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Standard XII

Physics

Conservative Force as Gradient of Potential

A particle, w...

Question

A particle, which is constrained to move along the x - axis the x - axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as $F (x) = - k x + a x^{3}$ . Here k and a are positive energy U(x) of the particle is

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Solution

The correct option is D
$d U_{(x)} = - F d x ∴ U_{x} = - \int_{0}^{x} F d x = \frac{k x^{2}}{2} - \frac{a x^{4}}{4}$
U = 0 at x = 0 and at $x = \sqrt{\frac{2 k}{a}}$
U(x) is negative for $x > \sqrt{\frac{2 k}{a}}$
F = 0 at x = 0. It means Slope of U - x graph is zero at x = 0.

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A particle, which is constrained to move along the x - axis the x - axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x)=−kx+ax3. Here k and a are positive energy U(x) of the particle is

The correct option is D dU(x)=−Fdx∴Ux=−∫x0Fdx=kx22−ax44 U = 0 at x = 0 and at x=√2ka U(x) is negative for x>√2ka F = 0 at x = 0. It means Slope of U - x graph is zero at x = 0.

A particle, which is constrained to move along the x - axis the x - axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as $F (x) = - k x + a x^{3}$ . Here k and a are positive energy U(x) of the particle is

The correct option is D
$d U_{(x)} = - F d x ∴ U_{x} = - \int_{0}^{x} F d x = \frac{k x^{2}}{2} - \frac{a x^{4}}{4}$
U = 0 at x = 0 and at $x = \sqrt{\frac{2 k}{a}}$
U(x) is negative for $x > \sqrt{\frac{2 k}{a}}$
F = 0 at x = 0. It means Slope of U - x graph is zero at x = 0.