The velocity (v) of a particle of mass m moving along x-axis is given by `v=alphasqrt(x)`, where `alpha` is a constant. Find work done by force acting on particle during its motion from x=0 to x=2m :-

Options

A. `malpha^(2)`
B. `malpha`
C. `(malpha)/(2)`
D. None

Answer

To find the work done by the force acting on a particle moving along the x-axis, we can use the work-energy theorem, which states that the net work done on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the given information**:
- The velocity of the particle is given by \( v = \alpha \sqrt{x} \).
- The mass of the particle is \( m \).
- We need to find the work done as the particle moves from \( x = 0 \) to \( x = 2 \, \text{m} \).

2. **Calculate the initial velocity**:
- At \( x = 0 \):
\[ v(0) = \alpha \sqrt{0} = 0 \]
- Therefore, the initial velocity \( u = 0 \).

3. **Calculate the final velocity**:
- At \( x = 2 \):
\[ v(2) = \alpha \sqrt{2} \]
- Therefore, the final velocity \( v = \alpha \sqrt{2} \).

4. **Calculate the initial kinetic energy**:
- The initial kinetic energy \( KE_i \) when \( x = 0 \):
\[ KE_i = \frac{1}{2} m u^2 = \frac{1}{2} m (0)^2 = 0 \]

5. **Calculate the final kinetic energy**:
- The final kinetic energy \( KE_f \) when \( x = 2 \):
\[ KE_f = \frac{1}{2} m v^2 = \frac{1}{2} m (\alpha \sqrt{2})^2 = \frac{1}{2} m (2\alpha^2) = m \alpha^2 \]

6. **Calculate the work done**:
- According to the work-energy theorem:
\[ W = KE_f - KE_i = m \alpha^2 - 0 = m \alpha^2 \]

7. **Conclusion**:
- The work done by the force acting on the particle during its motion from \( x = 0 \) to \( x = 2 \, \text{m} \) is:
\[ W = m \alpha^2 \]

### Final Answer:
The work done by the force is \( m \alpha^2 \).

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