If the work done by a force field along a closed path is zero, the force is said to be conservative otherwise it is called non-conservative. Under such forces, the work done depends only upon the initial and final positions and is path independent.

Conservative forces are non-dissipative and we store work (or energy ) whereas the non-conservative forces are dissipative where in we do not store work.

Example of conservative force : Gravitational force

Examples of non-conservative force : friction , viscous force etc.

Example : A particle is taken from point P to point Q via the path PAQ and then placed back to point P via the path QBP. Find the work done by gravity on the body over this closed path. Vertical separation between P and Q is h as shown.

Solution: Here, displacement of the particle is PQ, gravity is acting vertically downward.

The vertical component of PQ is h (say) upward. Hence

W_{(PAQ)} = -mgh . . . (1)

For the path QBP, component of the displacement along vertical is h(downward)

In this case, W_{(QBP)} = mgh

Total work done = W_{PAQ} + W_{QBP} = 0

Example : A block of mass m is moving down with constant velocity along an inclined plane of inclination θ. What is the work done in pulling the block along the inclined plane through a height h with constant velocity.

Solution : Since the block moves down along the inclined plane with constant velocity, its net acceleration is zero. Therefore the net external force acting on the block is zero.

=> ΣF_{x} = 0 and ΣF_{y} = 0

=> mg Sinθ − f_{k} = 0

Where , f_{k} = kinetic friction developed at the interface of the block and the inclined plane (since the block is sliding).

=> f_{k} = mg sinθ …..(1)

Let the applied force F be parallel to the inclined plane. When it pulls the block up, kinetic friction acts down to oppose the relative motion. The work done by the force to displace the block through a distance l along the inclined plane is given as

$ \displaystyle W= \vec{F}.\vec{S}$ = F S = F l

where l = h cosec θ

=> W = F h cosecθ ……..(2)

Since the block slides up with constant velocity, the net acceleration, therefore the net force acting on the block,

ΣF_{x} = 0 and ΣF_{y} = 0

⇒ F – mg sinθ – f_{k} = 0 ….(3)

Using (1) and (3)

F = 2mg sinθ ….(4)

Using (2) and (3), we obtain

W = 2 mgh

Examle: Find the work done by the frictional force in drawing a circle of radius r by a pencil of negligible mass with a normal pressing force N (coefficient of friction μ_{k})

Solution: The kinetic frictional force at any instantaneous position is f_{k} = μ_{k}N

f_{k} , acts tangentially opposite to the motion of the pencil.

For elementary displacement dl^{→} the work done by f_{k} ^{→} is

$ \displaystyle dW = \vec{f_k}.\vec{dl} $

= − f_{k} dl

$ \displaystyle W = -\int_{0}^{2\pi r}f_k dl$

$ \displaystyle W = -\mu_k N\int_{0}^{2\pi r} dl$

= − μ_{k}N ( 2πr )

= − 2πr μ_{k}N

Exercise : In the figure an inextensible string that connects two bodies of mass M and m, passing over a fixed smooth pulley. The body M slides along a smooth vertical rigid bar. If the body M is released from the given position, find the maximum distance raised by body m. (M < m)