Work Energy and Power
4.0 Conservative & Non-conservative forces
4.0 Conservative & Non-conservative forces
Conservative forces: The work done due to conservative forces are independent of path.
Example: Gravitational forces, Columb's force
The work done due to these forces depends on the initial and final position only.
The work done by a conservative force is equal ${W_1} = {W_2} = {W_3}$ to displace a particle from point $A$ to $B$ irrespective of the path followed.
Since the work done due to conservative forces depends on the initial and final position of the particle.
So, the total work done by the conservative forces in a closed path is zero.
Non-conservative forces: The work done due to nonconservative forces are path dependent.
Example: Frictional forces, viscous forces
Also, ${W_1} \ne {W_2} \ne {W_3}$, as the work done by non-conservative forces depends on the path followed between initial and final position.
Since the work done by the non-conservative forces depends on the path followed.
So, the total work done by non-conservative forces is not zero in a closed path.
Question 9. Find the work done by gravity in displacing a block of mass $m$ as shown in the figure for the following cases.
- Along the wedge from point $A$ to $B$
- Vertically from point $A$ to $B$
Is the work done by gravity in both the cases are same or different and why?
Solution: Let us solve both the cases individually.
Case 1:
Work done by gravity in displacing a block of mass $m$ along the wedge from point $A$ to $B$, $$\begin{equation} \begin{aligned} {W_g} = \overrightarrow F .\overrightarrow s \\ {W_g} = \left( {mg.H\cos ec\theta } \right)\cos \left( {90^\circ + \theta } \right) \\ {W_g} = - mgH \\\end{aligned} \end{equation} $$
Case 2:
Work done by gravity in displacing the same block from point $A$ to $B$ vertically,$$\begin{equation} \begin{aligned} {W_g} = \overrightarrow F .\overrightarrow s \\ {W_g} = mgH\cos \pi \\ {W_g} = - mgH \\\end{aligned} \end{equation} $$
Work done by gravity $(W_g)$ is same in both the cases because the gravitational force is a conservative force, as it only depends on the initial & the final position and also irrespective of the path followed.
Question 10. Find the work done by frictional force in displacing a block of mass $m$ from point $A$ to $B$ as shown in the figure for the following cases
Along the curve path $ACB$ of length $L$
Along the straight line path $AB$ of length $l$
Is the work done by frictional force in both the cases are same or different and why?
Solution: From FBD, $$\begin{equation} \begin{aligned} N = mg \\ f = \mu N \\ f = \mu mg \\\end{aligned} \end{equation} $$
Case 1:
Work done by frictional force in displacing a block of mass $m$ along the curve path $ACB$, $$\begin{equation} \begin{aligned} {W_f} = \overrightarrow F .\overrightarrow s \\ {W_f} = \mu mgL\cos \pi \\ {W_f} = - \mu mgL \\\end{aligned} \end{equation} $$
Case 2:
Work done by frictional force in displacing a block of mass $m$ along the straight path $AB$,$$\begin{equation} \begin{aligned} {W_f} = \overrightarrow F .\overrightarrow s \\ {W_f} = \mu mgl\cos \pi \\ {W_f} = - \mu mgl \\\end{aligned} \end{equation} $$
The work done by frictional force is different in both the cases because frictional force is non- conservative in nature, and it is path dependent.
