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Physics · LAWS OF MOTION
हिन्दीA time-dependent force \( F(t) = (12t^2 - 6t)\text{ N} \) acts on a particle of mass \( m = 0.5\text{ kg} \). If the particle starts from rest at \( t = 0\text{ s} \), calculate its velocity at \( t = 2\text{ s} \) and the total work done by this force on the particle during the time interval from \( t = 0\text{ s} \) to \( t = 2\text{ s} \).
Correct answer · Option 4
\( v = 40\text{ m/s} \) and \( W = 400\text{ J} \)
Explanation
Impulse \( J = \int_0^2 (12t^2 - 6t) dt = [4t^3 - 3t^2]_0^2 = 32 - 12 = 20\text{ N s} \). Velocity \( v = \frac{J}{m} = \frac{20}{0.5} = 40\text{ m/s} \). Work done \( W = \Delta K = \frac{1}{2} m v^2 = \frac{1}{2} (0.5) (1600) = 400\text{ J} \).
