Step 1: Understanding the Concept:
Work done is equal to the change in kinetic energy (\( W = \Delta KE \)). If the speed at the beginning and the end of the time interval is the same, the net work done is zero.
: Key Formula or Approach:
1. Express \( x \) in terms of \( t \).
2. Find velocity \( v = \frac{dx}{dt} \).
3. Compare \( v \) at \( t = 0 \) and \( t = 6 \).
Step 2: Detailed Explanation:
Given \( t = \sqrt{x} + 3 \).
Rearranging for \( x \):
\[ \sqrt{x} = t - 3 \]
\[ x = (t - 3)^2 \]
To find velocity, differentiate with respect to \( t \):
\[ v = \frac{dx}{dt} = 2(t - 3) \cdot 1 \]
\[ v = 2t - 6 \]
Now, check velocity at the initial and final times:
At \( t = 0 \text{ s} \): \( v_1 = 2(0) - 6 = -6 \text{ m/s} \).
At \( t = 6 \text{ s} \): \( v_2 = 2(6) - 6 = 12 - 6 = 6 \text{ m/s} \).
Change in Kinetic Energy:
\[ \Delta KE = \frac{1}{2}m(v_2^2 - v_1^2) \]
\[ \Delta KE = \frac{1}{2}m((6)^2 - (-6)^2) \]
\[ \Delta KE = \frac{1}{2}m(36 - 36) = 0 \]
Since change in kinetic energy is zero, the work done is zero.
Step 3: Final Answer:
The work done in the first 6 seconds is Zero.