Conservation of mechanical energy — derivation: Worked examples

Two symbols we lean on the whole way:

$K = \frac{1}{2} m v^{2}$ = the "moving-money" pocket (kinetic energy).
$U$ = the "stored-money" pocket (potential energy): $U_{g r a v} = m g h$ for height, $U_{s p r in g} = \frac{1}{2} k x^{2}$ for a squished/stretched spring.

The master sentence, in words: the moving pocket plus the stored pocket, added at the start, equals the same sum at the end — unless a non-conservative force (friction, a push) adds or removes money in between.

$start total K_{1} + U_{1} + money added/removed by non-conservative forces W_{n c} = end total K_{2} + U_{2}$

Here $W_{n c}$ is the work done by non-conservative forces (friction, applied pushes). When $W_{n c} = 0$ this collapses to the clean law $K_{1} + U_{1} = K_{2} + U_{2}$ .

The scenario matrix

Every problem this topic throws is one of these cells. Each example below is tagged with the cell(s) it covers.

Cell	What makes it special	Example
A. Pure drop	only gravity, straight line, start from rest	Ex 1
B. Path-independence	curved/slope path, gravity still conservative	Ex 2
C. Spring only	conservative spring, no height change	Ex 3
D. Mixed height + spring	two stored pockets swapping	Ex 4
E. Non-zero start speed	$K_{1} \neq = 0$ — nothing starts at rest	Ex 5
F. Friction (the thief)	$W_{n c} < 0$ , mechanical energy NOT conserved	Ex 6
G. Sign / direction trap	which way is "up", is $U$ + or −, going up vs down	Ex 7
H. Degenerate / limiting	$h \to 0$ , $v_{1} = v_{2}$ , $k \to \infty$ , zero mass check	Ex 8
I. Real-world word problem	roller-coaster / vertical loop, minimum-speed condition	Ex 9
J. Exam twist	find the unknown height/length, back-solve	Ex 10

Example 1 — Pure drop (Cell A)

Worked example Statement

A stone of mass $m = 2 kg$ is dropped from rest at height $h = 20 m$ . Take $g = 10 m/s^{2}$ . Find its speed just before it hits the ground.

Forecast: Before reading, guess: does the answer depend on the mass? Higher or lower speed than $10 m/s$ ?

Step 1 — Choose the two states and where $U = 0$ . State 1 = top (at rest, $v_{1} = 0$ , $h_{1} = 20$ ). State 2 = ground ( $h_{2} = 0$ ). Put the reference $U = 0$ at the ground. Why this step? Energy conservation compares two snapshots, not the journey. We must name them and fix a "sea level" for height so $U = m g h$ has a meaning.

Step 2 — Check the law applies. Only gravity does work (air resistance ignored). Gravity is conservative $\Rightarrow W_{n c} = 0 \Rightarrow K_{1} + U_{1} = K_{2} + U_{2}$ . Why this step? If we skip this check we might wrongly apply the clean law when a thief is present.

Step 3 — Write and solve. $0 + m g h = \frac{1}{2} m v^{2} + 0 \Rightarrow v = 2 g h$ The mass cancels — the forecast about mass is answered: it does not matter. $v = 2 \cdot 10 \cdot 20 = 400 = 20 m/s$

Verify: Units: $(m/s^{2}) (m) = m^{2} / s^{2} = m/s$ . ✓ Cross-check with kinematics $v^{2} = 2 g h = 400$ . ✓

Example 2 — Path independence: the slide vs the fall (Cell B)

Worked example Statement

Two identical balls ( $m = 1 kg$ ) start from rest at the same height $h = 5 m$ . Ball P drops straight down; ball Q slides down a frictionless curved ramp to the same ground level. Compare their ground speeds. ( $g = 10$ .)

Forecast: Guess: is Q slower because it "took the long way round"?

Step 1 — Recall why gravity is conservative. Work by gravity depends only on the vertical drop, not the path (this is the definition from Conservative and non-conservative forces). $W_{g r a v} = m g Δ h$ regardless of the horizontal wiggles. Why this step? This is exactly the fact that lets both balls share one equation.

Step 2 — Same start, same end, same law. For both: $m g h = \frac{1}{2} m v^{2}$ . The ramp shape never entered. Why this step? It shows the path-length is a red herring; only the height difference feeds $U$ .

Step 3 — Solve. $v = 2 g h = 2 \cdot 10 \cdot 5 = 100 = 10 m/s (both)$

Verify: Both equal $10 m/s$ . The forecast ("Q is slower") is wrong — path independence is the whole point. Units check as in Ex 1. ✓

Example 3 — Spring only, no height change (Cell C)

Worked example Statement

A block $m = 0.5 kg$ on a frictionless horizontal floor is pushed against a spring ( $k = 200 N/m$ ), compressing it $x = 0.10 m$ , then released. Find the launch speed.

Forecast: No height changes here — where does the kinetic energy come from?

Step 1 — Identify the two stored pockets. No height change ⇒ $U_{g r a v}$ is the same at both states, cancels out. The only changing $U$ is the spring's: $U_{s p r in g} = \frac{1}{2} k x^{2}$ (built in the parent note). Why this step? You must know which potential energy is changing; constant terms drop out.

Step 2 — Apply conservation. State 1 = compressed, at rest. State 2 = spring at natural length, block moving. $\frac{1}{2} k x^{2} = \frac{1}{2} m v^{2} \Rightarrow v = x \frac{k}{m}$ Why this step? The spring's stored money becomes the block's moving money.

Step 3 — Numbers. $v = 0.10 \frac{200}{0.5} = 0.10 400 = 0.10 \cdot 20 = 2 m/s$

Verify: Energy stored $= \frac{1}{2} (200) (0.10)^{2} = 1 J$ ; kinetic $= \frac{1}{2} (0.5) (2)^{2} = 1 J$ . Equal. ✓ Units: $m (N/m) / kg = m s^{- 2} = m/s$ . ✓ See Spring potential energy (Hooke's law).

Example 4 — Mixed: ramp then spring (Cell D)

Worked example Statement

A ball $m = 1 kg$ rolls down a frictionless ramp from height $h = 0.8 m$ and compresses a horizontal spring by a maximum of $x = 0.20 m$ . Find $k$ . ( $g = 10$ .)

Forecast: At maximum compression, is the ball moving or momentarily still?

Step 1 — The key snapshot: maximum compression. At maximum compression the ball is momentarily at rest ( $K = 0$ ): it has stopped before bouncing back. Why this step? Choosing this special instant makes $K_{2} = 0$ , so all gravitational money has become spring money.

Step 2 — Two stored pockets swap. $m g h = \frac{1}{2} k x^{2} \Rightarrow k = \frac{2 m g h}{x ^{2}}$ Why this step? $U_{g r a v}$ at the top fully converts to $U_{s p r in g}$ at the bottom — kinetic is zero at both ends.

Step 3 — Numbers. $k = \frac{2 \cdot 1 \cdot 10 \cdot 0.8}{( 0.20 ) ^{2}} = \frac{16}{0.04} = 400 N/m$

Verify: $U_{g r a v} = m g h = 8 J$ ; $U_{s p r in g} = \frac{1}{2} (400) (0.20)^{2} = 8 J$ . Equal. ✓ Units of $k$ : $J/m^{2} = N/m$ . ✓

Example 5 — Non-zero starting speed (Cell E)

Worked example Statement

A ball $m = 1 kg$ is thrown downward at $v_{1} = 3 m/s$ from height $h = 5 m$ . Find its ground speed. ( $g = 10$ .)

Forecast: Should the answer be larger than the $10 m/s$ we got for a dropped ball (Ex 2)?

Step 1 — Keep the full $K_{1}$ term. Nothing starts at rest here, so $K_{1} = \frac{1}{2} m v_{1}^{2}$ must be carried, not dropped to zero. Why this step? The most common slip is auto-setting $K_{1} = 0$ . Only do that when told "from rest."

Step 2 — Conservation with both kinetic terms. $\frac{1}{2} m v_{1}^{2} + m g h = \frac{1}{2} m v_{2}^{2} \Rightarrow v_{2} = v_{1}^{2} + 2 g h$ Why this step? Notice the direction of the throw ( $3 m/s$ down) never matters — energy is a scalar; only the magnitude enters through $v_{1}^{2}$ .

Step 3 — Numbers. $v_{2} = 3^{2} + 2 \cdot 10 \cdot 5 = 9 + 100 = 109 \approx 10.44 m/s$

Verify: Larger than $10 m/s$ (dropped case) as forecast — the extra $9 J$ of initial KE shows up. ✓ Units of $v_{1}^{2} + 2 g h$ : $m/s$ . ✓

Example 6 — The thief: friction (Cell F)

Worked example Statement

A block $m = 2 kg$ slides down a $3 0^{\circ}$ ramp of length $L = 4 m$ , starting from rest. Kinetic friction coefficient $μ = 0.25$ . Find the speed at the bottom. ( $g = 10$ .)

Forecast: Will the speed be more or less than the frictionless answer $2 g h$ ?

Step 1 — Height dropped. Vertical drop $h = L sin 3 0^{\circ} = 4 \cdot 0.5 = 2 m$ . Why this step? Only the vertical component of displacement changes $U_{g r a v}$ .

Step 2 — Friction work (the money the thief removes). Normal force $N = m g cos 3 0^{\circ}$ . Friction $f = μ N$ , opposing motion, over distance $L$ : $W_{n c} = - μ m g cos 3 0^{\circ} \cdot L = - 0.25 \cdot 2 \cdot 10 \cdot \frac{3}{2} \cdot 4 \approx - 17.32 J$ Why this step? Friction is non-conservative — its work must appear as $W_{n c}$ , NOT as a potential energy. See Energy lost to friction.

Step 3 — Modified conservation. $K_{1} + U_{1} + W_{n c} = K_{2} + U_{2} \Rightarrow 0 + m g h + W_{n c} = \frac{1}{2} m v^{2} + 0$ $m g h + W_{n c} = 40 + (- 17.32) = 22.68 J = \frac{1}{2} m v^{2}$ $v = \frac{2 \cdot 22.68}{2} = 22.68 \approx 4.76 m/s$

Verify: Frictionless would give $v = 2 g h = 40 \approx 6.32 m/s$ . Our answer is smaller as forecast — the thief took $17.32 J$ . ✓ Units of $W_{n c}$ : $N \cdot m = J$ . ✓

Example 7 — Sign / direction trap: throwing UP (Cell G)

Worked example Statement

A ball $m = 0.5 kg$ is thrown straight up at $v_{1} = 8 m/s$ . Find the maximum height reached. ( $g = 10$ .)

Forecast: At the highest point, is the ball moving? What is $K$ there?

Step 1 — Read the top correctly. At maximum height the ball is momentarily at rest ( $v_{2} = 0$ , $K_{2} = 0$ ). This is the trap: many keep a kinetic term at the top. Why this step? The special instant "highest point" is defined by zero velocity.

Step 2 — Conservation, choosing $U = 0$ at the launch point. $\frac{1}{2} m v_{1}^{2} + 0 = 0 + m g h_{ma x} \Rightarrow h_{ma x} = \frac{v _{1}^{2}}{2 g}$ Why this step? Going up, kinetic money converts into stored (height) money — opposite direction to a fall, but the same equation. The sign is handled automatically because $U = m g h$ grows as $h$ grows.

Step 3 — Numbers. $h_{ma x} = \frac{8 ^{2}}{2 \cdot 10} = \frac{64}{20} = 3.2 m$

Verify: Kinematics $v^{2} = v_{1}^{2} - 2 g h_{ma x} \Rightarrow 0 = 64 - 20 h_{ma x} \Rightarrow h_{ma x} = 3.2 m$ . ✓ Mass cancelled — correct, gravity accelerates all masses equally. Units: $\frac{( m/s ) ^{2}}{m/s ^{2}} = m$ . ✓

Example 8 — Degenerate and limiting cases (Cell H)

Worked example Statement

Check the master equation in three "edge" situations to make sure it behaves: (a) $h \to 0$ : drop from zero height. (b) $v_{1} = v_{2}$ : same speed at both states. (c) $k \to \infty$ : a rigid wall spring.

Forecast: Should any of these give a finite, sensible answer or blow up?

Step 1 — (a) Zero height. $v = 2 g h$ with $h = 0$ gives $v = 0$ . Sensible: no drop, no speed gained. Why this step? A correct law must return the trivial result at the trivial input.

Step 2 — (b) Equal speeds. If $v_{1} = v_{2}$ then $Δ K = 0$ , so $Δ U = 0$ : no net height change (e.g. ball returns to launch height with same speed magnitude). Consistent with a full up-then-down flight ignoring friction. Why this step? Confirms energy conservation predicts symmetric flight.

Step 3 — (c) Rigid spring. For a spring launch $v = x k / m$ . Compression needed for a target $v$ : $x = v m / k$ . As $k \to \infty$ , $x \to 0$ : an infinitely stiff spring needs (almost) no compression to deliver the same energy — but stored energy $\frac{1}{2} k x^{2}$ stays finite only if $x$ shrinks. Concretely with $v = 2, m = 0.5$ : $x = 2 0.5/ k$ . At $k = 800$ : $x = 2 0.000625 = 0.05 m$ . Why this step? Limiting behaviour reveals no division-by-zero or blow-up — the physics stays finite.

Verify: (a) $0$ . (b) $Δ K = 0$ . (c) $x = 0.05 m$ at $k = 800$ ; check energy $\frac{1}{2} (800) (0.05)^{2} = 1 J = \frac{1}{2} (0.5) (2)^{2}$ . ✓

Example 9 — Real-world: vertical loop minimum speed (Cell I)

Worked example Statement

A cart on a frictionless track must complete a vertical circular loop of radius $R = 2 m$ . From what minimum height $h$ above the bottom of the loop must it start (from rest) to just make it round the top? ( $g = 10$ .)

Forecast: At the very top of the loop, what provides the centripetal force — and what is the smallest speed that still keeps the cart on the track?

Step 1 — Minimum-speed condition at the top. "Just makes it" means gravity alone supplies the centripetal pull, track force $= 0$ : $m g = \frac{m v _{t o p}^{2}}{R} \Rightarrow v_{t o p}^{2} = g R$ . Why this step? This is the extra physics (circular motion) that sets the speed we then feed into energy conservation.

Step 2 — Energy from start (rest, height $h$ ) to loop top (height $2 R$ ). $m g h = \frac{1}{2} m v_{t o p}^{2} + m g (2 R)$ Why this step? The loop top sits at height $2 R$ above the bottom; conservation links the two heights.

Step 3 — Solve for $h$ . $g h = \frac{1}{2} g R + 2 g R = \frac{5}{2} g R \Rightarrow h = \frac{5}{2} R = 2.5 \cdot 2 = 5 m$

Verify: $v_{t o p}^{2} = g R = 20$ , so $v_{t o p} = 20 \approx 4.47 m/s$ . Energy: start $m g h = 100 m$ ; top $\frac{1}{2} m (20) + m g (4) = 10 m + 40 m = 50 m$ ... wait recompute: $m g (2 R) = m \cdot 10 \cdot 4 = 40 m$ , $\frac{1}{2} m (20) = 10 m$ , total $50 m$ . Hmm start must equal $50 m$ , giving $h = 5$ . ✓ Consistent ( $100 m \neq =$ ... let me restate: $m g h = 100 \cdot 0.5$ ? mass cancels; $g h = 10 \cdot 5 = 50 = \frac{5}{2} (10) (2) = 50$ ). ✓

Example 10 — Exam twist: back-solve an unknown (Cell J)

Worked example Statement

A block is launched up a frictionless incline and travels a distance $d$ along the slope before stopping. It started with speed $v_{1} = 6 m/s$ and the incline is at $3 7^{\circ}$ ( $sin 3 7^{\circ} = 0.6$ ). Find $d$ . ( $g = 10$ .)

Forecast: The unknown is a distance along the slope, not a height. How do the two relate?

Step 1 — Relate slope distance to height. Height gained $h = d sin 3 7^{\circ} = 0.6 d$ . Why this step? $U_{g r a v}$ depends on vertical rise, so we must convert the along-slope distance into height.

Step 2 — Conservation (block stops ⇒ $K_{2} = 0$ ). $\frac{1}{2} m v_{1}^{2} = m g h = m g (0.6 d) \Rightarrow d = \frac{v _{1}^{2}}{2 g ( 0.6 )}$ Why this step? All initial kinetic money is spent climbing; setting final $K = 0$ closes the equation.

Step 3 — Numbers. $d = \frac{6 ^{2}}{2 \cdot 10 \cdot 0.6} = \frac{36}{12} = 3 m$

Verify: Height gained $h = 0.6 \cdot 3 = 1.8 m$ ; $m g h = m \cdot 10 \cdot 1.8 = 18 m$ ; initial $\frac{1}{2} m (36) = 18 m$ . Equal. ✓ Units: $\frac{( m/s ) ^{2}}{m/s ^{2}} = m$ . ✓

Recall Quick self-test on the matrix

Which cell forbids using $K_{1} + U_{1} = K_{2} + U_{2}$ directly? ::: Cell F — friction, because $W_{n c} \neq = 0$ . In a spring-only problem why does $U_{g r a v}$ vanish from the equation? ::: The height does not change, so $U_{g r a v}$ is equal at both states and cancels. At maximum spring compression (or peak of a throw), what is the kinetic energy? ::: Zero — the object is momentarily at rest. Does the mass matter for the speed of a dropped ball? ::: No — it cancels, giving $v = 2 g h$ .

Worked examples — Conservation of mechanical energy — derivation

The scenario matrix

Example 1 — Pure drop (Cell A)

Example 2 — Path independence: the slide vs the fall (Cell B)

Example 3 — Spring only, no height change (Cell C)

Example 4 — Mixed: ramp then spring (Cell D)

Example 5 — Non-zero starting speed (Cell E)

Example 6 — The thief: friction (Cell F)

Example 7 — Sign / direction trap: throwing UP (Cell G)

Example 8 — Degenerate and limiting cases (Cell H)

Example 9 — Real-world: vertical loop minimum speed (Cell I)

Example 10 — Exam twist: back-solve an unknown (Cell J)

Connections