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Physics · WORK, ENERGY AND POWER
हिन्दीA particle of mass \( m = 0.1\text{ kg} \) oscillates in a potential well given by \( V(x) = \frac{1}{2} k x^2 \) with \( k = 0.5\text{ N m}^{-1} \). If its total mechanical energy is \( 1\text{ J} \), its speed when passing through \( x = \sqrt{2}\text{ m} \) is:
Correct answer · Option 1
\( \sqrt{10}\text{ m s}^{-1} \)
Explanation
Potential energy at \( x = \sqrt{2}\text{ m} \) is \( V = \frac{1}{2}(0.5)(\sqrt{2})^2 = 0.5\text{ J} \). Kinetic energy is \( K = E - V = 1 - 0.5 = 0.5\text{ J} \). Therefore, \( \frac{1}{2} m v^2 = 0.5 \implies 0.5 \times 0.1 \times v^2 = 0.5 \implies 0.05 v^2 = 0.5 \implies v^2 = 10 \implies v = \sqrt{10}\text{ m s}^{-1} \).
