A body of mass 0.5 kg travels in a straight line with velocity `v= kx^(3//2)` where `k=5m^(-1//2)s^(-1)`. The work done by the net force during its displacement from x=0 to x=2 m is

Options

A. A. 1.5 J
B. B. 50 J
C. C. 10 J
D. D. 100 J

Explanation

To solve the problem step by step, we need to calculate the work done by the net force on a body of mass 0.5 kg as it moves from \( x = 0 \) to \( x = 2 \) m, given the velocity function \( v = kx^{3/2} \) where \( k = 5 \, \text{m}^{-1/2}\text{s}^{-1} \).

### Step 1: Determine the initial and final velocities
First, we will calculate the initial and final velocities of the body at the given positions.

- At \( x = 0 \):
\[ v(0) = k \cdot (0)^{3/2} = 5 \cdot 0 = 0 \, \text{m/s} \]

- At \( x = 2 \):
\[ v(2) = k \cdot (2)^{3/2} = 5 \cdot (2\sqrt{2}) = 5 \cdot 2\sqrt{2} = 10\sqrt{2} \, \text{m/s} \]

### Step 2: Calculate the initial and final kinetic energies
Next, we will calculate the initial and final kinetic energies using the formula for kinetic energy:
\[ KE = \frac{1}{2}mv^2 \]

- Initial kinetic energy \( KE_i \) at \( x = 0 \):
\[ KE_i = \frac{1}{2} \cdot 0.5 \cdot (0)^2 = 0 \, \text{J} \]

- Final kinetic energy \( KE_f \) at \( x = 2 \):
\[ KE_f = \frac{1}{2} \cdot 0.5 \cdot (10\sqrt{2})^2 \]
\[ = \frac{1}{2} \cdot 0.5 \cdot (100 \cdot 2) = \frac{1}{2} \cdot 0.5 \cdot 200 = 50 \, \text{J} \]

### Step 3: Calculate the work done
According to the work-energy theorem, the work done \( W \) is equal to the change in kinetic energy:
\[ W = KE_f - KE_i \]
\[ W = 50 \, \text{J} - 0 \, \text{J} = 50 \, \text{J} \]

### Conclusion
The work done by the net force during the displacement from \( x = 0 \) to \( x = 2 \) m is \( 50 \, \text{J} \).

---