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 coordinates `(3, 3)`, only under the action of potential field force. Then its kinetic energy T at the instant when the particle is at a point with the coordinates `(1,1)` is

Options

A. (a) `8alpha`
B. (b) `24alpha`
C. (c) `16alpha`
D. (d) Zero

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy of the system remains constant when only conservative forces are acting on the particle.

### Step-by-Step Solution:

1. **Identify the Potential Energy Function**:
The potential energy \( U \) of the particle is given by:
\[ U = \alpha (x^2 + y^2) \]
where \( \alpha \) is a positive constant.

2. **Calculate Initial Potential Energy**:
The particle starts from the point \( (3, 3) \). We can calculate the initial potential energy \( U_i \):
\[ U_i = \alpha (3^2 + 3^2) = \alpha (9 + 9) = 18\alpha \]

3. **Calculate Final Potential Energy**:
The particle moves to the point \( (1, 1) \). We will calculate the final potential energy \( U_f \):
\[ U_f = \alpha (1^2 + 1^2) = \alpha (1 + 1) = 2\alpha \]

4. **Apply Conservation of Mechanical Energy**:
According to the conservation of mechanical energy:
\[ \Delta T + \Delta U = 0 \]
This implies:
\[ \Delta T = -\Delta U \]
where \( \Delta U = U_f - U_i \).

5. **Calculate Change in Potential Energy**:
Now we can find \( \Delta U \):
\[ \Delta U = U_f - U_i = 2\alpha - 18\alpha = -16\alpha \]

6. **Calculate Change in Kinetic Energy**:
Substituting \( \Delta U \) into the equation for \( \Delta T \):
\[ \Delta T = -(-16\alpha) = 16\alpha \]

7. **Conclusion**:
The kinetic energy \( T \) at the point \( (1, 1) \) is:
\[ T = 16\alpha \]

### Final Answer:
The kinetic energy \( T \) at the instant when the particle is at the point \( (1, 1) \) is \( 16\alpha \).

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