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. If 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 agent is - 10 J, then the change in potential energy between point A and B is

Options

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

Answer

To find the change in potential energy between points A and B, we can use the work-energy principle, which states that the work done by the external agent is equal to the change in total mechanical energy (kinetic + potential energy) of the body.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the body, \( m = 2 \, \text{kg} \)
- Velocity at point A, \( v_A = 5 \, \text{m/s} \)
- Velocity at point B, \( v_B = 3 \, \text{m/s} \)
- Work done by the external agent, \( W = -10 \, \text{J} \)

2. **Calculate the initial kinetic energy (KE) at point A:**
\[ KE_A = \frac{1}{2} m v_A^2 = \frac{1}{2} \times 2 \times (5)^2 = \frac{1}{2} \times 2 \times 25 = 25 \, \text{J} \]

3. **Calculate the final kinetic energy (KE) at point B:**
\[ KE_B = \frac{1}{2} m v_B^2 = \frac{1}{2} \times 2 \times (3)^2 = \frac{1}{2} \times 2 \times 9 = 9 \, \text{J} \]

4. **Use the work-energy principle:**
The work done by the external agent is equal to the change in mechanical energy:
\[ W = KE_B + PE_B - (KE_A + PE_A) \]
Rearranging gives:
\[ \Delta U = W + (KE_A - KE_B) \]

5. **Substituting the values:**
\[ \Delta U = -10 + (25 - 9) \]
\[ \Delta U = -10 + 16 = 6 \, \text{J} \]

6. **Conclusion:**
The change in potential energy between point A and B is \( \Delta U = 6 \, \text{J} \).