Potential energy and conservative forces

Last week you calculated the work of each force one by one. For some forces you can think of this work beforehand as “stored” energy: potential energy. When you lift a book onto a shelf, the work you did is not lost; it is stored in the position. When the book falls, it is given back. This week you will learn two types of potential energy (gravitational and spring), for which forces a potential energy can be defined, and finding the force from $U(x)$.

Golden rules:

1. $U_g = mgy$, $U_s = \tfrac12 kx^2$.
2. The work of a conservative force is $W = -\Delta U$ and is independent of the path.
3. Friction is nonconservative: its work depends on the length of the path, and it has no potential energy.
4. $F_x = -dU/dx$. The minus sign says that the force points in the direction in which $U$ decreases.

Gravitational potential energy

When you lift an object of mass $m$ at constant speed from $y_i$ to $y_f$, gravity does $-mg(y_f - y_i)$ of work. We define the negative of this work as the change in potential energy:

$U_g = mgy \qquad \Delta U_g = mg\,\Delta y = -W_g$

$y$ is measured upward from the reference level you choose. You can choose the reference as the ground, a table, the ceiling, wherever you like. The value of $U$ changes, $\Delta U$ does not. Physics cares only about differences.

Worked example 1. A 0.500 kg ball sits on a roof 30.0 m high.

Reference = ground:  U = 0.500 · 9.80 · 30.0 = 147 J
Reference = roof:    U = 0
If the ball falls to the ground: ΔU = -147 J   (the same in both references)
The work of gravity: W_g = -ΔU = +147 J

PITFALL · height is measured from the reference. If the shelf is 1.50 m above the table and the table is 0.900 m above the floor, the height of the shelf from the floor is 2.40 m. Read carefully which height the question is asking about.

Spring potential energy

The work you do to stretch a spring by $x$ was $\tfrac12 kx^2$ (week 10). This work is stored in the spring as elastic potential energy:

$U_s = \tfrac12 kx^2$

$x$ is the extension or compression from the equilibrium position. Since it is squared, both give positive energy.

Worked example 2. A spring with $k = 250\ \text{N/m}$ is compressed 8.00 cm.

$U_s = \tfrac12(250\ \text{N/m})(0.0800\ \text{m})^2 = 0.800\ \text{J}$

If you double the extension, the energy becomes four times: $U \propto x^2$.

Unlike gravity, the reference of spring energy is not free. $x$ is always measured from the natural length of the spring, because $U_s = 0$ when the spring is neither stretched nor compressed. In a vertically hanging spring the two energies appear together: as the block goes down $U_g$ decreases and $U_s$ increases. In problems write the two as separate terms and do not mix up the reference of each: a height you choose for $U_g$, and the natural length of the spring for $U_s$.

Conservative and nonconservative forces

There are two equivalent conditions for a force to be conservative:

1. The work it does from one point to another is independent of the path (path independence); it depends only on the endpoints.
2. The work it does along a closed path is zero.

Why is friction nonconservative? Because at every step it is opposite to the motion. Both going and coming back do negative work. When you push a box 3 m forward and bring it back, the total work of friction is $-f \cdot 6\ \text{m}$, not zero. This energy is converted to heat (internal energy) and does not come back mechanically.

PITFALL · for friction it is the path, not the displacement. A box going from $(0,0)$ to $(4,3)$ along an L shape travels 7.00 m of path, and its displacement is 5.00 m. The work of friction is $-f \cdot 7.00$. The work of gravity on the horizontal path is zero in both cases.

The relation between force and potential energy

For a conservative force $dU = -F_x\,dx$. From this a two-way relation:

$F_x = -\frac{dU}{dx} \qquad U(x) = -\int F_x\,dx + C$

Check: $U_s = \tfrac12 kx^2 \Rightarrow F = -kx$ (Hooke). $U_g = mgy \Rightarrow F_y = -mg$ (downward). Both hold.

Worked example 3. $U(x) = 4x^2 - x^4$ (J, m). What are the force at $x = 1.00$ m and the equilibrium points?

dU/dx = 8x - 4x³
F(1) = -(8 - 4) = -4.00 N   (in the minus x direction)
Equilibrium: 8x - 4x³ = 0 → 4x(2 - x²) = 0 → x = 0, x = ±√2 = ±1.41 m
d²U/dx² = 8 - 12x²
  x = 0:      +8  > 0 → U minimum → stable
  x = ±1.41:  -16 < 0 → U maximum → unstable

Worked example 4. $F(x) = -2x$ (N, m), $U(0) = 0$. What is $U(3)$?

$U(3) = -\int_0^3 (-2x)\,dx = \left[x^2\right]_0^3 = 9.00\ \text{J}$

Reading a potential energy graph

The $U(x)$ graph is like a landscape of hills and valleys. The force always points downhill (opposite to the slope):

• If the slope is positive ($U$ is increasing), the force is in the $-x$ direction.
• If the slope is zero the force is zero: equilibrium.
• The bottom of a valley (minimum): stable equilibrium. If you push slightly, it comes back.
• The top of a hill (maximum): unstable equilibrium. If you push slightly, it moves away.
• A flat region: neutral equilibrium.

PITFALL · $U = 0$ and $F = 0$ are not the same thing. Equilibrium is where the slope of $U$ is zero. The point where $U$ itself is zero carries no special meaning; it depends on the reference.

Mixed exam examples

Worked example 5 (force and equilibrium from U). $U(x) = 2x^3 - 6x$ (J, m).

F(x) = -dU/dx = -(6x² - 6)
F(2.00) = -(24 - 6) = -18.0 N
Equilibrium: 6x² - 6 = 0  →  x = +1.00 m and x = -1.00 m
d²U/dx² = 12x
  x = +1.00:  +12 > 0  → minimum, stable,   U(1) = -4.00 J
  x = -1.00:  -12 < 0  → maximum, unstable, U(-1) = +4.00 J

If a particle with total energy $E = 0$ is near $x = 1$, how far can it go? The places where $U(x) = 0$: $2x(x^2 - 3) = 0$, that is, $x = 0$ and $x = \sqrt3 = 1.73$ m. The particle oscillates between these two turning points. You will see this idea as an energy diagram in week 13.

Worked example 6 (path independence). A 4.00 kg box is raised to a height of 3.00 m by two routes: by a 10-step staircase or by an 8.00 m ramp. What is the work of gravity?

$W_g = -mg\,\Delta y = -(4.00)(9.80)(3.00) = -118\ \text{J}$

It is the same on both routes. On the staircase, gravity does no work on the horizontal steps, and the vertical steps add up to 3.00 m. On the ramp the force component is small but the path is long: $mg\sin\theta \cdot L = mg\,h$. The length of the path changes only the work of friction.

TIP · the place of this topic on the exam. Pure potential energy questions are easy ($mgh$, $\tfrac12 kx^2$). The real points are in the $F = -dU/dx$ derivative questions and the conservative force concept questions. Be sure to practice the sign of the derivative and the stability test.

Common mistakes