An alternate method to solve many dynamic problems is the energy method. This method is particularly helpful if the problem involves finding position or velocity when two different positions along its motion path are known.

Basically, energy methods analyze all energy added or deleted as an object moves between two points. The difference in energy between the two points must be accounted for with an increase or decrease in kinetic energy (i.e. velocity). In equation form, this is stated as,

T_{1} + ΣU_{1-2} = T_{2}

where
T_{1} = kinetic energy at position
1 = ½mv_{1}^{2}
T_{2} = kinetic energy at position
2 = ½mv_{2}^{2} ΣU_{1-2} =
total change in energy (i.e. work)
between positions 1 and 2

Energy methods are a convenient method to solve for velocity or position. However, energy methods do not work for acceleration.

Notice, energy terms are scalars which do not have direction. The energy equation is not a vector equation. This means that only one unknown can be determined.

The most common energy generated in a dynamics problem is a force, F, acting through a distance (or more acurately, a change in position, r). This written in equation form as

ΔU = F•dr

The dot product is needed since only the force in the direction of the motion will produce work.

To calculate the total work, integrate the work over the motion path. The total work done from position 1 to position 2 is

Notice that F and the angle θ can change over the path.

When F and θ are constant, the work equation simplifies to

Positive work means the force is acting in the direction of the motion (i.e. increasing its velocity). Negative work is when the force is acting opposite the motion (i.e. reducing its velocity).

The motion of an object is also stored energy and is called kinetic energy. This can be illustrated by applying a simple force, F, on an object over a distance, s, and finding its final velocity. In other words, the change in work (FΔs) will produce a change in kinetic energy, or simply

ΔU_{kinetic} = F Δs

dU_{kinetic} = Fds

Using the relationships, "F = ma" and, "a ds = v dv", gives,

dU_{kinetic} = m(v dv/ds) ds = mv dv

Integrating both sides gives,

U_{kinetic} = ½mv^{2}

Thus the kinetic energy, T, for any object in motion is