2.3.2.2. Work-Energy Theorem

Table of Contents

Core Idea

The work-energy theorem connects force and motion through energy. It states that the net work done on an object equals the change in its kinetic energy.

If an object speeds up, its kinetic energy increases, so the net work on it is positive. If it slows down, its kinetic energy decreases, so the net work is negative. If its speed stays the same, the net work is zero.

Work-energy theorem
$$
W_{\text{net}} = \Delta K = K_f - K_i
$$
For translational motion,
$$
K = \frac{1}{2}mv^2
$$
so
$$
W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2
$$

Meaning of the Theorem

This theorem is powerful because it links the total effect of all forces to the change in speed, without always needing to find the acceleration and time separately.

The word net is important. It is not the work done by just one force, unless that is the only force acting. It is the sum of the work done by all forces on the object.

If several forces act on an object, then

$$
W_{\text{net}} = W_1 + W_2 + W_3 + \cdots
$$

and this total equals the change in kinetic energy.

A Simple Derivation

For motion along a straight line, start with Newton's second law:

$$
F_{\text{net}} = ma
$$

Using the kinematic relation

$$
a = \frac{dv}{dt}
$$

and the fact that

$$
v = \frac{dx}{dt}
$$

we can write

$$
a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}
$$

So,

$$
F_{\text{net}} = m v \frac{dv}{dx}
$$

Multiply both sides by $dx$:

$$
F_{\text{net}}\,dx = mv\,dv
$$

Integrate from the initial state to the final state:

$$
\int_{x_i}^{x_f} F_{\text{net}}\,dx = \int_{v_i}^{v_f} mv\,dv
$$

The left side is the net work, and the right side becomes

$$
W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2
$$

which is exactly

$$
W_{\text{net}} = \Delta K
$$

Interpreting the Sign of Work

The theorem helps you read what forces are doing to motion.

Net work	Change in kinetic energy	Effect on speed
Positive	$\Delta K > 0$	Speed increases
Negative	$\Delta K < 0$	Speed decreases
Zero	$\Delta K = 0$	Speed unchanged

An object can still be moving when the net work is zero. Zero net work means its kinetic energy does not change, not that its velocity must be zero.

How to Use the Theorem

In many problems, the main steps are simple. First identify the object. Then determine the initial and final speeds. Then find the net work done on the object. Finally set that equal to the change in kinetic energy.

If the object has constant mass, the theorem becomes

$$
W_{\text{net}} = \frac{1}{2}m\left(v_f^2 - v_i^2\right)
$$

This often lets you solve for speed directly.

Practical form
$$
W_{\text{net}} = \frac{1}{2}m\left(v_f^2 - v_i^2\right)
$$
If you know the net work and the initial speed, you can find the final speed.

Example, Object Speeding Up

Suppose a $2.0\,\text{kg}$ object starts from rest and the net work done on it is $18\,\text{J}$.

Since $v_i = 0$,

$$
18 = \frac{1}{2}(2.0)v_f^2
$$

$$
18 = v_f^2
$$

$$
v_f = \sqrt{18} \approx 4.2\,\text{m/s}
$$

The positive net work increases the object's kinetic energy.

Example, Object Slowing Down

A $1.5\,\text{kg}$ object moves initially at $6.0\,\text{m/s}$. The net work done on it is $-18\,\text{J}$. Then

$$
-18 = \frac{1}{2}(1.5)v_f^2 - \frac{1}{2}(1.5)(6.0)^2
$$

First compute the initial kinetic energy:

$$
K_i = \frac{1}{2}(1.5)(36) = 27\,\text{J}
$$

$$
-18 = K_f - 27
$$

$$
K_f = 9\,\text{J}
$$

Now solve for the final speed:

$$
9 = \frac{1}{2}(1.5)v_f^2
$$

$$
9 = 0.75v_f^2
$$

$$
v_f^2 = 12
$$

$$
v_f \approx 3.46\,\text{m/s}
$$

The object is still moving, but more slowly.

Work-Energy Theorem and Multiple Forces

Often several forces act at once. For example, on a box moving across a floor, an applied force may do positive work while friction does negative work. The theorem says that the sum of these works determines the change in kinetic energy.

$$
W_{\text{applied}} + W_{\text{friction}} + W_{\text{other}} = \Delta K
$$

So even if one force adds energy, another may remove some of it.

Rendering scientific sketch…

Forces and net work on a moving box

If the applied work is larger in magnitude than the negative work of friction, the box speeds up. If friction's negative work is larger, the box slows down.

Geometric View from a Force-Position Graph

When force acts along the direction of motion, work is the area under the force versus position graph. Therefore, the net area under the net force curve equals the change in kinetic energy.

$$
W_{\text{net}} = \int_{x_i}^{x_f} F_{\text{net}}(x)\,dx = \Delta K
$$

This form is especially useful when the force changes with position.

Limits of What the Theorem Tells You

The work-energy theorem gives information about speed, because kinetic energy depends on $v^2$. By itself, it does not directly tell you the direction of motion. It also does not directly give the time taken.

This is why it is often used together with other ideas in mechanics, depending on what the problem asks.

Important limitation
The work-energy theorem determines the change in kinetic energy, not directly the direction of velocity or the elapsed time.

Common Mistakes

A frequent mistake is to use the work done by only one force and set it equal to $\Delta K$ when several forces act. Another common mistake is to forget that negative work reduces kinetic energy. It is also important to remember that kinetic energy depends on the square of speed, so a doubling of speed does not mean a doubling of kinetic energy.

Mistake	Correction
Using one force's work instead of total work	Use $W_{\text{net}}$
Forgetting negative work	Negative work means kinetic energy decreases
Confusing velocity with kinetic energy	Kinetic energy depends on $v^2$
Assuming zero net work means zero motion	Zero net work means constant kinetic energy

Final Statement

The work-energy theorem is one of the most useful results in mechanics because it turns a force problem into an energy relation:

$$
W_{\text{net}} = \Delta K
$$
It means that the total work done on an object changes its kinetic energy.

Whenever you want to know how forces change an object's speed, this theorem is often the fastest path.

Comments

Please login to add a comment.

Don't have an account? Register now!