Conservative And Non-conservative Forces

Conservative and non-conservative forces play a crucial role in the study of physics, particularly in understanding energy conservation and the behaviour of objects in motion. Conservative forces, such as gravitational and elastic forces, are those that do not dissipate energy; the work done by these forces is path-independent, meaning it only depends on the initial and final positions of the object. On the other hand, non-conservative forces, like friction and air resistance, cause energy dissipation, often converting mechanical energy into thermal energy, and their work is path-dependent.

In real life, the distinction between these forces is evident in everyday experiences. For example, when you lift an object and then lower it back to the ground, the energy you expended is recovered as potential energy due to gravity—a conservative force. However, if you slide a book across a table, friction, a non-conservative force, will dissipate some of the book's kinetic energy as heat, making it impossible to recover all the initial energy. Understanding these concepts is essential in engineering, mechanics, and even in designing systems that optimize energy efficiency.

1. Conservative Field

• In the conservative field, work done by the force depends only upon the initial and final position.

• In the conservative field, work done by the force does not depend on the path.

• In the conservative field, work done by the force along a closed path is zero.

2. Conservative Force

• The forces of these types of fields are known as conservative forces.

Examples: Electrostatic forces, gravitational forces, the spring force

3. Non-Conservative Field

• In Non- conservative field, work done by the force depends on the path followed between any two positions/points.

• In the Non-conservative field, work done by the force along a closed path is non-zero.

4. Non-Conservative Force

• The forces corresponding to the Non-Conservative field are known as non-conservative forces.

• Non-Conservative Force is dissipative.

Example: Frictional force, Viscous force

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Solved Examples Based on Conservative And Non-conservative Forces

Example 1: If W₁, W₂ and W₃ represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively (as shown) in the gravitational field of a point mass m, find the correct relation

1) $W_1>W_2>W_3$
2) $W_1=W_2=W_3$
3) $W_1<W_2<W_3$
4) $W_2>W_1>W_3$

Solution:

Work Done in Conservative and Non-Conservative Fields

1. Conservative field

• In the conservative field, work done by the force depends only upon the initial and final position.

• In the conservative field, work done by the force does not depend on the path.

• In the conservative field, work done by the force along a closed path is zero.

2. Conservative force

• The forces of these types of fields are known as conservative forces.

Examples: Electrostatic forces, gravitational forces, the spring force

So,

As gravitational field is conservative in nature. So work done in moving a particle from A to B does not depend upon the path followed by the body. It always remains the same.

Example 2: If a block moves from a height h above the ground then the work done by the gravitational force is given by_______.

1) $-\int_0^h W d y$
2) $\int_0^h W d y$
3) $-\int_{-h}^h W d y$
4) $\int_{-h}^h W d y$

Solution

When a block of mass m moves from a height h above the ground, the force acting on a body is in the upward direction against the force of gravity is F.

Work done by the force F,

$W=\int_0^h F d y$

The force of gravity is in the downward direction, but the displacement of the body is in the upward direction. So, the work done by the force of gravity is negative.

The block moves from the height h to the ground, and then work is done by the force of gravity- $W_{\text {gravity }}=-W=-\int_0^h F d y-\int_0^h W d y$

The negative sign is because the work done is in the opposite direction of the motion of the body.

Hence, the answer is the option (1).

Example 3: A particle moves from point (0,0) to point (9,9) by two paths under a constant force. Path 1 is OP and path 2 is OQP as shown in the diagram. If work done by both paths is $W_1$ and $W_2$ then

1) $W_1=W_2$
2) $W_1=2 W_2$
3) $W_2=2 W_1$
4) $W_2=4 W_1$

Solution

Work done is path-independent

It depends upon the initial and final position

wherein

When force is constant.

For path OP
$
\begin{aligned}
& W_1=\vec{F} \cdot \vec{S} \\
& \vec{F}=(3 \hat{i}+4 \hat{j}) \quad s=a \hat{i}+a \hat{j} \\
& F=3 a+4 a=7 a J
\end{aligned}
$
For path OQP

$
\begin{aligned}
W_2 & =W_{O Q}+W_{Q P} \\
W_2 & =\vec{F} \cdot \overrightarrow{O Q}+\overrightarrow{F Q P} \\
& =((3 \hat{i}+4 \hat{j}) \cdot \hat{a i})+((3 \hat{i}+4 \hat{j}) \cdot a \hat{j}) \\
= & 3 a+4 a=7 a \\
W_1 & =W_2
\end{aligned}
$

Example 4: Find the horizontal distance (in meters) covered by the block when it hits the ground. Assume all surfaces are frictionless and it is horizontal at Q.(height of point P from the ground is 1m )

1) 1

2) 2

3) 1.5

4) 3

Solution:

Conservative Force

Work done by the force along a closed path is zero.

wherein

Work done by the Conservative Force, i.e. Gravitational Force, depends only upon the initial and final position.

Work done from P to Q
$
\begin{aligned}
& W=m g(1-0.5)=m g / 2 \\
& \quad \Delta W=(\text { change in } K . E) \\
& \text { so } \\
& \frac{m g}{2}=\frac{1}{2} m v^2 \Rightarrow v=\sqrt{g}
\end{aligned}
$
Time to reach the ground

$
\begin{gathered}
t=\sqrt{\frac{2 h}{g}} \\
t=\sqrt{\frac{2 \times 0.5}{g}}=\frac{1}{\sqrt{g}}
\end{gathered}
$
Horizontal Distance $=v t=\sqrt{g} \times \frac{1}{\sqrt{g}}=1 \mathrm{~m}$

Example 5: A ball of mass m = 3 Kg is being pulled from the ground on a fixed rough hemispherical surface up to the top of the hemisphere with the help of a light inextensible string. Find work (in Joule) done by tension M in the string if the radius of the hemisphere is R = 1 m and the friction coefficient is $\mu=0.6$

Assume that the block is pulled with negligible velocity

1) -40

2) 48

3) 35

4) -50

Solution:

$\begin{aligned} & T=f_r+m g \sin \theta \\ & T=\mu N+m g \sin \theta \\ & N=m g \cos \theta\end{aligned}$

$\begin{aligned} & T=m g(\sin \theta+\mu \cos \theta) \\ & \quad W=\int_0^{\pi / 2} m g(\sin \theta+\mu \cos \theta) R d \theta \\ & =m g(-\cos \theta+\mu \sin \theta)_0^{\frac{\pi}{2}} \\ & =3 \times 10 \times[0.6-(-1)]=+48 J\end{aligned}$

Summary

Conservative forces, such as gravitational and elastic forces, do not dissipate energy and depend only on the initial and final positions of an object, making the work done path-independent. In contrast, non-conservative forces like friction and air resistance dissipate energy, and the work done by these forces depends on the path taken. Understanding these forces helps solve problems related to energy conservation, as demonstrated in the examples of objects moving under gravitational and frictional forces.