A particle of mass m starts moving from origin along a horizontal x-y plane under the influence of a force of constant magnitude F ( which is always parallel to the x-axis ) and a constraint force. The trajectory of its motion is `f ( x) = ( x)/( x^(2) + 1)`. Then,

Options

A. work down by the force F during the motion from x = 0 to x = 1 m is `( F //2) Nm`
B. work done by the force F during the motion from x = 0 to x = 1 m is ( F ) Nm
C. final velocity of the particle is `sqrt((2F)/(m))`
D. final velocity of the particle is `sqrt((F)/(m))`

Answer

To solve the problem, we need to determine the work done by the force \( F \) and the final velocity of the particle.

### Step 1: Understanding the Work Done by the Force
The work done \( W \) by a constant force \( F \) when moving through a displacement \( dx \) is given by the formula:
\[ dW = F \cdot dx \]
Since the force \( F \) is parallel to the x-axis, the angle between \( F \) and \( dx \) is \( 0^\circ \). Thus, we can simplify the expression:
\[ dW = F \cdot dx \cdot \cos(0) = F \cdot dx \]

### Step 2: Integrating the Work Done
To find the total work done as the particle moves from \( x = 0 \) to \( x = 1 \), we integrate:
\[ W = \int_{0}^{1} F \, dx = F \int_{0}^{1} dx = F[x]_{0}^{1} = F(1 - 0) = F \]
Thus, the work done by the force \( F \) during the motion from \( x = 0 \) to \( x = 1 \) is:
\[ W = F \, \text{(in Newton meter)} \]

### Step 3: Applying the Work-Energy Theorem
The work-energy theorem states that the work done by the force is equal to the change in kinetic energy:
\[ W = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]
Given that the particle starts from rest, the initial kinetic energy \( KE_{\text{initial}} = 0 \). Therefore:
\[ W = KE_{\text{final}} = \frac{1}{2} mv^2 \]
Substituting the work done:
\[ F = \frac{1}{2} mv^2 \]

### Step 4: Solving for Final Velocity
Rearranging the equation to solve for \( v \):
\[ v^2 = \frac{2F}{m} \implies v = \sqrt{\frac{2F}{m}} \]
Thus, the final velocity of the particle is:
\[ v = \sqrt{\frac{2F}{m}} \]

### Conclusion
The correct answers are:
- The work done by the force \( F \) from \( x = 0 \) to \( x = 1 \) is \( F \) (Option B).
- The final velocity of the particle is \( \sqrt{\frac{2F}{m}} \) (Option C).