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

Options

A. 50J
B. 45J
C. 25J
D. none of these

Explanation

To solve the problem, we will use the work-energy theorem, which states that the work done by the net force on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the body, \( m = 0.5 \, \text{kg} \)
- Velocity function, \( v = 5x^{3/2} \)
- Displacement from \( x = 0 \) to \( x = 2 \, \text{m} \)

2. **Calculate Initial Velocity (\( v_i \)):**
- At \( x = 0 \):
\[ v_i = 5(0)^{3/2} = 0 \, \text{m/s} \]

3. **Calculate Initial Kinetic Energy (\( KE_i \)):**
- Using the formula for kinetic energy, \( KE = \frac{1}{2} mv^2 \):
\[ KE_i = \frac{1}{2} \times 0.5 \times (0)^2 = 0 \, \text{J} \]

4. **Calculate Final Velocity (\( v_f \)):**
- At \( x = 2 \):
\[ v_f = 5(2)^{3/2} = 5 \times 2 \sqrt{2} = 5 \times 2 \times 1.414 \approx 14.14 \, \text{m/s} \]

5. **Calculate Final Kinetic Energy (\( KE_f \)):**
- Using the final velocity:
\[ KE_f = \frac{1}{2} \times 0.5 \times (14.14)^2 \]
- First, calculate \( (14.14)^2 \):
\[ (14.14)^2 \approx 200 \]
- Now calculate \( KE_f \):
\[ KE_f = \frac{1}{2} \times 0.5 \times 200 = 0.25 \times 200 = 50 \, \text{J} \]

6. **Calculate Change in Kinetic Energy (\( \Delta KE \)):**
- The change in kinetic energy is given by:
\[ \Delta KE = KE_f - KE_i = 50 \, \text{J} - 0 \, \text{J} = 50 \, \text{J} \]

7. **Determine Work Done (\( W \)):**
- According to the work-energy theorem:
\[ W = \Delta KE = 50 \, \text{J} \]

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

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