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JEE PHYSICS MCQ: WORK-ENERGY-POWER

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A certain spring is found not to obey Hooke's law, it exerts a restoring force F(x)=−αx−βx2F(x)=-\alpha x-\beta x^2 if it is stretched or compressed, where α=48 N/m\alpha=48 \mathrm{~N} / \mathrm{m} and β=24 N/m2.\beta=24 \mathrm{~N} / \mathrm{m}^2. The mass of the spring is negligible. An object with mass 1 kg1 \mathrm{~kg} on a frictionless, horizontal surface is attached to the spring, pulled a distance 1 m1 \mathrm{~m} to the right to stretch the spring and released. The speed of the object when it is 0.5 m0.5 \mathrm{~m} to the right of the x=0x=0 equilibrium position is
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Correct option: A
52 m/s5 \sqrt{2} \text{ m/s}
Option A (52 m/s5 \sqrt{2} \text{ m/s}) is correct because it is calculated using conservation of energy, considering the non-Hookean restoring force.

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