Using Work-Energy Theorem:
\[ W = \Delta K = K_2 - K_1 \]
Step 1: Find velocity by differentiating displacement
\[ x(t) = \alpha t^2 + \beta t + \gamma \]
\[ v(t) = \frac{dx}{dt} \]
\[ v(t) = 2\alpha t + \beta \]
Substitute given values:
\[ v(t) = 2(1)t + 1 \]
\[ v(t) = 2t + 1 \]
Step 2: Find velocity at t = 2 s
\[ v(2) = 2(2) + 1 \]
\[ v(2) = 5 \text{ m/s} \]
Step 3: Find velocity at t = 3 s
\[ v(3) = 2(3) + 1 \]
\[ v(3) = 7 \text{ m/s} \]
Step 4: Calculate kinetic energies
\[ K = \frac{1}{2}mv^2 \]
\[ K_1 = \frac{1}{2}(2)(5^2) \]
\[ K_1 = 25 \text{ J} \]
\[ K_2 = \frac{1}{2}(2)(7^2) \]
\[ K_2 = 49 \text{ J} \]
Step 5: Apply Work-Energy Theorem
\[ W = 49 - 25 \]
\[ W = 24 \text{ J} \]
Final Answer: 24 J
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.