A certain spring is found not to obey Hooke’s law, it exerts a restoring force $F (x) = - α x - β x^{2}$ if it is stretched (or) compressed, where $α = 48 \frac{N}{m} & β = \frac{24 N}{m^{2}}$ . The mass of the spring is negligible. An object with mass (2 kg) on a friction less -horizontal surface is attached to the spring, pulled a distance of 1 m to the right to stretch the spring and released. The speed of the object when it is 0.5 m to the right of the x = 0 equilibrium position is

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6 m/s

3 m/s

5 m/s

4 m/s

answer is C.

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Detailed Solution

$F = - α - β x^{2}$

$F = - \frac{d v}{d x}$

$d v = - F d x$ $\Rightarrow Δ U = - \int F d x = (\frac{α x^{2}}{2} + \frac{β x^{3}}{3})$

$M \cdot E_{i} = M \cdot E_{f}$

${(K + U)}_{i} = {(K + U)}_{f}$

$0 + (\frac{α}{2} + \frac{β}{3}) = \frac{1}{2} m v^{2} + [\frac{α}{2} {(\frac{1}{2})}^{2} + \frac{β}{3} {(\frac{1}{2})}^{3}] \Rightarrow \frac{3 α}{8} + \frac{7 β}{24} = \frac{1}{2} m v^{2}$

$\frac{3}{8} \times \overset{6}{48} + \frac{7}{24} \times 24 = \frac{1}{2} \times 2 \times v^{2} \Rightarrow v = 5 m / s$

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