A point mass of `0.5` kg is moving along `x-` axis as `x=t^(2)+2t` , where, `x` is in meters and `t` is in seconds. Find the work done (in J) by all the forces acting on the body during the time interval `[0,2s]`.

Options

A. `6J`
B. `7J`
C. `8J`
D. `9J`

Explanation

To find the work done by all the forces acting on a point mass of 0.5 kg moving along the x-axis, we can use the work-energy theorem, which states that the work done by all forces is equal to the change in kinetic energy of the mass.

### Step-by-Step Solution:

1. **Identify the given information:**
- Mass (m) = 0.5 kg
- Position as a function of time (x) = t² + 2t
- Time interval = [0, 2 seconds]

2. **Find the velocity as a function of time:**
- Velocity (v) is the derivative of position (x) with respect to time (t):
\[ v = \frac{dx}{dt} = \frac{d}{dt}(t^2 + 2t) = 2t + 2 \]

3. **Calculate the initial and final velocities:**
- At \( t = 0 \):
\[ v(0) = 2(0) + 2 = 2 \, \text{m/s} \]
- At \( t = 2 \):
\[ v(2) = 2(2) + 2 = 4 + 2 = 6 \, \text{m/s} \]

4. **Calculate the initial and final kinetic energies:**
- Kinetic energy (KE) is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- Initial kinetic energy (KE_initial) at \( t = 0 \):
\[ KE_{initial} = \frac{1}{2} \times 0.5 \times (2)^2 = \frac{1}{2} \times 0.5 \times 4 = 1 \, \text{J} \]
- Final kinetic energy (KE_final) at \( t = 2 \):
\[ KE_{final} = \frac{1}{2} \times 0.5 \times (6)^2 = \frac{1}{2} \times 0.5 \times 36 = 9 \, \text{J} \]

5. **Calculate the change in kinetic energy:**
- Change in kinetic energy (ΔKE):
\[ \Delta KE = KE_{final} - KE_{initial} = 9 \, \text{J} - 1 \, \text{J} = 8 \, \text{J} \]

6. **Conclusion:**
- According to the work-energy theorem, the work done by all forces is equal to the change in kinetic energy:
\[ W = \Delta KE = 8 \, \text{J} \]

### Final Answer:
The work done by all the forces acting on the body during the time interval [0, 2s] is **8 Joules**.