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Physics · WORK, ENERGY AND POWER
हिन्दीThe potential energy of a particle of mass \( 1\text{ kg} \) moving along the \( x \)-axis is given by \( V(x) = (x^3 - 6x^2 + 9x + 5)\text{ J} \). If the total mechanical energy of the particle is \( 15\text{ J} \), its maximum speed is:
Correct answer · Option 2
\( 2\sqrt{5}\text{ m s}^{-1} \)
Explanation
\( \frac{dV}{dx} = 3x^2 - 12x + 9 = 0 \implies x=1, 3 \). At \( x=3 \), \( V_{\min} = 27 - 54 + 27 + 5 = 5\text{ J} \). \( K_{\max} = E - V_{\min} = 15 - 5 = 10\text{ J} \implies v_{\max} = \sqrt{2(10)/1} = \sqrt{20} = 2\sqrt{5}\text{ m s}^{-1} \).
