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Potential energy of a body of mass 1kg free to move along X-axis is given by $U (x) = \{\frac{x^{2}}{2} - x\} J .$ If the total mechanical energy of the body is 2J, then the maximum speed of the body is (Assume only conservative force acts on the body)

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$5 {ms}^{- 1}$

$3.5 {ms}^{- 1}$

$\sqrt{8} {ms}^{- 1}$

$\sqrt{5} {ms}^{- 1}$

answer is A.

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Detailed Solution

Given,

m = 1kg

$U (x) = \{\frac{x^{2}}{2} - x\} J .$

Total energy = 2J

For maximum speed

$\frac{d u}{d x} = 0 \frac{d u}{d x} = \frac{d}{d x} (\frac{x^{2}}{2} - x) = \frac{2 x}{2} - 1 \frac{d^{2} u}{d x^{2}} = \frac{2}{2} - 1 \frac{d^{2} u}{d x^{2}} = 0 x = 1$

Hence,

$u (x) = \frac{x^{2}}{2} - x u (x) = \frac{1}{2} - 1 u (x) = \frac{- 1}{2} J$

Now,

$\frac{1}{2} m v^{2} + (\frac{- 1}{2}) = 2 (\frac{1}{2}) \times 1 \times v^{2} - \frac{1}{2} = 2 u = \sqrt{5} m / s e c$

Hence the correct answer is $\sqrt{5} m / s e c .$

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