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Physics · LAWS OF MOTION
हिन्दीThe vertical displacement of a particle of mass \( 0.5\text{ kg} \) moving under gravity and variable resistive forces is given by \( y(t) = 5 t + 3 t^2 - t^3 \) (where \( y \) is in meters and \( t \) in seconds). The net vertical force acting on the particle at time \( t = 2\text{ s} \) has a magnitude of:
Correct answer · Option 1
3.0 N
Explanation
Vertical velocity \( v_y = \frac{dy}{dt} = 5 + 6t - 3t^2 \). Acceleration \( a_y = \frac{dv_y}{dt} = 6 - 6t \). At \( t = 2\text{ s} \), \( a_y = 6 - 12 = -6\text{ m/s}^2 \). Force magnitude \( |F| = m |a_y| = 0.5 \times 6 = 3.0\text{ N} \).
