A person brings a mass of `1 kg` from infinity to a point `A`. Initially the mass was at rest but it moves with a speed of `2 m//s` as it reaches `A`. The work done by the person on a mass is `-3 J`. The potential of `A` is:

Answer

To find the potential at point A, we can use the work-energy principle, which states that the work done by an external agent is equal to the change in kinetic energy plus the change in potential energy.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass (m) = 1 kg
- Initial speed (u) = 0 m/s (at infinity)
- Final speed (v) = 2 m/s (at point A)
- Work done by the person (W) = -3 J

2. **Calculate the Change in Kinetic Energy (ΔKE):**
The formula for kinetic energy (KE) is:
\[ KE = \frac{1}{2} m v^2 \]
- Initial kinetic energy at infinity (KE_initial) = 0 (since the mass is at rest)
- Final kinetic energy at point A (KE_final) = \(\frac{1}{2} \times 1 \times (2)^2 = \frac{1}{2} \times 1 \times 4 = 2 \, \text{J}\)

Therefore, the change in kinetic energy (ΔKE) is:
\[ \Delta KE = KE_{final} - KE_{initial} = 2 \, \text{J} - 0 \, \text{J} = 2 \, \text{J} \]

3. **Apply the Work-Energy Principle:**
According to the work-energy principle:
\[ W = \Delta PE + \Delta KE \]
where ΔPE is the change in potential energy.

Rearranging the equation gives us:
\[ \Delta PE = W - \Delta KE \]

4. **Substitute the Known Values:**
Substitute the values of W and ΔKE into the equation:
\[ \Delta PE = -3 \, \text{J} - 2 \, \text{J} = -5 \, \text{J} \]

5. **Determine the Potential at Point A (VA):**
The potential energy at infinity is considered to be zero. Therefore, the potential at point A (VA) is equal to the change in potential energy:
\[ VA = \Delta PE = -5 \, \text{J} \]

### Final Answer:
The potential at point A is \( VA = -5 \, \text{J} \).