Kinetic vs Potential Energy: The Difference

Kinetic energy is the energy an object has because it is moving, one half m v squared. Potential energy is not a property of one object at all: it belongs to a system of two or more objects that interact through a conservative force, and it depends on where those objects sit relative to each other.

AP Physics: Unit 3 (topics 3.1 Translational Kinetic Energy, 3.3 Potential Energy, 3.4 Conservation of Energy). This comparison sits across three topics of AP Physics 1 Unit 3, Work, Energy, and Power, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods. Topic 3.1 has one learning objective, 3.1.A, describe the translational kinetic energy of an object in terms of the object's mass and velocity, supported by essential knowledge 3.1.A.1 through 3.1.A.3, with suggested skills 1.C, 2.B, 3.B and 3.C. Topic 3.3 has one learning objective, 3.3.A, describe the potential energy of a system, supported by essential knowledge 3.3.A.1 through 3.3.A.5, with suggested skills 1.C, 2.C, 2.D and 3.B. The CED lists no boundary statement under either Topic 3.1 or Topic 3.3. The system rules come from Topic 3.4: EK 3.4.A.1 says a system composed of only a single object can only have kinetic energy, EK 3.4.A.2 says a system whose objects interact via conservative forces or that can change shape reversibly may have both, EK 3.4.B.1 defines mechanical energy as the sum of a system's kinetic and potential energies, and EK 3.4.C.2 and 3.4.C.3 give the conditions under which a chosen system's total energy is or is not constant. The Topic 3.4 boundary statement says AP Physics 1 expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces.

One belongs to an object, the other to a system

Almost every wrong sentence about potential energy starts the same way: "the ball has potential energy." It does not. A single object cannot have potential energy. The AP Physics 1 CED says so twice, in two different topics.

Essential knowledge 3.3.A.1: a system composed of two or more objects has potential energy if the objects within that system only interact with each other through conservative forces. Count the requirements in that sentence. Two or more objects. Interacting. Through conservative forces.

EK 3.4.A.1 states the other half: a system composed of only a single object can only have kinetic energy.

Kinetic energy has no such requirement. EK 3.1.A.1 gives an object's translational kinetic energy as

K = \frac{1}{2}mv^2

which needs one mass and one speed. Hand it a lone object and it returns a number.

So the ball held above the floor does not have potential energy. The ball and the Earth have it, together, because of how far apart they are. The spring does not have elastic potential energy; the block and the spring have it, because of how far the spring is from its relaxed length. That is not pedantry you can skip: the CED asks you to identify the system before you write an energy term, and a free-response answer that assigns potential energy to a lone object has misidentified the physics that stores it.

A one-line test: if you cannot name the two things whose separation changed, you do not have a potential energy term. Kinetic energy needs no such pair. It needs a mass, a speed, and a frame to measure the speed in.

Kinetic vs potential energy, side by side

Question you are asking	Kinetic energy	Potential energy
What owns it	One object, from its motion	A system of two or more interacting objects, from their arrangement
Symbol	$K$	$U$
SI unit	joule	joule
Vector or scalar	Scalar (EK 3.1.A.2)	Scalar (EK 3.3.A.2)
What it depends on	Mass and speed	The positions of the objects within the system
Direction of motion matters	No, $v$ is squared	No, only the configuration
Can be negative	No	Yes, and the general gravitational form always is
Needs a reference choice	A reference frame for the speed	A zero of potential energy
Who makes that choice	The observer's frame (EK 3.1.A.3)	The observer, to simplify the analysis (EK 3.3.A.3)
Universal formula	Yes, $\frac{1}{2}mv^2$ for translation	No, one form per interaction
Which forces produce it	Any force that changes the speed	Conservative forces only (EK 3.2.A.1.iii)
Zero when	The object is at rest in your frame	The system sits at the configuration you called zero
CED essential knowledge	3.1.A.1, 3.1.A.2, 3.1.A.3	3.3.A.1 through 3.3.A.5

Two rows in that table carry the page.

The "can be negative" row. Kinetic energy is $\frac{1}{2}mv^2$ with $v$ squared, so it cannot come out below zero however you set up your axes. Potential energy can be negative, and the general gravitational form on the equation sheet, $U_G = -\frac{Gm_1m_2}{r}$ , is negative for every finite separation. A negative potential energy is not an error and it does not mean the system has less than nothing; it means the configuration sits below whatever configuration you chose to call zero.

The "needs a reference choice" row. Both quantities need a human decision before they have a number, and the two decisions are different in kind. Kinetic energy needs a frame. Potential energy needs a zero. Getting those two confused produces the error in the next section.

Kinetic energy needs a frame, potential energy needs a zero

This is the asymmetry students never get told, and the CED states both halves.

EK 3.1.A.3: different observers may measure different values of the translational kinetic energy of an object, depending on the observer's frame of reference. A suitcase resting on a train's luggage rack has zero kinetic energy to a passenger and a large kinetic energy to someone standing on the platform. Neither is wrong. They measured different speeds because they were in different frames, and $v$ is the only thing $\frac{1}{2}mv^2$ takes.

EK 3.3.A.3: the definition of zero potential energy for a given system is a decision made by the observer considering the situation to simplify or otherwise assist in analysis. Put the zero at the floor, at the table top, at the release point, or at infinity. The number for $U$ changes with that choice.

What neither choice can move is the physics.

Changing your frame changes $K$ , but the object still arrives at the same place at the same time.
Changing your zero changes $U$ , but $\Delta U$ between two configurations is the same for every choice of zero, because the zero cancels in the subtraction.

That second point is why $\Delta U_g = mg\Delta y$ is written as a change on the equation sheet rather than as a value. There is no near-Earth formula for $U_g$ itself, only for the difference between two heights, and the difference is the quantity that does not depend on your bookkeeping.

Worked example one below takes one falling ball, measures it from two frames, and shows the kinetic energies disagreeing while the change in potential energy does not move at all.

Kinetic energy has one formula, potential energy has one per interaction

There is a single expression for translational kinetic energy and you use it everywhere. Potential energy has no single expression, and that is a consequence of the definition rather than an inconvenience. Potential energy is stored in an interaction, so each interaction gets its own expression, and EK 3.3.A.4 says the potential energy of common physical systems can be described using the physical properties of that system.

The AP Physics 1 sheet prints three potential energy expressions and one kinetic energy expression for translation.

Expression on the sheet	What kind of system it describes	The two objects
$K = \frac{1}{2}mv^2$	Any single moving object	Not applicable, one object is enough
$U_s = \frac{1}{2}k(\Delta x)^2$	An object and an ideal spring	The object and the spring
$U_G = -\frac{Gm_1m_2}{r}$	Two approximately spherical masses at any separation	The two masses, labeled $m_1$ and $m_2$
$\Delta U_g = mg\Delta y$	An object near a planet's surface	The object and the planet

Read the right-hand column down. Every potential energy row names two objects. The kinetic energy row cannot, because there is nothing to pair the object with.

A notation detail worth knowing before the exam: the sheet writes the general gravitational form with a capital subscript, $U_G$ , and the near-surface change with a lowercase subscript, $\Delta U_g$ . They are the same physics at two levels of approximation. EK 3.3.A.4.iii explains the link: because the gravitational field near the surface of a planet is nearly constant, the change in gravitational potential energy in a system consisting of an object with mass $m$ and a planet with gravitational field of magnitude $g$ when the object is near the surface of the planet may be approximated by $\Delta U_g = mg\Delta y$ .

One further rule for systems with more than two objects, EK 3.3.A.5: the total potential energy of a system containing more than two objects is the sum of the potential energy of each pair of objects within the system. Pairs, again. There is no other way to count it.

AP Physics 1 also has a second kinetic energy, rotational, $K = \frac{1}{2}I\omega^2$ , which arrives in Unit 6. It changes nothing about this page: it is still a property of one rotating system's motion, and it still needs no partner object.

Only conservative forces get a potential energy

EK 3.2.A.1.iii is one sentence long and it rules out half the forces you have met: potential energies are associated only with conservative forces.

That is why there is no "friction potential energy" and never will be. EK 3.2.A.1.iv says the work done by a nonconservative force is path-dependent, and EK 3.2.A.1.v names friction and air resistance as examples. A path-dependent force cannot have a potential energy, because a potential energy would have to assign one number to a configuration, and a path-dependent force gives you a different answer depending on how the system got there.

The positive version of the same rule, EK 3.2.A.1.i: the work done by a conservative force exerted on a system is path-independent and only depends on the initial and final configurations of that system. That path independence is exactly what lets a configuration carry a number.

EK 3.2.A.1.ii adds the closing test, and it is worth having in the form the CED gives it: the work done by a conservative force on a system, or the change in the potential energy of the system, will be zero if the system returns to its initial configuration. Take a book from the table to the shelf and back to the table and $\Delta U$ for the book and Earth is exactly zero. Push a crate across the floor and back to its starting spot and the thermal energy produced by friction is not zero, and no potential energy term can account for it.

EK 3.4.A.2 puts the two conditions together for you: a system that contains objects that interact via conservative forces or that can change its shape reversibly may have both kinetic and potential energies. Note may. Whether it does depends on the system you drew a boundary around, which is the subject of the next section.

The full split between the two force types, including what happens to the missing energy, is in conservative vs nonconservative force.

The same motion has potential energy or not, depending on your system

Here is the part that makes the object-against-system distinction practical rather than philosophical. Whether a potential energy term exists in your calculation is your choice, because it depends on where you drew the system boundary.

Drop a box. Two legitimate ways to account for it:

System: the box and the Earth. They interact gravitationally, which is a conservative force, so the system has gravitational potential energy. Nothing outside the system does work on it, so EK 3.4.C.2 applies and the total mechanical energy is constant. Potential energy falls, kinetic energy rises, the sum holds still.
System: the box alone. EK 3.4.A.1 says a single-object system can only have kinetic energy, so there is no potential energy term to write. Gravity is now an external force doing work on the system, EK 3.4.C.3 applies, and energy is transferred in from the environment. The kinetic energy rises by exactly the work gravity did.

Both give the same speed at the bottom. The CED spells this out in its own appendix, in a discussion titled "Constant or conserved?": in the box and Earth system, total mechanical energy is both conserved and constant, while in a system consisting only of the box, the total amount of energy in that system is not constant, but energy is conserved, because the increase in the box's kinetic energy is due to the transfer of energy into the box system by the external force of gravity doing work on the box.

Read the two words carefully, because the CED is drawing a line between them. Conserved means no energy vanished from the universe. Constant means the total inside your chosen system did not change. Those are different claims, and the second one depends on your system while the first never does.

Worked example three below runs one falling box through both routes and gets the same answer twice, which is the point.

One practical consequence: the moment you write $U$ into an equation, you have committed to a system that contains at least two interacting objects, and you must not then also count the work that the same interaction does. Counting gravity as a potential energy and as an external work in the same equation is double counting, and it doubles your answer.

What they share, and why that hides the difference

The two agree on enough to feel like one topic.

They are both energies, both measured in joules, and both scalars. EK 3.1.A.2 says translational kinetic energy is a scalar quantity and EK 3.3.A.2 says potential energy is a scalar quantity associated with the position of objects within a system. Neither has a direction, so neither gets resolved into components, and adding them is straightforward addition with signs.

EK 3.4.B.1 then adds them by name: mechanical energy is the sum of a system's kinetic and potential energies. Once you are inside a conservation-of-mechanical-energy problem, the two behave as a single budget with a fixed total, and every $U$ that leaves shows up as a $K$ that arrives. That is precisely the situation in which nobody needs the distinction, which is why it can go unnoticed for a whole unit.

The distinction reappears in four places, all of them exam-shaped.

A question that names the system. "For the block and spring system" is doing work in that sentence, and an answer that assigns the elastic energy to the spring alone has not read it.
A question that changes the reference. Move the zero height, or ask what an observer on a moving train measures, and one of the two quantities changes while the other does not.
A question that asks whether mechanical energy is constant. The answer depends on the system boundary and on whether a nonconservative force acts inside it.
A question about a single object. The moment the system is one object, potential energy is off the table by EK 3.4.A.1, and only kinetic energy and external work remain.

The energy accounting method is the routine for tracking both across a problem, and the work-energy theorem is the route that uses no potential energy term at all.

Where the confusion costs a mark

Each of these is a specific scoring error, not general caution.

Writing that an object has potential energy. EK 3.3.A.1 requires two or more objects. Say the ball and Earth system, or the block and spring system. On a question that asks you to identify the system, this is the mark.
Writing a potential energy term for a single-object system. EK 3.4.A.1 forbids it outright, and this error usually arrives together with counting gravity's work as well, which doubles the answer.
Counting gravity twice. Either gravity is inside the system and appears as $\Delta U_g$ , or it is outside and appears as work. Never both in one equation.
Calling a negative potential energy impossible. $U_G = -\frac{Gm_1m_2}{r}$ is negative at every finite separation with the zero at infinity, and that is the form the equation sheet prints.
Calling a negative kinetic energy possible. It is not. If your $K$ comes out negative you squared a velocity wrong or you subtracted in the wrong order. $\frac{1}{2}(2.0)(-3.0)^2$ is $9.0$ J.
Forgetting to state the zero of potential energy. EK 3.3.A.3 makes it your decision, which means an unstated choice is an incomplete answer. Declare it in words before the first energy line and hold it to the end.
Changing the zero halfway through. A height measured from the floor in one line and from the table in the next produces a wrong $\Delta U$ and it is invisible in the arithmetic.
Assuming kinetic energy is frame-independent. EK 3.1.A.3 says different observers may measure different values. A relative-motion question can turn on exactly this.
Inventing a potential energy for friction. EK 3.2.A.1.iii restricts potential energies to conservative forces. Friction dissipates mechanical energy; the Topic 3.4 boundary statement says AP Physics 1 expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces.
Treating $U_s = \frac{1}{2}k(\Delta x)^2$ as if $\Delta x$ were a position. EK 3.3.A.4.i defines $\Delta x$ as the distance the spring has been stretched or compressed from its equilibrium length, so it is measured from the relaxed length, not from your coordinate origin.

What the CED asks, and how the exam frames it

These two ideas sit in AP Physics 1 Unit 3, Work, Energy, and Power, which is weighted at 18 to 23 percent of the multiple-choice section and estimated at about 22 to 27 class periods. That makes Unit 3 one of two units at the top of the weighting range, alongside Unit 2.

Topic 3.1, Translational Kinetic Energy, has one learning objective, 3.1.A: describe the translational kinetic energy of an object in terms of the object's mass and velocity. It carries exactly three pieces of essential knowledge, 3.1.A.1 through 3.1.A.3: the formula, the fact that it is a scalar, and the frame dependence. The CED lists no boundary statement under Topic 3.1. Suggested skills are 1.C, 2.B, 3.B and 3.C.

Topic 3.3, Potential Energy, has one learning objective, 3.3.A: describe the potential energy of a system. Note the last word. It carries five pieces of essential knowledge, 3.3.A.1 through 3.3.A.5, and 3.3.A.4 has three sub-points for the spring, general gravitational, and near-surface gravitational forms. The CED lists no boundary statement under Topic 3.3 either. Suggested skills are 1.C, 2.C, 2.D and 3.B.

Topic 3.4, Conservation of Energy, is where the two are combined, and it supplies the system rules quoted above at 3.4.A.1, 3.4.A.2, 3.4.B.1, 3.4.C.2 and 3.4.C.3. Its boundary statement is short and worth having exactly: AP Physics 1 expects students to know that mechanical energy can be dissipated as thermal energy or sound by nonconservative forces.

One more boundary matters for what you will and will not be asked. Under Topic 3.2 the CED says AP Physics 1 only expects students to analyze the transfer of mechanical energy, as defined in Unit 3 Topic 4, although students should be aware that mechanical energy may be dissipated in the form of thermal energy or sound, and that in AP Physics 2 students will also study how thermal energy can be transferred between systems through heating or cooling. So in AP Physics 1 you account for where the mechanical energy went; you do not compute a temperature rise.

For the CED framing topic by topic, see Topic 3.1, Topic 3.3 and Topic 3.4. The kinetic energy calculator will check a $\frac{1}{2}mv^2$ by hand, and the work, energy and power practice set has problems that name the system on purpose.

Two observers, two kinetic energies, one change in potential energy

A $0.40$ kg ball is released from rest and falls $1.8$ m. Observer A stands on the ground. Observer B rides past on a cart moving horizontally at a steady $3.0$ m/s. Find the ball's kinetic energy before and after the fall as each observer measures it, and find the change in gravitational potential energy of the ball and Earth system. Use $g = 9.8\ \text{m/s}^2$ and take upward as positive.

Speed at the bottom, from the ground frame: starting from rest, $v^2 = 2g h = 2(9.8)(1.8) = 35.28\ \text{m}^2/\text{s}^2$ , so $v = 5.94$ m/s downward to three significant figures. Keep $35.28$ unrounded, because the next lines use $v^2$ rather than $v$ .
Observer A, on the ground. Before: the ball is at rest in this frame, so $K_i = 0$ . After: $K_f = \frac{1}{2}(0.40)(35.28) = 7.056\ \text{J}$ . So $\Delta K = 7.056\ \text{J}$ .
Observer B, on the cart. In B's frame the ball carries a constant horizontal velocity component of $3.0$ m/s throughout, in addition to whatever it is doing vertically. Before: $K_i' = \frac{1}{2}(0.40)(3.0)^2 = 1.8\ \text{J}$ , and the ball was never at rest as far as B is concerned. After: the speed squared is the sum of the squares of the components, $3.0^2 + 35.28 = 9.0 + 35.28 = 44.28$ , so $K_f' = \frac{1}{2}(0.40)(44.28) = 8.856\ \text{J}$ .
Compare the kinetic energies. A measures $0$ then $7.056$ J. B measures $1.8$ then $8.856$ J. Four different numbers for one fall, and neither observer is wrong, which is EK 3.1.A.3 in action.
Now the potential energy. Take the release point as $y = 0$ , a choice EK 3.3.A.3 leaves to you. The ball falls, so $\Delta y = -1.8$ m and $\Delta U_g = mg\Delta y = (0.40)(9.8)(-1.8) = -7.056\ \text{J}$ . Both observers get this same number, because $m$ , $g$ and $\Delta y$ are the same in both frames and no speed appears anywhere in that line.
Check the energy books in A's frame: $\Delta K + \Delta U_g = 7.056 + (-7.056) = 0$ , so the mechanical energy of the ball and Earth system was constant, as EK 3.4.C.2 requires when nothing outside does work and no nonconservative force acts inside.
Now the subtle part, and the reason this example is worth running. In B's frame $\Delta K' = 8.856 - 1.8 = 7.056\ \text{J}$ , the same as A's. That agreement is not a general law: it happens here because B's frame moves horizontally while the ball's velocity change is vertical, so the cross term between the frame velocity and the velocity change is zero. The values of $K$ always depend on the frame; the change in $K$ happens to agree in this particular geometry. A frame moving vertically would not give $7.056$ J.
Note also what did not change under either choice. Moving the potential energy zero from the release point to the landing point would make $U$ read $+7.056$ J then $0$ instead of $0$ then $-7.056$ J, and $\Delta U_g$ would still be $-7.056$ J.

Ground observer: $K$ goes from $0$ to $7.056$ J. Cart observer: $K$ goes from $1.8$ J to $8.856$ J. Both measure $\Delta U_g = -7.056$ J for the ball and Earth system. Kinetic energy is a frame-dependent property of the moving object; the potential energy change belongs to the two-object system and no choice of frame or zero moves it.

A spring launcher: naming the two objects that hold the energy

A spring of stiffness $k = 240$ N/m is compressed $0.15$ m against a $0.30$ kg block on a frictionless horizontal surface, then released. The block leaves the spring, crosses the flat surface, and runs up a frictionless ramp. Find the elastic potential energy stored, the block's speed as it leaves the spring, and the maximum height it reaches on the ramp. State which system holds each energy at each stage.

Set the system and the zeros before touching a number. System: the block, the spring and the Earth. Zero of elastic potential energy: the spring at its relaxed length. Zero of gravitational potential energy: the flat surface. Upward positive.
Elastic potential energy stored, with $\Delta x$ measured from the relaxed length as EK 3.3.A.4.i requires: $U_s = \frac{1}{2}k(\Delta x)^2 = \frac{1}{2}(240)(0.15)^2 = \frac{1}{2}(240)(0.0225) = 2.7\ \text{J}$ . That $2.7$ J belongs to the block and spring together, not to the spring, because EK 3.3.A.1 needs two objects.
Release. The spring returns to its relaxed length, so $U_s$ drops to $0$ , and with no friction and no height change the whole $2.7$ J appears as kinetic energy of the block: $K = 2.7\ \text{J}$ . This kinetic energy belongs to the block alone, and it needed no partner object.
Speed on leaving: $\frac{1}{2}mv^2 = 2.7$ , so $v = \sqrt{\frac{2(2.7)}{0.30}} = \sqrt{18} = 4.24\ \text{m/s}$ to three significant figures.
Up the ramp. At the highest point the block is momentarily at rest, so $K = 0$ and the $2.7$ J has become gravitational potential energy of the block and Earth system: $mg h = 2.7$ , giving $h = \frac{2.7}{(0.30)(9.8)} = \frac{2.7}{2.94} = 0.918\ \text{m}$ .
Check by substitution rather than by trusting the algebra: $(0.30)(9.8)(0.918) = 2.70\ \text{J}$ , matching the stored energy to three significant figures.
Now read the three stages as a change of ownership. Stage one: $2.7$ J in the block and spring pair. Stage two: $2.7$ J in the block's motion, owned by one object. Stage three: $2.7$ J in the block and Earth pair. The same joules changed which system held them three times, while the kinetic energy stage was the only one that a single object could own.
One thing the ramp angle did not affect: the height. No angle appeared in any line, so a $20^\circ$ ramp and a $70^\circ$ ramp both stop the block at $0.918$ m, at different distances along the slope.

$U_s = 2.7$ J stored in the block and spring system, $v = 4.24$ m/s on leaving the spring, and a maximum height of $0.918$ m, where the $2.7$ J now sits in the block and Earth system. The kinetic energy stage is the only one owned by a single object; both potential energy stages name a pair.

One falling box, two system boundaries, one answer

A $2.5$ kg box is released from rest and falls $4.0$ m. Find its speed just before landing twice: once with the system taken as the box and the Earth, and once with the system taken as the box alone. Use $g = 9.8\ \text{m/s}^2$ and take downward displacement as a fall of $4.0$ m.

Route one, system is the box and the Earth. Two objects interacting through gravity, a conservative force, so EK 3.3.A.1 is satisfied and the system has gravitational potential energy. Nothing outside the system does work on it and no nonconservative force acts inside it, so by EK 3.4.C.2 the total mechanical energy is constant.
Set the zero of gravitational potential energy at the landing point and take upward as positive. The box falls, so $\Delta y = -4.0$ m and $\Delta U_g = mg\Delta y = (2.5)(9.8)(-4.0) = -98\ \text{J}$ .
Mechanical energy constant means $\Delta K + \Delta U_g = 0$ , so $\Delta K = +98\ \text{J}$ . The box started at rest, so $K = 98\ \text{J}$ and $v = \sqrt{\frac{2(98)}{2.5}} = \sqrt{78.4} = 8.85\ \text{m/s}$ .
Route two, system is the box alone. EK 3.4.A.1 says a system composed of only a single object can only have kinetic energy, so there is no potential energy term available and none may be written. Gravity is now an external force, and it does work on the system.
Work done by gravity, using $W = F_{\parallel}d$ from EK 3.2.A.3.i with the force and the displacement both downward so $\cos\theta = 1$ : $W = (mg)(d) = (2.5)(9.8)(4.0) = 98\ \text{J}$ , positive because the force and the displacement point the same way.
Apply the work-energy theorem, EK 3.2.A.4: $\Delta K = \sum W_i = 98\ \text{J}$ . Same $98$ J, so the same $v = 8.85$ m/s.
Compare the two ledgers. Route one: no energy crossed the system boundary, and $98$ J moved from potential to kinetic inside it. Route two: $98$ J crossed the boundary from outside, and the system's energy is not constant. Both are correct, and the CED names the distinction in its appendix: energy is conserved in both, but it is constant only in the box and Earth system.
The error this example is built to prevent: writing both ledgers in one equation. If you include $\Delta U_g = -98$ J and also credit gravity with $+98$ J of work, you get $\Delta K = 196$ J and a speed of $12.5$ m/s, too large by a factor of $\sqrt{2}$ . Choose one boundary and stay inside it.

$v = 8.85$ m/s by both routes. The box and Earth system stores $98$ J as gravitational potential energy and converts it, with mechanical energy constant. The box-alone system has no potential energy term at all and receives $98$ J of work from gravity as an external force. Using both accounts at once doubles the answer to $196$ J and is the error to watch for.

Frequently asked questions

What is the difference between kinetic and potential energy?

Kinetic energy is the energy an object has because it is moving, one half times mass times speed squared. Potential energy is stored in the arrangement of a system of two or more objects that interact through a conservative force, such as an object and the Earth or a block and a spring. The AP Physics 1 CED makes this a hard rule: essential knowledge 3.3.A.1 says a system composed of two or more objects has potential energy, and essential knowledge 3.4.A.1 says a system composed of only a single object can only have kinetic energy. So a lone object can have kinetic energy but never potential energy.

Can a single object have potential energy?

No. Potential energy is stored in an interaction, so it takes at least two objects. The AP Physics 1 CED states this at essential knowledge 3.3.A.1, which requires a system composed of two or more objects interacting only through conservative forces, and again at essential knowledge 3.4.A.1, which says a system composed of only a single object can only have kinetic energy. The everyday phrase, that a ball held above the floor has potential energy, is shorthand for the fact that the ball and Earth system has it, because of how far apart they are. If you cannot name the two objects whose separation would change, you do not have a potential energy term.

Can potential energy be negative?

Yes, and one standard form always is. The general gravitational potential energy on the AP Physics 1 equation sheet is minus G times m one times m two divided by r, which is negative at every finite separation because the zero of that expression sits at infinite separation. A negative value does not mean the system holds less than no energy; it means the current arrangement lies below the arrangement you chose to call zero. Kinetic energy, by contrast, can never be negative, because the speed is squared. If a kinetic energy comes out negative, the arithmetic is wrong.

Does kinetic energy depend on your reference frame?

Yes. The AP Physics 1 CED states it at essential knowledge 3.1.A.3: different observers may measure different values of the translational kinetic energy of an object, depending on the observer's frame of reference. A suitcase resting on a moving train has zero kinetic energy for a passenger and a large kinetic energy for someone on the platform, and both measurements are correct, because kinetic energy depends only on the speed and the two observers measure different speeds. Potential energy does not work this way. What it needs instead is a chosen zero configuration, which is a separate decision the observer makes under essential knowledge 3.3.A.3.

Where do you put the zero of potential energy?

Wherever makes the problem easiest, because it is your choice. The AP Physics 1 CED says at essential knowledge 3.3.A.3 that the definition of zero potential energy for a given system is a decision made by the observer considering the situation to simplify or otherwise assist in analysis. Common choices are the floor, the lowest point of the motion, the release point, or the spring's relaxed length for elastic energy. The value of the potential energy changes with the choice, but the change in potential energy between two configurations does not, because the zero cancels in the subtraction. State your choice in words before the first energy line and do not move it partway through.

Why is there no potential energy for friction?

Because potential energies exist only for conservative forces, which the AP Physics 1 CED states outright at essential knowledge 3.2.A.1.iii. A potential energy has to assign one number to each configuration of the system, and that is only possible when the work done depends on the initial and final configurations alone. Essential knowledge 3.2.A.1.iv says the work done by a nonconservative force is path-dependent, and 3.2.A.1.v names friction and air resistance as examples, so no single number can be attached to a configuration. Friction dissipates mechanical energy as thermal energy or sound instead, which the Topic 3.4 boundary statement expects AP Physics 1 students to know.

Is mechanical energy the same as kinetic plus potential energy?

Yes. The AP Physics 1 CED defines it that way at essential knowledge 3.4.B.1: mechanical energy is the sum of a system's kinetic and potential energies. Whether that sum stays constant is a separate question, and it depends on the system you chose. Essential knowledge 3.4.C.2 gives the condition: if the work done on a selected system is zero and there are no nonconservative interactions within the system, the total mechanical energy of the system is constant. If work is done on the system from outside, essential knowledge 3.4.C.3 says energy is transferred between the system and the environment, so the total inside changes even though energy overall is still conserved.