Part 1 · Chapter 08

Potential Energy and Conservation of Energy

Energy stored in arrangement — and the great bookkeeping law that nothing is ever lost

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 40 min
i What you'll learn
  • What potential energy is — energy stored in the configuration of a system — and how it relates to work via \(\Delta U = -W\).
  • The difference between conservative and nonconservative forces, and the path-independence test.
  • The formulas for gravitational (\(U = mgy\)) and elastic (\(U = \tfrac{1}{2}kx^{2}\)) potential energy.
  • The conservation of mechanical energy, and how to read a potential-energy curve.
  • How friction produces thermal energy, and the all-encompassing law of conservation of energy.
Section 8-1

What Is Physics?

One job of physics is to catalog the kinds of energy in the world. This chapter introduces potential energy \(U\) — energy stored in the arrangement of a system of objects that exert forces on one another. A bungee jumper plunging toward Earth converts gravitational potential energy into kinetic energy, and then the stretching cord stores it as elastic potential energy. Getting those calculations right is exactly what keeps the jump thrilling instead of fatal.

Section 8-2

Work and Potential Energy

Throw a tomato up: as it rises, gravity does negative work, draining its kinetic energy into the gravitational potential energy of the tomato–Earth system. As it falls, the transfer reverses. We define the change in potential energy as the negative of the work done by the (conservative) force:

Potential energy and work
\[ \Delta U = -W \]
This holds for gravity (tomato–Earth) and for a spring (block–spring). Energy the force removes from kinetic energy is stored as potential energy, and vice versa.
♻ Conservative force

The energy it transfers can be fully recovered — it round-trips with zero net loss. Gravity and the spring force are conservative, which is why each has a potential energy.

🔥 Nonconservative force

Friction and drag are not conservative: they transfer kinetic energy into thermal energy, which cannot be transferred back. So thermal energy is not a potential energy.

Section 8-3

Path Independence of Conservative Forces

🔁
The test for a conservative force
The net work a conservative force does on a particle around any closed path is zero.

Equivalently, the work it does between two points does not depend on the path taken: \(W_{ab,1} = W_{ab,2}\). This is enormously useful — when the actual path is hard, swap in an easier path between the same two endpoints and get the same work.

Section 8-4

Determining Potential Energy Values

Generally, \(\Delta U = -\int F(x)dx\). Apply this to the two forces of this chapter. For gravity (substituting \(-mg\) for the force) and for a spring (substituting \(-kx\)):

Gravitational & elastic potential energy
\[ U(y) = m g y \qquad U(x) = \tfrac{1}{2} k x^{2} \]
Both use a convenient reference where U = 0 (at y = 0 for gravity, at the relaxed length x = 0 for the spring). Gravitational PE depends only on height, not on horizontal position.
Only changes in U matter. You may put the \(U = 0\) reference wherever it's convenient — at the ground, a tabletop, anywhere. The physics depends on \(\Delta U\), which is the same no matter where you set the zero.
Section 8-5

Conservation of Mechanical Energy

The mechanical energy of a system is \(E_{\text{mec}} = K + U\). When the system is isolated and only conservative forces act, a gain in one is exactly a loss in the other (\(\Delta K = -\Delta U\)), so the sum can't change:

⚖️
Conservation of mechanical energy
In an isolated system with only conservative forces, K + U is constant.

\(K_{2} + U_{2} = K_{1} + U_{1}\), or \(\Delta E_{\text{mec}} = \Delta K + \Delta U = 0\). This relates two instants without tracking the messy motion in between — a pendulum trades K and U back and forth, but their sum never wavers.

Section 8-6

Reading a Potential Energy Curve

From a graph of \(U(x)\), you can read off the force — it's the negative slope — and the kinetic energy at any point:

Force from the curve, and kinetic energy
\[ F(x) = -dU(x)/dx \qquad K(x) = E_{\text{mec}} - U(x) \]
Draw the constant Emec as a horizontal line; the gap down to the U(x) curve is the kinetic energy.
↩ Turning point

Where the line meets the curve, \(K = 0\) and the particle reverses. It can never enter a region where \(E_{\text{mec}} \lt U\) (that needs negative K).

⚖ Equilibrium

Where the slope is zero, the force is zero. A valley is stable equilibrium, a hilltop is unstable, and a flat stretch is neutral.

Section 8-7

Work Done on a System by an External Force

An external force can transfer energy into or out of a whole system. With no friction, that work just changes the mechanical energy. With friction, sliding also raises the thermal energy:

External work, with and without friction
\[\begin{gathered} W = \Delta E_{\text{mec}} \qquad \text{(no friction)} \\ W = \Delta E_{\text{mec}} + \Delta E_{\text{th}}, \qquad \text{where} \Delta E_{\text{th}} = f_{k} d \qquad \text{(friction)} \end{gathered}\]
The thermal energy generated equals the kinetic friction force times the sliding distance — the warmth you feel rubbing your hands together.
Section 8-8

Conservation of Energy

🌐
The law of conservation of energy
The total energy of a system changes only by the energy transferred to or from it.

\(W = \Delta E = \Delta E_{\text{mec}} + \Delta E_{\text{th}} + \Delta E_{\text{int}}\). For an isolated system (\(W = 0\)), the total energy cannot change at all: \(\Delta E_{\text{mec}} + \Delta E_{\text{th}} + \Delta E_{\text{int}} = 0\). No exception to this law has ever been found.

A rock climber rappelling bleeds the gravitational potential energy of the climber–gear–Earth system not into dangerous kinetic energy but into thermal energy of the rope and friction rings — controlled, deliberately. The total energy of the isolated system stays put; he just chooses which account it lands in.
Worked Examples

Putting It to Work

1 Slippery cheese — using an easier path

Problem. A 2.0 kg block of cheese slides on a frictionless track from \(a\) to \(b\), covering 2.0 m of track but only 0.80 m of net vertical drop. How much work does gravity do?

Solution. Gravity is conservative, so replace the complicated track with an easy two-leg path: a horizontal leg (where \(\varphi = 90^{\circ}\), so \(W = 0\)) and a vertical leg of 0.80 m (where \(\varphi = 0^{\circ}\)).

Work along the substitute path
\[ W = W_{h} + W_{v} = 0 + mgd = (2.0)(9.8)(0.80) \approx 16 J \]

Because gravity is path-independent, this 16 J is exactly the work done along the real, curved track — without ever needing the track's shape.

2 A hanging sloth — the reference level is your choice

Problem. A 2.0 kg sloth hangs 5.0 m above the ground. Find the gravitational potential energy for various choices of the \(y = 0\) reference, and the change in U when it drops to the ground.

U depends on the reference; ΔU does not
\[\begin{gathered} ground ref: U = mgy = (2.0)(9.8)(5.0) \approx 98 J \\ limb ref: U = mg(0) = 0 J \qquad above-limb ref: U = mg(-1.0) \approx -20 J \\ on falling to ground: \Delta U = mg\Delta y = (2.0)(9.8)(-5.0) \approx -98 J \qquad \text{(any reference)} \end{gathered}\]

The value of U slides up or down with your choice of zero, but the change on the fall is always −98 J. Only ΔU has physical meaning.

3 A frictionless water slide — conservation of mechanical energy

Problem. A child is released from rest at the top of a frictionless water slide, height \(h = 8.5 m\). Find her speed at the bottom.

Solution. Only gravity does work (the normal force is perpendicular to the motion), so mechanical energy is conserved: \(K_{b} + U_{b} = K_{t} + U_{t}\). With \(v_{t} = 0\) and a drop of \(h\):

Speed at the bottom
\[ \tfrac{1}{2}mv_{b}^{2} = mgh \qquad \;\Longrightarrow\; \qquad v_{b} = \sqrt{2gh} = \sqrt{2\cdot 9.8\cdot 8.5} \approx 13 m/s \]

Exactly the speed of a free 8.5 m fall — the slide's shape and slope never enter. (Note: energy methods give the speed but not the time; for that you'd need the geometry.)

4 Cabbage crate on a rough floor — friction makes heat

Problem. A shipper pushes a 14 kg crate with a constant 40 N over 0.50 m; the speed drops from 0.60 m/s to 0.20 m/s. Find the work done and the thermal energy generated.

External work, then the heat
\[\begin{gathered} W = Fd \cos 0^{\circ} = (40)(0.50) = 20 J \\ \Delta E_{\text{mec}} = \Delta K = \tfrac{1}{2}m(v^{2} - v_{0}^{2}) = \tfrac{1}{2}(14)(0.20^{2} - 0.60^{2}) = -2.2 J \\ \Delta E_{\text{th}} = W - \Delta E_{\text{mec}} = 20 - (-2.2) \approx 22 J \end{gathered}\]

The push delivers 20 J, but the crate is slowing — so friction is bleeding energy faster than the push adds it. About 22 J ends up as warmth in the crate and floor.

5 Tamales into a spring with friction — full energy accounting

Problem. A 2.0 kg package slides at \(v_{1} = 4.0 m/s\) into a spring (\(k = 10 000 N/m\)). The approach is frictionless, but while compressing the spring a 15 N kinetic friction force acts. How far does the spring compress before the package stops?

Solution. Treat package–spring–floor as one isolated system: \(\Delta E_{\text{mec}} = -\Delta E_{\text{th}}\). Initial mechanical energy is all kinetic; final is all spring potential. With \(\Delta E_{\text{th}} = f_{k}d\):

Energy balance → a quadratic in d
\[\begin{gathered} \tfrac{1}{2}kd^{2} = \tfrac{1}{2}mv_{1}^{2} - f_{k}d \qquad \;\Longrightarrow\; \qquad 5000d^{2} + 15d - 16 = 0 \\ d \approx 0.055 m = 5.5 \mathrm{cm} \end{gathered}\]

Friction steals some of the kinetic energy on the way in, so the spring compresses a bit less than it would on a frictionless floor.

Review

Chapter Summary

✓ Potential energy

Energy of configuration: \(\Delta U = -W\). Defined only for conservative forces (gravity, springs).

✓ Path independence

A conservative force does zero net work around any closed loop; work between two points is path-independent.

✓ PE formulas

Gravity: \(U = mgy\). Spring: \(U = \tfrac{1}{2}kx^{2}\). Only ΔU is physical.

✓ Mechanical energy

\(E_{\text{mec}} = K + U\) is conserved in an isolated system with only conservative forces.

✓ Energy curves

\(F = -dU/dx\); \(K = E_{\text{mec}} - U\). Turning points and equilibria read straight off the graph.

✓ Energy conservation

Friction adds \(\Delta E_{\text{th}} = f_{k}d\). Total energy of an isolated system never changes.

Practice

Problems

Energy bookkeeping is the theme: pick a system, set a convenient \(U = 0\) reference, and decide whether mechanical energy is conserved (only conservative forces) or whether friction is siphoning energy into heat. Take \(g = 9.8 m/s^{2}\).

  1. What is the spring constant of a spring that stores 25 J of elastic potential energy when compressed by 7.5 cm?
  2. A 2.00 kg book is dropped to a friend 10.0 m below whose hands are 1.50 m above the ground. Find (a) the work gravity does on the book during the drop and (b) ΔU of the book–Earth system, taking U = 0 at ground level.
  3. A 2.0 g ice flake is released from the rim of a frictionless hemispherical bowl of radius 22.0 cm. Find its speed at the bottom.
  4. A runaway 1.2×10⁴ kg truck moving at 130 km/h coasts up a frictionless escape ramp inclined at 15°. What minimum ramp length stops it?
  5. A 700 g block is dropped from rest onto a vertical spring (k = 400 N/m), compressing it 19.0 cm before stopping. Find the height above the spring from which it was released.
  6. A 0.341 kg ball on a 0.452 m rod swings from horizontal down to its lowest point. Find the work gravity does and the change in the ball–Earth potential energy.
  7. A 25 kg bear slides 12 m down a pine tree, reaching 5.6 m/s at the bottom. Find (a) ΔU of the bear–Earth system and (b) the average frictional force.
  8. In a potential-energy graph, a 2.0 kg particle at x = 2.0 m has speed 1.5 m/s with mechanical energy fixed. Using \(F = -dU/dx\), explain how to locate its turning points and the force direction. (Sketch a plausible U(x).)
  9. A 2.5 kg block slides into a spring (k = 320 N/m), compressing it 7.5 cm before stopping, with μk = 0.25 between block and floor. Find (a) the spring's work and (b) the thermal energy generated.
  10. A 60 kg skier leaves a ramp at 24 m/s, 25° above horizontal, and lands 14 m below moving at 22 m/s. By how much did air drag reduce the skier–Earth mechanical energy?
Tip: deciding what to include in "the system" is the key move. Put the spring, floor, and Earth inside the system and friction's work becomes an internal ΔEth term — then total energy is simply conserved, and a hard problem becomes one tidy equation.