Classical-Mechanics · Unit 16 · Video 4 · Interactive Practice

Three Potential Energy Functions and Where You Put the Zero

IKey Formulas

Formula	Name	What you need
$\Delta U = -W_c = -\displaystyle\int_A^B \vec{F}_c \cdot d\vec{r}$	Potential energy of an internal conservative force	The force, and the two configurations $A$ and $B$
$U^g(y) = mgy, \qquad U^g(0) \equiv 0$	Gravity near the earth's surface	Height $y$ above the chosen zero
$U^s(x) = \tfrac{1}{2}kx^2, \qquad U^s(0) \equiv 0$	Ideal spring	Displacement $x$ from the relaxed length
$U^G(r) = -\dfrac{Gm_1m_2}{r}, \qquad U^G(\infty) \equiv 0$	Universal gravitation	Center-to-center separation $r$

Key Insight: only differences are physical. Replacing $U$ by $U + C$ leaves every $U_f - U_i$ untouched, so the zero of potential energy is a bookkeeping choice — pick the reference that deletes a term or marks a natural configuration, and state which one you picked.

IIVisualization 1 — One Recipe, Three Forces

The same five steps, run on three different internal forces, yield three unrelated-looking functions.

System Recipe step 1

1 · The internal conservative force

\vec{F}^g = -mg\,\hat{j}

2 · Dot it with the displacement

d\vec{r} = dy\,\hat{j}

\vec{F}^g \cdot d\vec{r} = -mg\,dy

3 · Integrate — the work

W^g = \int_{y_i}^{y_f} (-mg)\,dy

= -mg\,(y_f - y_i)

4 · Negate

\Delta U^g = -W^g = mg\,(y_f - y_i)

5 · Choose the zero

y_i = 0,\quad U^g(0) \equiv 0

U^g(y) = mgy\ \checkmark

1 · The internal conservative force

\vec{F}^s = -kx\,\hat{i}

2 · Dot it with the displacement

d\vec{r} = dx\,\hat{i}

\vec{F}^s \cdot d\vec{r} = -kx\,dx

3 · Integrate — the work

W^s = \int_{x_i}^{x_f} (-kx)\,dx

= -\tfrac{1}{2}k\,(x_f^2 - x_i^2)

4 · Negate

\Delta U^s = -W^s = \tfrac{1}{2}k\,(x_f^2 - x_i^2)

5 · Choose the zero

x_i = 0,\quad U^s(0) \equiv 0

U^s(x) = \tfrac{1}{2}kx^2\ \checkmark

1 · The internal conservative force

\vec{F}^G_{2,1} = -\frac{Gm_1m_2}{r^2}\,\hat{r}_{2,1}

2 · Dot it with the displacement

d\vec{r}_{2,1} = dr\,\hat{r}_{2,1}

\vec{F}^G \cdot d\vec{r} = -\frac{Gm_1m_2}{r^2}\,dr

3 · Integrate — the work

W^G = \int_{r_i}^{r_f} -\frac{Gm_1m_2}{r^2}\,dr

= Gm_1m_2\left(\frac{1}{r_f} - \frac{1}{r_i}\right)

4 · Negate

\Delta U^G = -\frac{Gm_1m_2}{r_f} + \frac{Gm_1m_2}{r_i}

5 · Choose the zero

r_i \to \infty,\quad U^G(\infty) \equiv 0

U^G(r) = -\frac{Gm_1m_2}{r}\ \checkmark

IIIVisualization 2 — Sliding the Zero

Move the reference configuration: every value of $U$ changes, but does any difference?

System Zero reference 0

💡 Because nothing measurable changes, a bare number is not yet an answer: " $U = 12\ \text{J}$ " means nothing until you say where $U = 0$ sits.

IVVisualization 3 — Why Infinity Is the Natural Zero

Push the reference separation outward and one whole term of $\Delta U^G$ dies away.

Reference separation 2.0

VQuiz Questions

Problem 1 · Running the Recipe for Near-Earth Gravity

Given: a $3.0\ \text{kg}$ book is raised $1.2\ \text{m}$ from the floor to a shelf, with $U^g = 0$ chosen at the floor and $g = 9.8\ \text{m/s}^2$ — find $\Delta U^g$ of the earth–book system.

✅ Correct! Raising the book stores energy in the earth–book configuration, so

\Delta U^g

is positive.

❌ That is $W^g$ , not $\Delta U^g$ . Gravity does

-35.3\ \text{J}

of work on the way up; step 4 of the recipe negates it:

\Delta U^g = -W^g = +35.3\ \text{J}

❌ Not quite. The recipe gives

\Delta U^g = mg\,\Delta y

— no factor of

\tfrac12

, and

g

is not optional.

Show solution

Steps 1–3 give the work of the constant force $\vec{F}^g = -mg\,\hat{j}$ over the rise:

W^g = \int_{y_i}^{y_f}(-mg)\,dy = -mg\,(y_f - y_i) = -(3.0)(9.8)(1.2) = -35.28\ \text{J}

Step 4 negates it:

\Delta U^g = -W^g = mg\,\Delta y = (3.0)(9.8)(1.2) = 35.28 \approx 35.3\ \text{J}

The factor $\tfrac12$ belongs to the spring, whose force grows with the displacement; gravity near the surface is constant, so the integral is just $mg$ times the height gained.

Problem 2 · The Zero Is Wherever You Declare It

Given: a $2.0\ \text{kg}$ block rests on the floor, and you declare $U^g = 0$ at the ceiling, $3.0\ \text{m}$ above the floor ( $g = 9.8\ \text{m/s}^2$ ) — find $U^g$ of the block on the floor.

✅ Correct! Below the chosen zero,

U^g

is negative — and a negative potential energy is perfectly ordinary.

❌ The floor is not automatically the zero. Here the zero was declared at the ceiling, and the block sits

3.0\ \text{m}

below it.

❌ Right magnitude, wrong sign.

U^g = mg(y - y_{\text{ref}})

with

y - y_{\text{ref}} = 0 - 3.0 = -3.0\ \text{m}

❌ Not quite. Measure the height from the declared zero:

U^g = mg\,(y - y_{\text{ref}})

Show solution

Choosing the zero at $y_{\text{ref}} = 3.0\ \text{m}$ means $U^g(y) = mg\,(y - y_{\text{ref}})$ . On the floor, $y = 0$ :

U^g(0) = (2.0)(9.8)(0 - 3.0) = -58.8\ \text{J}

Compare with the usual choice $y_{\text{ref}} = 0$ , which gives $U^g(0) = 0$ and $U^g(3.0) = +58.8\ \text{J}$ . The two conventions differ by the constant $C = -58.8\ \text{J}$ everywhere — and every difference is identical: lifting the block to the ceiling costs $+58.8\ \text{J}$ either way.

Problem 3 · From Compressed to Stretched

Given: a spring with $k = 250\ \text{N/m}$ starts compressed $0.12\ \text{m}$ and ends stretched $0.20\ \text{m}$ , with $U^s = 0$ at the relaxed length — find $\Delta U^s$ and the work $W^s$ done by the spring force.

What is the change in potential energy?

What work did the spring force do?

✅ Correct! The spring stores

3.2\ \text{J}

, and step 4 of the recipe says the spring force did exactly

-3.2\ \text{J}

of work.

❌ $\tfrac12 k(\Delta x)^2$ is not $\tfrac12 k(x_f^2 - x_i^2)$ .

U^s

is quadratic, so the two squares must be taken separately and then subtracted.

❌ Compression does not store negative energy.

U^s = \tfrac12 kx^2

is even in

x

: at

x = -0.12\ \text{m}

the spring already holds

+1.8\ \text{J}

, so that energy is subtracted, not added.

❌ Check the endpoints. Use

x_i = -0.12\ \text{m}

and

x_f = +0.20\ \text{m}

\Delta U^s = \tfrac12 k(x_f^2 - x_i^2)

❌ Check the sign. Step 4 of the recipe is

\Delta U = -W

, so

W^s = -\Delta U^s

Show solution

With the zero at the relaxed length, $U^s(x) = \tfrac12 kx^2$ . Compression counts as $x_i = -0.12\ \text{m}$ , and the sign is irrelevant because $x$ is squared:

U^s_i = \tfrac12(250)(-0.12)^2 = 1.8\ \text{J}, \qquad U^s_f = \tfrac12(250)(0.20)^2 = 5.0\ \text{J}

\Delta U^s = 5.0 - 1.8 = 3.2\ \text{J}

Equivalently, straight from the recipe:

\Delta U^s = \tfrac12 k\,(x_f^2 - x_i^2) = 125\,(0.0400 - 0.0144) = 3.2\ \text{J}

And step 4 read backwards gives the work of the spring force itself:

W^s = -\Delta U^s = -3.2\ \text{J}

Negative, as it must be: the spring pushed back against this whole move.

Problem 4 · Separating Two Gravitating Masses

Given: two masses at center-to-center separation $r$ are pulled apart to $3r$ , using $U^G(r) = -\dfrac{Gm_1m_2}{r}$ with $U^G(\infty) \equiv 0$ — find $\Delta U^G$ , and then decide what a different zero would do to it.

What is the change in potential energy?

Now move the zero to a different separation — what happens to that change?

✅ Correct! Separating attracting masses stores energy, and moving the zero shifts both

U

values by the same constant, so the difference survives untouched.

❌ That is $U^G(3r)$ , not the change. A potential energy value is not a potential energy difference — you still have to subtract

U^G(r)

❌ Close — the subtraction ran backwards. You computed

U_i - U_f

. Pulling attracting masses apart must store energy, so

\Delta U^G > 0

❌ The two terms were added as magnitudes. Keep the signs: both

U

values are negative, and

-\tfrac13 - (-1) = +\tfrac23

in units of

Gm_1m_2/r

❌ Not quite. Evaluate

U^G

at both separations and subtract:

\Delta U^G = U^G(3r) - U^G(r)

❌ Not quite. A new zero adds one constant

C

U

everywhere, and

C

cancels out of

U_f - U_i

Show solution

Part 1 — the change with the zero at infinity:

U^G(r) = -\frac{Gm_1m_2}{r}, \qquad U^G(3r) = -\frac{Gm_1m_2}{3r}

\Delta U^G = U^G(3r) - U^G(r) = -\frac{Gm_1m_2}{3r} + \frac{Gm_1m_2}{r} = \frac{2Gm_1m_2}{3r}

Positive, as it must be: you had to pull against an attraction, and that energy is now stored in the pair's configuration.

Part 2 — a different zero: declaring $U^G = 0$ at $2r$ replaces the function by

\tilde{U}^G(\rho) = -\frac{Gm_1m_2}{\rho} + \frac{Gm_1m_2}{2r} = U^G(\rho) + C

with the single constant $C = Gm_1m_2/(2r)$ . Then

\tilde{U}^G(3r) - \tilde{U}^G(r) = \big(U^G(3r) + C\big) - \big(U^G(r) + C\big) = \frac{2Gm_1m_2}{3r}

The constant cancels. Individual values change — under this convention $U^G(2r) = 0$ and $U^G(r)$ is negative — but no measurable prediction moves, which is exactly why choosing the zero is allowed at all.

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