Lesson 10: Conservation of Mechanical EnergyAP

The work–energy theorem becomes far more powerful when some forces are conservative — gravity and springs chief among them — because their work can be pre-computed and stored as potential energy. For a conservative force, work depends only on endpoints, never on path; friction, by contrast, burns mechanical energy in proportion to path length. This lesson (CED Unit 5, Energy) builds the bookkeeping: E_mech = K + U, and W_nc = ΔE_mech for everything non-conservative.

The potential energies you need

• Gravitational: U_g = mgh, with h measured from a zero level you choose. Only changes in U matter, so put the zero wherever simplifies the problem (usually the lowest point).
• Elastic: U_s = ½kx², with x the stretch or compression from the spring's natural (unstretched) length — not from any other reference.

Choose the system deliberately: gravity's potential energy belongs to the object–Earth system. If Earth is outside your system, there is no U_g — instead gravity does work. Both views are correct; mixing them mid-problem (counting mgh and gravity's work) is the classic double count.

The conservation law and when it bends

If only conservative forces do work, E_mech is constant: K_i + U_i = K_f + U_f. If friction or a push also acts, W_nc = ΔE_mech patches the account: K_i + U_i + W_nc = K_f + U_f. Energy bar charts — bars for K, U_g, U_s at two instants, with a work column between — are the College Board's preferred representation and the fastest way to avoid sign errors: the heights must reconcile.

Speeds from energy are path-independent: a block sliding down a frictionless hill of height h reaches √(2gh) whether the slope is steep, gentle, or a curved slide — dynamics alone could not give that so cheaply. What energy can never tell you is time; for that you need kinematics or momentum methods.

Worked example 1 — pendulum speed

A pendulum bob is released from rest 0.45 m above its lowest point (g = 10 m/s²). Find its speed at the bottom, and at the point where it has risen back to 0.20 m above the lowest point. Given: no friction, tension does no work (perpendicular to motion). Method: energy between instants.

At bottom: mgh = ½mv² → v = √(2 × 10 × 0.45) = √9 = 3.0 m/s
At 0.20 m: ½mv² = mg(0.45 − 0.20) → v = √(2 × 10 × 0.25) = √5 ≈ 2.2 m/s

Check: the mass cancels — Galileo's result that fall speed ignores mass. The speed at 0.20 m is less than at the bottom, and would be the same on the way up as on the way down at that height: energy arguments are time-symmetric.

Worked example 2 — spring launcher

A spring with k = 200 N/m is compressed 0.15 m and launches a 0.10 kg ball vertically (g = 10 m/s²). Find (a) the ball's speed as it leaves the spring (measured at the spring's natural length, at the launch level), (b) the maximum height reached above the launch level. Given: U_s = ½kx²; treat the small height change during decompression as negligible (state the approximation).

Energy stored: U_s = ½(200)(0.15)² = 2.25 J
(a) ½mv² = 2.25 → v = √(2 × 2.25 / 0.10) = √45 ≈ 6.7 m/s
(b) mgh = 2.25 → h = 2.25 / (0.10 × 10) = 2.25 m

Check: energy conservation between (a) and (b) self-audits: ½(0.10)(6.7²) ≈ 2.25 J = (0.10)(10)(2.25) ✓. A stiffer spring or bigger compression scales the energy with x² — double the compression, four times the launch energy.

Real-world application: coasters, dams and vaults

Roller-coaster design is energy bookkeeping with a safety margin: the lift hill sets the total budget E = mgh, and every later hill must be lower, because friction and air resistance keep withdrawing from the account. Hydroelectric plants sell exactly U_g → K → electrical: each kilogram of water falling 100 m yields about 1,000 J, so a flow of 1,000 kg/s is a megawatt before losses. The pole vaulter chains K (run) → U_s (bent pole) → U_g (height) — three systems, one ledger.

Experimental design note

Test conservation honestly: release a cart from height h on a track and measure its speed at the bottom with a photogate. Plot v² against h — energy predicts a straight line through the origin of slope 2g. Real data fall slightly below and curve away at larger h (friction and air resistance); quantifying that shortfall as "percentage of energy lost" and proposing its mechanism is stronger AP science than pretending the data are perfect.

Common mistakes (from College Board scoring commentary)

• Double counting gravity: including U_g and the work done by gravity in the same equation.
• Measuring spring stretch from the wrong reference — U_s uses displacement from the natural length.
• Applying plain conservation when friction acts; the fix is the W_nc term, not hope.
• Believing energy methods can yield times or accelerations directly.

AP Exam Experience

MCQ 1. A 2.0 kg ball drops from rest from a height of 5.0 m (g = 10 m/s²). Its speed just before impact is:

• (A) 5.0 m/s
• (B) 7.1 m/s
• (C) 10 m/s
• (D) 14 m/s

Answer: (C). v = √(2gh) = √(2 × 10 × 5.0) = √100 = 10 m/s, independent of the 2.0 kg mass.

MCQ 2. A spring (k = 100 N/m) is compressed 0.20 m. The stored elastic potential energy is:

• (A) 1.0 J
• (B) 2.0 J
• (C) 4.0 J
• (D) 20 J

Answer: (B). U_s = ½kx² = ½ × 100 × 0.04 = 2.0 J. (D) is the answer from forgetting both the ½ and the square.

MCQ 3. A block slides down a frictionless ramp from height h, then up a second frictionless ramp. Ignoring losses, the greatest height it can reach on the second ramp is:

• (A) h/2
• (B) h
• (C) 2h
• (D) dependent on the ramps' angles

Answer: (B). Mechanical energy is conserved and speed at equal heights is equal; reaching height h makes K = 0 again. Ramp angles change the journey, not the energy ledger — the path-independence point of this lesson.

FRQ. A 0.50 kg block is released from rest at the top of a track 2.0 m above a level section where it meets a spring (k = 400 N/m) fixed at the end. The track is frictionless except for a rough patch 1.0 m long on the level section where μ_k = 0.30 (g = 10 m/s²). (a) Calculate the block's speed just before it reaches the rough patch (at the bottom of the slope). (b) Calculate the work done by friction crossing the patch. (c) Calculate the maximum compression of the spring. (d) The block rebounds. Explain whether it will climb back to its original height. Scoring guideline. (a) 1 point: mgh = ½mv² → v = √(2 × 10 × 2.0) ≈ 6.3 m/s. (b) 1 point: W_f = −μ_k mg × d = −0.30 × 5.0 × 1.0 = −1.5 J. (c) 2 points: total energy mgh = 10 J; after friction 8.5 J converts entirely to U_s at maximum compression: ½(400)x² = 8.5 → x = √0.0425 ≈ 0.21 m (1 point set-up, 1 point value). (d) 1 point: no — friction acts on both crossings, removing another 1.5 J on the return, so the block climbs to (10 − 3.0)/(mg) = 7.0/5.0 = 1.4 m; each pass through the patch permanently converts mechanical energy to thermal energy.

Worked example 3 — dropped onto a spring (the quadratic in disguise)

A 0.50 kg mass is dropped from 0.40 m above a vertical spring (k = 200 N/m) and compresses it before momentarily stopping. Find the maximum compression x (g = 10 m/s²). Method: energy from release to the stop point — the mass falls a total of (0.40 + x), and at maximum compression it is (instantly) at rest.

mg(0.40 + x) = ½kx²
5.0(0.40 + x) = 100x²
100x² − 5.0x − 2.0 = 0
x = [5.0 + √(25 + 800)]/200 = (5.0 + 28.7)/200 ≈ 0.17 m

Check: the compression (0.17 m) exceeds the static equilibrium compression mg/k = 0.025 m by a lot — dropping adds the fall energy, which is precisely why a gentle placement and a drop of the same mass load a spring so differently. Substituting back: spring stores ½ × 200 × 0.17² ≈ 2.9 J; gravity lost 5.0 × 0.57 ≈ 2.9 J ✓.

Reading an energy bar chart like a marker

Construct one for the drop: at release, a single bar U_g (measuring from the stop point) of 2.9 J... better practice: draw bars at three instants — release (U_g only), first contact with the spring (K + reduced U_g), maximum compression (U_s plus the zeroed remainder) — with widths reconciling to the same total. If your bars do not reconcile, a term is missing (usually the gravitational drop during compression, the x inside (0.40 + x) above). Scorers award the reconciliation, not the artwork: equal totals, labelled forms, and a visible W_nc column whenever friction acts.

Worked example 4 — counting a reservoir

A pumped-storage reservoir holds 1.0 × 10⁶ kg of water 50 m above its turbines. (a) How much energy can it release? (b) For how long could it supply a town drawing 2.0 MW, at 100% efficiency (g = 10 m/s²)?

(a) E = mgh = 1.0 × 10⁶ × 10 × 50 = 5.0 × 10⁸ J
(b) t = E/P = 5.0 × 10⁸/2.0 × 10⁶ = 250 s — about 4 minutes

Check: the humbling answer is the lesson — gravitational storage is feeble per kilogram (≈500 J/kg here) against chemical storage (petrol ≈ 4.6 × 10⁷ J/kg), which is why pumped hydro needs mountains and lakes, and why real stations (this is roughly 1/7,000 of the UK's Dinorwig) matter at grid scale only in bulk. It is also a memorable unit conversion: 5.0 × 10⁸ J ≈ 140 kWh — one reservoir, one lesson in magnitudes.

Does the zero of height matter? A two-line proof

Drop a ball from a table 1.0 m high; it falls to the floor. Choose zero at the floor: U_i = mg(1.0), U_f = 0, ΔU = −mg. Choose zero at the tabletop: U_i = 0, U_f = −mg(1.0), ΔU = −mg. Identical — because only differences of height enter any energy equation. Negative potential energies (below your zero) frighten students unnecessarily; they are bookkeeping artefacts. What you may not do is switch zeros mid-problem: that is the one unforgivable move, and it is always a choice, never a necessity.