An object of mass 2 Kg is moved from infinity to a poitn P. Initially that object was st rest but on reching P its speed is 2m/s. The workdone in moving that object is -4J. Then potential at P is …………….. J/kg

Options

A. 8
B. -8
C. 4
D. -4

Explanation

To find the potential at point P, we can use the work-energy principle and the relationship between kinetic energy, potential energy, and work done. Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Identify Given Data**:
- Mass of the object (m) = 2 kg
- Initial velocity (V_initial) = 0 m/s (at infinity)
- Final velocity (V_final) = 2 m/s (at point P)
- Work done (W) = -4 J

2. **Calculate Initial Kinetic Energy (KE_initial)**:
\[ KE_{\text{initial}} = \frac{1}{2} m V_{\text{initial}}^2 = \frac{1}{2} \times 2 \, \text{kg} \times (0 \, \text{m/s})^2 = 0 \, \text{J} \]

3. **Calculate Final Kinetic Energy (KE_final)**:
\[ KE_{\text{final}} = \frac{1}{2} m V_{\text{final}}^2 = \frac{1}{2} \times 2 \, \text{kg} \times (2 \, \text{m/s})^2 = \frac{1}{2} \times 2 \times 4 = 4 \, \text{J} \]

4. **Use the Work-Energy Principle**:
The work done on the object is equal to the change in total mechanical energy (kinetic + potential):
\[ W = KE_{\text{final}} + PE_{\text{final}} - (KE_{\text{initial}} + PE_{\text{initial}}) \]
Since the object starts from infinity, we assume:
\[ PE_{\text{initial}} = 0 \, \text{J} \quad \text{(potential energy at infinity)} \]
Thus, the equation simplifies to:
\[ W = KE_{\text{final}} + PE_{\text{final}} - KE_{\text{initial}} \]
Substituting the known values:
\[ -4 = 4 + PE_{\text{final}} - 0 \]

5. **Solve for Potential Energy at Point P (PE_final)**:
Rearranging the equation gives:
\[ PE_{\text{final}} = -4 - 4 = -8 \, \text{J} \]

6. **Calculate Potential at Point P**:
The potential at point P (V_P) is given by the potential energy per unit mass:
\[ V_P = \frac{PE_{\text{final}}}{m} = \frac{-8 \, \text{J}}{2 \, \text{kg}} = -4 \, \text{J/kg} \]

### Final Answer:
The potential at point P is **-4 J/kg**.

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