In the figure the variation of potential cnergy of a...

In the figure the variation of potential cnergy of a particle of mass $m=2 \mathrm{~kg}$ is represented $w . r . f$, its $x$ -coordinate. The particle moves under the effect of this conservative force along the $x$ -axis, 1. If the particle is released at the origin then: (a) it will move towards positive $x$ -axis. (b) it will move towards negative $x$ -axis. (c) it will remain stationary at the origin. (d) its subsequent motion cannot be decided due to lack of information, 2. If the particle is released at $x=2+A$ where $A \rightarrow 0$ (it is positive) then its maximum speed in subsequent motion will be: (a) $\sqrt{10} \mathrm{~m} / \mathrm{s}$ (b) $5 \mathrm{~m} / \mathrm{s}$ (c) $5 \sqrt{2} \mathrm{~m} / \mathrm{s}$ (d) $7.5 \mathrm{~m} / \mathrm{s}$ 3. $x=5 \mathrm{~m}$ and $x=10 \mathrm{~m}$ positions of the particle are respectively of (a) neutral and stable equilibrium. (b) neutral and unstable equilibrium. (c) unstable and stable equilibrium. (d) stable and unstable equilibrium.

In the figure the variation of potential cnergy of a particle of mass $m=2 \mathrm{~kg}$ is represented $w . r . f$, its $x$ -coordinate. The particle moves under the effect of this conservative force along the $x$ -axis,
1. If the particle is released at the origin then:
(a) it will move towards positive $x$ -axis.
(b) it will move towards negative $x$ -axis.
(c) it will remain stationary at the origin.
(d) its subsequent motion cannot be decided due to lack of information,
2. If the particle is released at $x=2+A$ where $A \rightarrow 0$ (it is positive) then its maximum speed in subsequent motion will be:
(a) $\sqrt{10} \mathrm{~m} / \mathrm{s}$
(b) $5 \mathrm{~m} / \mathrm{s}$
(c) $5 \sqrt{2} \mathrm{~m} / \mathrm{s}$
(d) $7.5 \mathrm{~m} / \mathrm{s}$
3. $x=5 \mathrm{~m}$ and $x=10 \mathrm{~m}$ positions of the particle are respectively of
(a) neutral and stable equilibrium.
(b) neutral and unstable equilibrium.
(c) unstable and stable equilibrium.
(d) stable and unstable equilibrium.

Key Concepts

Equilibrium and Stability

Equilibrium points occur where the net force on a particle is zero, corresponding to points where the derivative of the potential energy is zero. The nature of the equilibrium—whether it is stable, unstable, or neutral—is determined by the curvature of the potential energy function. A local minimum indicates stable equilibrium (small displacements result in restoring forces), a local maximum indicates unstable equilibrium (small displacements lead to forces that drive the particle away), and a flat potential (constant around the point) may indicate neutral equilibrium.

Conservation of Mechanical Energy

In systems influenced only by conservative forces, the total mechanical energy, which is the sum of kinetic and potential energies, remains constant. This conservation principle is essential for analyzing the motion of objects, as it allows the conversion of potential energy into kinetic energy (and vice versa) to determine quantities such as the maximum speed of a particle as it moves through regions of varying potential energy.

Conservative Force and Potential Energy

A conservative force is one that can be expressed as the negative gradient of a potential energy function. This means that the work done by the force on a particle moving between any two points depends only on the initial and final positions, not on the path taken. The variation of potential energy with position helps to identify points where the particle’s state of energy changes, influencing its motion.

Relationship Between Potential Energy and Force

The force acting on a particle in a conservative field is given by the negative derivative (or gradient in higher dimensions) of the potential energy function. This relationship implies that a particle will accelerate in the direction of decreasing potential energy, which is key to understanding its initial motion and subsequent dynamics when released from a given point.

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