The displacement x of a particle at time t moving under a constant force is `t=sqrtx+3`, x in metres, t in seconds. Find the work done by the force in the interval from `t=0` to `t=6` second.

Options

A. 8J
B. 4J
C. 0J
D. 6J

Explanation

To solve the problem step by step, we need to find the work done by the force on a particle whose displacement \( x \) is given as a function of time \( t \) by the equation \( t = \sqrt{x} + 3 \). We will follow these steps:

### Step 1: Rearranging the equation
We start with the given equation:
\[ t = \sqrt{x} + 3 \]
To express \( x \) in terms of \( t \), we rearrange the equation:
\[ \sqrt{x} = t - 3 \]
Now, squaring both sides gives:
\[ x = (t - 3)^2 \]

### Step 2: Finding the velocity
Velocity \( v \) is defined as the rate of change of displacement with respect to time, which can be expressed as:
\[ v = \frac{dx}{dt} \]
Differentiating \( x = (t - 3)^2 \) with respect to \( t \):
\[ \frac{dx}{dt} = 2(t - 3) \]
Thus, the velocity function is:
\[ v(t) = 2(t - 3) \]

### Step 3: Calculating initial and final velocities
Now, we need to find the velocities at \( t = 0 \) and \( t = 6 \).

- For \( t = 0 \):
\[ v(0) = 2(0 - 3) = 2 \times -3 = -6 \, \text{m/s} \]

- For \( t = 6 \):
\[ v(6) = 2(6 - 3) = 2 \times 3 = 6 \, \text{m/s} \]

### Step 4: Applying the Work-Energy Theorem
According to the work-energy theorem, the work done \( W \) by the force is equal to the change in kinetic energy \( \Delta KE \):
\[ W = KE_{\text{final}} - KE_{\text{initial}} \]
The kinetic energy \( KE \) is given by:
\[ KE = \frac{1}{2} mv^2 \]
Thus, we can express the work done as:
\[ W = \frac{1}{2} m v_{\text{final}}^2 - \frac{1}{2} m v_{\text{initial}}^2 \]
Factoring out \( \frac{1}{2} m \):
\[ W = \frac{1}{2} m (v_{\text{final}}^2 - v_{\text{initial}}^2) \]

### Step 5: Substituting the velocities
Substituting \( v_{\text{final}} = 6 \, \text{m/s} \) and \( v_{\text{initial}} = -6 \, \text{m/s} \):
\[ W = \frac{1}{2} m (6^2 - (-6)^2) \]
Calculating the squares:
\[ W = \frac{1}{2} m (36 - 36) = \frac{1}{2} m \cdot 0 = 0 \]

### Conclusion
Thus, the work done by the force in the interval from \( t = 0 \) to \( t = 6 \) seconds is:
\[ \boxed{0 \, \text{Joules}} \]