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Suppose there is a gravitational force acting on a particle, as shown below.

The gravitational force acting on the particle is pointing in the direction of gravity and its magnitude is equal to

The work done by gravity in moving the particle is dependent only on the change in vertical height (displacement) Δ

If Δ

A negative sign is present in the above equation because vertical displacement (Δ

The expression for

Now, if we have a body instead of a particle, the same analysis applies. We simply treat the center of mass of the body as a particle, and apply the above procedure to find the work done by gravity. In other words, if we want to find the work done by gravity on the body we look at the motion of its center of mass, and then apply the equation above for

The work done by gravity on the body is dependent only on the change in vertical height Δ

If Δ

Note that it doesn’t matter at all how the body moves. We only need to know the change in vertical position of its center of mass (as it moves from

Consider a spring force acting on a particle, as shown below. The spring is attached to a wall at point

Note that the dashed line represents the equilibrium position of the spring, where the spring is unstretched. Once again, the possible presence of other forces acting on the particle is irrelevant to this discussion, since we are only focusing on the work done by the spring.

The work done by the spring (

where

For the above equation we are assuming we have a spring that obeys Hooke's Law.

Therefore, the work done by the spring on the particle depends only on the position of

If we replace the particle with a point on a body (to which the end of the spring is attached), the work done by the spring on the body is also given by the above equation. The figure below illustrates this situation.

The work done by the spring on the body is dependent only on the amount the spring is stretched or compressed from its equilibrium position, as it moves from

The work done by the spring (

Note that it doesn’t matter at all how the body moves. We only need to know the amount the spring is stretched or compressed from its equilibrium position (as it moves from

Knowing if a force is conservative is very useful for solving problems using energy calculations, since you only have to know the initial and final position where the force is acting. You don’t need to know the path traveled in between.

Conservative forces do zero work in those cases where the points (or particles) upon which the forces act, return to their original position. As a result, if a system is acted on by only conservative forces (such as gravity, or a spring), and that system returns to its original position, then that system will experience no net loss or gain of energy, due to those forces. Energy will thus be conserved for the system.

A non conservative force is a force that acts on a particle (or point), such that the work done by this force in moving this particle from one point to another is

The same reasoning applies if friction is acting on a body. The work done by the friction force depends on the path traveled by that area on the body acted upon by the friction force.

Thus, if a system is acted on by a non conservative force (such as friction), and that system returns to its original position, then that system will experience a net loss of energy, due to those forces. Energy will thus not be conserved for the system. This makes sense intuitively since we know friction is a source of energy loss. This is why we always try to minimize friction in moving parts and machine components so as to minimize the energy wasted.

Using Calculus, we can set up a general expression for the work done by a force vector

Let |

Let

where

Using the vector dot product, the previous equation can be rewritten as:

where

Now, let

Where:

Substituting into the above equation for

The total work (

If

where

In other words, the work

The above equation for work is analogous to the work done by a conservative gravitational force, and a conservative spring force, that is dependent only on the positions

Without loss of generality, we can say that the position of the particle on the blue curve is represented by three functions for

Now,

Substituting these into equation (1) we get

Now if,

The above integral can be rewritten as

By the chain rule for partial derivatives, the term in brackets is equal to

Therefore,

This result is very important. It says that a force field

For example, the force field due to gravity acting on a body of mass

where we define gravity as pointing in the negative

Therefore, by comparison

These above three equations are satisfied by

where

If a function

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