Conservative vs. Non-conservative Forces

Moving an object from $A$ to $B$ under the influence of a force requires some work $W$. If at the end of the process the total energy of our particle is $W,$ then we can recapture the work done to the system. However, if the total energy is less than $W$, it implies that some energy was lost along the way, or dissipated. Thus, a force is conservative when the work it does on our system can be recovered fully.

Take for example gravity: when a force is applied against gravity, the work is stored as gravitational potential energy, such as when a box is lifted and then dropped. On the other hand, friction is a non-conservative force, as work is lost as heat.

When only conservative forces are acting, the energy of the system is conserved.

Consider a particle at rest, with mass $m$ at a height $H$ on Earth (supposing there is no friction wih air), the energy of the system is gravitational

$$\large E_{0}=mgH$$

When it falls on the ground, kinematics $(v^2 = u^2 + 2as)$ tells us that the speed of the particle is $v=\sqrt{2gH}$.
The potential energy is defined to be zero at the ground level.

$$\large \begin{aligned} E_{f}&=\frac{mv^2}{2} \\ &=mgH \end{aligned}$$

Since $E_{0}=E_{f}$ we conclude that every force acting in the situation is conservative, so Gravitational Force is an example of conservative force.

If a non-conservative force does work on an object as it moves, then the total mechanical energy (the sum of kinetic energy and its potential energy) changes along its trajectory. In many natural processes, an object's total mechanical energy decreases with time, so that the process is dissipative. Forces that oppose an object's motion without building up a potential—like friction and air resistance—dissipate energy.

Imagine a particle with mass $m$ on a fixed incline plane inclined with angle $\theta$ and coefficient of kinetic friction $\mu$.
Consider axes parallel and normal to the inclined plane. On the body, we have a normal reaction $F|\vec{N}|=mg\cos(\theta)$ so the magnitude of the friction force equals $f=\mu mg \cos \theta$.
The magnitude of the acceleration down the plane is then $a=g(\sin(\theta)-\mu \cos(\theta))$.
If initially the particle is at a height $H$, the initial energy is $E_{0}=mgH$, defining the potential energy at the ground level to be $0$.
The distance $s$ that it has travelled equals $\frac{H}{\sin(\theta)}$.
Its speed, by the relation from kinematics, $$v^2 = u^2 + 2as$$ is $v=\sqrt{\frac{2gH(\sin\theta-\mu\cos\theta)}{\sin(\theta)}}$ as $u = 0$.
The final energy is $E_{f}=mgH(1 - \mu \cot\theta)$.
As we can see, $E_{0} \ne E_{f}$, showing that friction is a non-conservative force.

The work done over any closed path (i.e. a trajectory that begins and ends at the same point in space) for a conservative force is zero, while it will generally be non-zero for dissipative, non-conservative forces. Note that the work done by a non-conservative force can be zero over some closed paths, while for a conservative force, the work is zero over any closed path.

Work done by any force, where $dx$ is an infinitesimal distance along the path $C$ between two points, is,

$$W = \int\limits_{C} \vec{F} \cdot d\vec{x}$$

where $\vec{F} \cdot {d}\vec{x} = |\vec{F}|(dx)\cos\theta\$

and $\theta$ is the angle between the force vector and the displacement vectors.

For conservative forces, this work is independent of the choice of path $C$.

Is tension conservative when a particle undergoes vertical circular motion? Can we comment about conservative nature of tension in general?

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Yes, during the motion, work done by tension is $0$ because it is perpendicular to the instantaneous velocity at every instant. $$dW = \vec{F} \cdot d\vec{s} = \vec{F} \cdot \vec{v} dt = 0$$ We cannot comment about conservative nature of tension, in general. For example, it will do some work when one of its components is along the velocity.

A particle of mass $m$ and charge $q$ is taken along a parabolic path with equation $y = x^2$, from origin to the point $P (x_0, x_0^2)$.
In this region, there exists a gravitational field $g$ in the $- y$ direction, and electric field $E$ in the $+ x$ direction. What is the work done by the external forces on this particle, assuming it is moved without dissipation?

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Both the gravitational and electrostatic fields are conservative.
Since we're moving the body slowly, there is no change in the kinetic energy, hence,

$$W_{all} = W_{ext} + W_{ele} + W_{grav} = \Delta KE = 0$$ by the work-energy theorem.
Now,

$$W_{ele} = \int_{O}^P q\vec{E} \cdot d\vec{s} = \int_{0}^{x_0} q\vec{E} d{x} = qEx_0$$ as the electric field is along the $x$ axis.
and,

$$W_{grav} = \int_{O}^P mg \cdot d\vec{s} = \int_{0}^{x_0^2} -q\vec{E} d{y} = -mgx_0^2$$ So, from the first relation,

$$W_{ext} =- (W_{ele} + W_{grav}) = mgx_0^2 - qEx_0$$

If work done by gravity in curved path is $W$, find the work done by gravity in moving the particle along a straight line if the distance of the curved path is $3$ times the distance between the particles.

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It will be $W$ again, as the work done by a conservative force is independent of path.

Try the following problems:

• Work
• Conservative Force
• Conservation of Mechanical Energy
• Conservative nature of Electric Fields