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The potential energy of a particle of mass m free to move along the x-axis is given by U = $\frac{1}{2}kx^{2}$ for x < 0 and U = 0 for x $\geq$ 0 (x denotes the x-coordinate of the particle and k is a positive constant). If the total mechanical energy of the particle is E, then its speed at x = $-\sqrt{\frac{2E}{k}}$ is
Option: 1
zero

Option: 2
$\sqrt{\frac{2E}{m}}$

Option: 3
$\sqrt{\frac{E}{m}}$

Option: 4
$\sqrt{\frac{E}{2m}}$

Answers (1)

best_answer

If only conservative forces act on a system, total mechnical energy remains constant -

$K+U=E\left ( constant \right )$

$\Delta K+\Delta U=0$

$\Delta K=-\Delta U$

-

From the conservation of energy

K.E. + P.E. = E or $K.E = E - \frac{1}{2}kx^{2}$

$\therefore K.E \ at \ x= -\sqrt{\frac{2E}{k}} is$

$KE=E - \frac{1}{2}k\left ( \frac{2E}{k} \right ) = 0$

$\therefore$ The speed of particle at x = $- \sqrt{\frac{2E}{k}}$ is zero.

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Nehul

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