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Physics · WORK, ENERGY AND POWER
हिन्दीA particle of mass \( 0.5\text{ kg} \) is subjected to a one-dimensional conservative force whose potential energy is \( V(x) = 20 + (x - 2)^2 \) (in Joules). When the particle is released from rest at \( x = 5\text{ m} \), the speed of the particle when it crosses \( x = 2\text{ m} \) is:
Correct answer · Option 3
\( 6\text{ m s}^{-1} \)
Explanation
\( V(5) = 20 + (3)^2 = 29\text{ J} \). At \( x=2 \), \( V(2) = 20\text{ J} \). By conservation of mechanical energy, \( K = V(5) - V(2) = 9\text{ J} \implies \frac{1}{2}(0.5)v^2 = 9 \implies v^2 = 36 \implies v = 6\text{ m s}^{-1} \).
