When asked to perform a physically difficult task, a typical person is likely to say either "That's too much work!" or "That takes too much energy!"

The fact that these expressions are used interchangeably, and that most people use “energy” and “work” to mean the same thing when it comes to their relationship to physical toil, is no coincidence; as is so often the case, physics terms are often extremely illuminating even when used colloquially by science-naïve folks.

Objects that possess internal energy by definition have the capacity to do *work*. When an object’s *kinetic energy* (energy of motion; various subtypes exist) changes as a result of work being done on the object to speed it up or slow it down, the change (increase or decrease) in its kinetic energy is equal to the work performed on it (which can be negative).

Work, in physical-science terms, is the result of a force displacing, or changing the position of, an object with mass. “Work is force times distance” is one way to express this concept, but as you’ll find, that’s an oversimplification.

Since a net force accelerates, or changes the velocity of, an object with mass, developing the relationships between the motion of an object and its energy is a critical skill for any high-school or college physics student. The **work-energy theorem** packages all of this together in a neat, easily assimilated and powerful way.

## Energy and Work Defined

Energy and work have the same basic units, kg ⋅ m^{2}/s^{2}. This mix is given an SI unit of its own, the **Joule**. But work is usually given in the equivalent **newton-meter** (**N ⋅m**). They are scalar quantities, meaning that they have a magnitude only; vector quantities such as **F**, **a**, **v** and **d** have both a magnitude and a direction.

Energy can be kinetic (KE) or potential (PE), and in each case it comes in numerous forms. KE can be translational or rotational and involve visible motion, but it can also include vibrational motion at the molecular level and below. Potential energy is most often gravitational, but it can be stored in springs, electrical fields and elsewhere in nature.

Net (total) work done is given by the following general equation:

**W _{net} = F_{net} ⋅ d cos θ,**

where **F _{net}** is the net force in the system,

**d**is the displacement of the object, and θ is the angle between the displacement and force vectors. Though both force and displacement are vector quantities, work is a scalar. If the force and the displacement are in opposite directions (as occurs during deceleration, or a decrease in velocity while an object continues on the same path), than cos θ is negative and W

_{net}has a negative value.

## Definition of the Work-Energy Theorem

Also known as the work-energy principle, the work-energy theorem states that the total amount of work done on an object is equal to its change in kinetic energy (the final kinetic energy minus the initial kinetic energy). Forces do work in slowing objects down as well as speeding them up, as well as moving objects at constant velocity when doing so requires overcoming an extant force.

If KE decreases, then net work W is negative. In words, this means that when an object slows down, "negative work" has been done on that object. An example is a skydiver's parachute, which (fortunately!) causes the skydiver to lose KE by slowing her down greatly. Yet the motion during this deceleration (loss of velocity) period is downward because of the force of gravity, opposite the direction of the drag force of the chute.

- Note that when
**v**is constant (that is, when ∆v = 0), ∆KE = 0 and W_{net}= 0. This is the case in uniform circular motion, such as satellites orbiting a planet or star (this is actually a form of free fall in which only the force of gravity accelerates the body).

## Equation for the Work-Energy Theorem

The most commonly encountered form of the theorem is probably

**W _{net} = (1/2)mv^{2}– (1/2)mv_{0}^{2}**,

Where **v**_{0} and **v** are the initial and final velocities of the object and *m* is its mass, and *W _{net}* is the net work, or total work.

#### Tips

The simplest way to envision the theorem is

**W**_{net}= ∆KE, or W_{net}= KE_{f}– KE_{i}.

As noted, work is usually in newton-meters, while kinetic energy is in joules. Unless otherwise specified, force is in newtons, displacement is in meters, mass is in kilograms and velocity is in meters per second.

## Newton’s Second Law and the Work-Energy Theorem

You already know that W_{net} = **F _{net}d cos** θ

**,**which is the same thing as W

_{net}= m|

**a||d| cos**θ (from Newton's second law,

**F**= m

_{net}**a**). This means that the quantity (

**ad**), acceleration times displacement, is equal to W/m. (We delete cos(θ) because the associated sign is taken care of by the product of

**a**and

**d**).

One of the standard kinematic equations of motion, which deals with situations involving constant acceleration, relates an object's displacement, acceleration, and final and initial velocities: **ad** = (1/2)(**v _{f}^{2} – v_{0}^{2}**). But because you just saw that

**ad**= W/m, then W = m(1/2)(

**v**), which is equivalent to W

_{f}^{2}– v_{0}^{2}_{net}= ∆KE = KE

_{f}

**–**KE

_{i}.

## Real-Life Examples of the Theorem in Action

**Example 1:** A car with a mass of 1,000 kg brakes to a stop from a velocity of 20 m/s (45 mi/hr) over a length of 50 meters. What is the force applied to the car?

∆KE = 0 – [(1/2)(1,000 kg)(20 m/s)^{2}] = –200,000 J

W = **–**200,000 Nm = (**F**)(50 m); **F = –4,000 N**

**Example 2:** If the same car is to be brought to rest from a velocity of 40 m/s (90 mi/hr) and the same braking force is applied, how far will the car travel before it stops?

∆KE = 0 – [(1/2)(1,000 kg)(40 m/s)^{2}] = –800,000 J

-800,000 = (–4,000 N)d; **d = 200 m**

Thus doubling speed causes the stopping distance to quadruple, all else held the same. If you have the perhaps intuitive idea in your mind that going from 40 miles an hour in a car to zero "only" results in twice as long a skid as going from 20 miles an hour to zero does, think again!

**Example 3:** Assume you have two objects with the same momentum, but m_{1} > m_{2} while v_{1} < v_{2}. Does it take more work to stop the more massive, slower object, or the lighter, speedier object?

You know that m_{1}v_{1} = m_{2}v_{2}, so you can express v_{2} in terms of the other quantities: v_{2} = (m_{1}/m_{2})v_{1.} Thus the KE of the heavier object is (1/2)m_{1}v_{1}^{2} and that of the lighter object is (1/2)m_{2}[(m_{1}/m_{2})v_{1}]^{2}. If you divide the equation for the lighter object by the equation for the heavier one, you find that the lighter object has (m_{2}/m_{1}) more KE than the heavier one. This means that when confronted with a bowling ball and marble with the same momentum, the bowling ball will take less work to stop.

#### References

- Physics LibreTexts: Work-Energy Theorem
- B.C. Open Textbooks: Kinetic Energy and the Work-Energy Theorem
- Union College: Kinetic Energy & the Work-Energy Theorem
- Texas Gateway by TEA: The Work-Energy Theorem
- Georgia State University: HyperPhysics: Work, Energy and Power
- University of Winnipeg: Kinetic Energy and the Work Energy Theorem