Step 1: Understanding the Question:
Power is the rate of doing work, which according to the Work-Energy Theorem, is equal to the rate of change of kinetic energy.
Given power as a function of time, we can find the total work done by integrating power over time. This total work equals the change in kinetic energy.
Step 2: Key Formula or Approach:
Work Done \(W = \int P dt = \Delta K.E.\)
\(\Delta K.E. = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2\).
Given \(m = 2 \text{ kg}\), \(v_i = 0\) at \(t = 0\).
Step 3: Detailed Explanation:
Calculate total work done from \(t = 0\) to \(t = 2\) s:
\[ W = \int_{0}^{2} P(t) dt = \int_{0}^{2} \frac{3t^2}{2} dt \]
\[ W = \frac{3}{2} \left[ \frac{t^3}{3} \right]_0^2 \]
\[ W = \frac{1}{2} [ t^3 ]_0^2 = \frac{1}{2} (8 - 0) = 4 \text{ Joules} \]
Set work equal to change in kinetic energy:
\[ W = K.E_{\text{final}} - K.E_{\text{initial}} \]
\[ 4 = \frac{1}{2} m v^2 - 0 \]
Substitute mass \(m = 2 \text{ kg}\):
\[ 4 = \frac{1}{2} (2) v^2 \]
\[ 4 = v^2 \]
\[ v = \sqrt{4} = 2 \text{ m/s} \]
Step 4: Final Answer:
The velocity of the 2 kg particle at \(t = 2\) s, after receiving power \(P = 1.5t^2\), is 2 m/s.