at the point (1, 1) is \( 4K\alpha \). Thus, we can set up the equation:
\[ 4K\alpha = 16\alpha \]

### Step 7: Solve for \( K \)
Dividing both sides by \( \alpha \) (since \( \alpha \) is positive and non-zero):
\[ 4K = 16 \]
Now, divide both sides by 4:
\[ K = 4 \]

### Final Answer
The value of \( K \) is:
\[ \boxed{4} \]