A particle moves along the X - axis with velocity `v = ksqrt(x)` . What will be work done by the forces acting on particle during interval it moves from x = 0 to x = d ? (Assume mass of the aprticle is m ) .

Options

A. `1/2mk^(2)d`
B. `1/4mk^(2)d`
C. `1/2mkd^(2)`
D. `1/4mkd^(2)`

Explanation

To find the work done by the forces acting on a particle moving along the X-axis with the velocity given by \( v = k\sqrt{x} \), we can use the work-energy theorem. The work done by the forces is equal to the change in kinetic energy of the particle.

### Step-by-Step Solution:

1. **Identify the Initial and Final Positions:**
- The particle moves from \( x = 0 \) to \( x = d \).

2. **Calculate the Initial Velocity:**
- At \( x = 0 \):
\[ v_{\text{initial}} = k\sqrt{0} = 0 \]
- Therefore, the initial kinetic energy \( KE_{\text{initial}} \) is:
\[ KE_{\text{initial}} = \frac{1}{2} m v_{\text{initial}}^2 = \frac{1}{2} m (0)^2 = 0 \]

3. **Calculate the Final Velocity:**
- At \( x = d \):
\[ v_{\text{final}} = k\sqrt{d} \]
- Therefore, the final kinetic energy \( KE_{\text{final}} \) is:
\[ KE_{\text{final}} = \frac{1}{2} m v_{\text{final}}^2 = \frac{1}{2} m (k\sqrt{d})^2 = \frac{1}{2} m k^2 d \]

4. **Calculate the Work Done:**
- According to the work-energy theorem:
\[ W = KE_{\text{final}} - KE_{\text{initial}} \]
- Substituting the values we found:
\[ W = \frac{1}{2} m k^2 d - 0 = \frac{1}{2} m k^2 d \]

### Final Answer:
The work done by the forces acting on the particle during the interval it moves from \( x = 0 \) to \( x = d \) is:
\[ W = \frac{1}{2} m k^2 d \]