A body of mass 2 kg is moved from a point A to a point B by an external agent in a conservative force field. It the velocity of the body at the points A and B are 5 m/s and 3 m/s respectively and the work done by the external agents is –10 J, then the change in potential energy between points A and B is-

Options

A. 6J
B. 36J
C. 16J
D. None of these

Answer

To find the change in potential energy (ΔU) between points A and B, we can use the work-energy principle, which states that the work done by external forces is equal to the change in kinetic energy plus the change in potential energy.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the body (m) = 2 kg
- Velocity at point A (v_A) = 5 m/s
- Velocity at point B (v_B) = 3 m/s
- Work done by the external agent (W) = -10 J

2. **Calculate the Initial Kinetic Energy (KE_A) at Point A:**
\[ KE_A = \frac{1}{2} m v_A^2 = \frac{1}{2} \times 2 \, \text{kg} \times (5 \, \text{m/s})^2 \]
\[ KE_A = 1 \times 25 = 25 \, \text{J} \]

3. **Calculate the Final Kinetic Energy (KE_B) at Point B:**
\[ KE_B = \frac{1}{2} m v_B^2 = \frac{1}{2} \times 2 \, \text{kg} \times (3 \, \text{m/s})^2 \]
\[ KE_B = 1 \times 9 = 9 \, \text{J} \]

4. **Calculate the Change in Kinetic Energy (ΔKE):**
\[ \Delta KE = KE_B - KE_A = 9 \, \text{J} - 25 \, \text{J} = -16 \, \text{J} \]

5. **Use the Work-Energy Principle:**
According to the work-energy principle:
\[ W = \Delta KE + \Delta U \]
Rearranging gives:
\[ \Delta U = W - \Delta KE \]

6. **Substitute the Values:**
\[ \Delta U = -10 \, \text{J} - (-16 \, \text{J}) = -10 \, \text{J} + 16 \, \text{J} = 6 \, \text{J} \]

### Final Answer:
The change in potential energy (ΔU) between points A and B is **6 J**.