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 the body during the time interval \([0, 2s]\), we will use the work-energy theorem, which states that the work done by all forces is equal to the change in kinetic energy of the body.

### Step 1: Determine the position function
The position of the point mass is given by the equation:
\[ x(t) = t^2 + 2t \]

### Step 2: Find the velocity function
To find the velocity, we need to differentiate the position function with respect to time \(t\):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^2 + 2t) = 2t + 2 \]

### Step 3: Calculate the initial and final velocities
Now we will calculate the velocities at \(t = 0s\) and \(t = 2s\):
- At \(t = 0s\):
\[ v(0) = 2(0) + 2 = 2 \, \text{m/s} \]
- At \(t = 2s\):
\[ v(2) = 2(2) + 2 = 4 + 2 = 6 \, \text{m/s} \]

### Step 4: Calculate the change in kinetic energy
The kinetic energy \(KE\) is given by the formula:
\[ KE = \frac{1}{2}mv^2 \]
Where \(m = 0.5 \, \text{kg}\).

- Initial kinetic energy at \(t = 0s\):
\[ KE_1 = \frac{1}{2} \times 0.5 \times (2)^2 = \frac{1}{2} \times 0.5 \times 4 = 1 \, \text{J} \]

- Final kinetic energy at \(t = 2s\):
\[ KE_2 = \frac{1}{2} \times 0.5 \times (6)^2 = \frac{1}{2} \times 0.5 \times 36 = 9 \, \text{J} \]

### Step 5: Calculate the work done
Now we can find the work done by all forces:
\[ W = KE_2 - KE_1 = 9 \, \text{J} - 1 \, \text{J} = 8 \, \text{J} \]

### Final Answer
The work done by all the forces acting on the body during the time interval \([0, 2s]\) is:
\[ \boxed{8 \, \text{J}} \]