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 step by step, we will analyze the motion of a particle of mass \( m \) moving under the influence of a constant force \( F \) along a trajectory defined by the function \( f(x) = \frac{x}{x^2 + 1} \). We will find the work done by the force and the final velocity of the particle.

### Step 1: Understanding the Work Done by the Force
The work done \( W \) by a force \( F \) when the particle moves from position \( x = 0 \) to \( x = 1 \) can be calculated using the formula:
\[ W = \int_{x_1}^{x_2} F \, dx \]
where \( x_1 = 0 \) and \( x_2 = 1 \).

### Step 2: Setting Up the Work Integral
Since the force \( F \) is always parallel to the x-axis, we can express the differential work done \( dW \) as:
\[ dW = F \, dx \]
Thus, the total work done from \( x = 0 \) to \( x = 1 \) is:
\[ W = \int_{0}^{1} F \, dx \]

### Step 3: Evaluating the Work Done
Integrating the expression, we have:
\[ W = F \int_{0}^{1} dx = F [x]_{0}^{1} = F (1 - 0) = F \]
So, the work done by the force when the particle moves from \( x = 0 \) to \( x = 1 \) is:
\[ W = F \text{ (in Joules)} \]

### Step 4: Applying the Work-Energy Theorem
According to the work-energy theorem, the work done on the particle is equal to the change in kinetic energy:
\[ W = \Delta KE = KE_f - KE_i \]
Since the particle starts from rest, the initial kinetic energy \( KE_i = 0 \). Therefore:
\[ W = KE_f - 0 = KE_f \]

### Step 5: Expressing Kinetic Energy
The kinetic energy of the particle can be expressed as:
\[ KE_f = \frac{1}{2} m v^2 \]
where \( v \) is the final velocity of the particle. Setting the work done equal to the final kinetic energy:
\[ F = \frac{1}{2} m v^2 \]

### Step 6: Solving for Final Velocity
Rearranging the equation to solve for \( v \):
\[ v^2 = \frac{2F}{m} \]
Taking the square root gives us the final velocity:
\[ v = \sqrt{\frac{2F}{m}} \]

### Summary of Results
1. The work done by the force \( F \) when moving from \( x = 0 \) to \( x = 1 \) is \( W = F \).
2. The final velocity of the particle is given by \( v = \sqrt{\frac{2F}{m}} \).