A person brings a mass of 1 kg from infinity to a point . Initally the mass was at rest but it moves at a speed of 2 `ms^-1` as it reaches A. The work done by the person on the mass is -3J. The potential at A is

Options

A. `-3 "J kg"^(-1)`
B. `-12 "J kg"^(-1)`
C. `-5 "J kg"^(-1)`
D. None of these

Explanation

To solve the problem, we need to find the gravitational potential at point A, given the work done by a person in bringing a mass from infinity to that point, along with the kinetic energy of the mass at point A.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Mass (m) = 1 kg
- Speed at point A (v) = 2 m/s
- Work done by the person (W) = -3 J (negative indicates that work is done against the gravitational field)
- Potential energy at infinity (U_infinity) = 0 J (since potential energy at infinity is conventionally taken as zero)

2. **Calculate the Kinetic Energy at Point A:**
The kinetic energy (K_A) at point A can be calculated using the formula:
\[ K_A = \frac{1}{2} m v^2 \]
Substituting the values:
\[ K_A = \frac{1}{2} \times 1 \, \text{kg} \times (2 \, \text{m/s})^2 = \frac{1}{2} \times 1 \times 4 = 2 \, \text{J} \]

3. **Apply the Work-Energy Principle:**
The work done on the mass can be expressed in terms of the change in total mechanical energy (potential energy + kinetic energy):
\[ W = U_A + K_A - (U_{\infty} + K_{\infty}) \]
Since \(U_{\infty} = 0\) and \(K_{\infty} = 0\), we have:
\[ W = U_A + K_A \]
Substituting the known values:
\[ -3 \, \text{J} = U_A + 2 \, \text{J} \]

4. **Solve for Potential Energy at Point A:**
Rearranging the equation to find \(U_A\):
\[ U_A = -3 \, \text{J} - 2 \, \text{J} = -5 \, \text{J} \]

5. **Conclusion:**
The potential energy at point A is:
\[ U_A = -5 \, \text{J} \]

### Final Answer:
The potential at point A is -5 J.