3.4 Conservation of energy

Four little picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. The half pipe: height turns into speed and back

2. Dropping in: why each pass is a little lower

Read the story as text

A skateboarder stands on the lip of a half-pipe, 4 m above the bottom. She leans forward and drops in without pushing. At the bottom she is flying. Up the other side she rises, but never quite back to 4 m. Each pass she gets a little lower, and the wheels and bearings get a little warm.

Where does her speed come from, and where does the "missing" height go? Her speed comes from her height: gravitational potential energy turns into kinetic energy. The "missing" height has become internal (thermal) energy in the warm wheels, bearings and ramp. The total energy stays the same.

A skater drops in from rest, with no friction. At the lowest point of the half pipe she has:

3. Heavy or light: the same speed at the bottom

A heavy ball and a light ball roll down the same slippery (frictionless) slope from the same height. At the bottom:

4. Check yourself

Think of your answer first, then tap to see it.

a) A skater starts from rest at the top of a half pipe with no friction at all. How high does she get on the other side?

Show answer

Back to the same height. All her speed energy turns back into height energy, and none turns into heat.

b) A real skater starts 4 m up and only reaches 3 m on the other side. Where did the "missing" energy go?

Show answer

Into internal (thermal) energy: the wheels, bearings and ramp got a little warmer. The total energy is the same; it just changed form.

c) A 50 kg skater drops 4.0 m with no friction. How fast is she going at the bottom?

Show answer

m g h = ½ m v², so v = √(2 g h) = √(2 × 9.8 × 4.0) = √78.4 = 8.9 m/s. The 50 kg cancels.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead.

A 2.0 kg ball is dropped from 5.0 m (no air resistance). Its kinetic energy just before it lands is about:

A block slides to a stop on a rough floor. For the block + floor system, the total energy:

A skater starts from rest 4 m up a frictionless half pipe. When she is 1 m above the bottom, what fraction of her energy is kinetic?

Energy is never made or destroyed. It moves between kinetic, potential and internal energy, and into or out of a system by work. Closed and open systems, energy bar charts, and where friction "loses" energy to.

The idea

In pictures

  • High up and still: all height energy. Low down and fast: all speed energy. The skater just swaps one for the other.
  • Rubbing (friction) turns some energy into heat: warm wheels. That is why each pass is a little lower.
  • Add up height energy + speed energy + heat: the total never changes, unless something outside pushes or pulls.
  • Heavy or light, a skater dropping the same height reaches the same speed (with no friction).

Below is the same idea in words, and then with numbers.

Energy does not appear or disappear. For any system you choose:

ΔEsystem = Wexternal

The energy inside the system changes only when a force from outside the system does work on it. Inside, energy just changes form:

Ki + Ui + Wext = Kf + Uf + ΔEth

Friction makes internal energy

When two surfaces rub, kinetic friction turns mechanical energy into internal (thermal) energy, Eth: the surfaces get slightly warmer. If both surfaces are inside the system, the energy is not lost, it is just in a form you cannot easily get back:

ΔEth = fk d

where d is the distance slid. So "energy lost to friction" really means "energy turned into internal energy".

Energy bar charts (LOL diagrams)

An energy bar chart shows the energy of a system at two moments. Left: the start. Middle: the system and any outside work. Right: the end. Each bar is a number of equal blocks. The rule: blocks at the start + outside work = blocks at the end.

Start (top of ramp) End (bottom) KUgUsEth KUgUsEth system:skater + Earth+ track W ext = 0

Here 4 blocks of Ug at the top become 3 blocks of K and 1 block of Eth at the bottom. 4 = 3 + 1. The sims below draw these bars live.

Choosing the system

The same event can be told two ways. Skater + Earth + track: closed, gravity and friction are inside, so Ug and Eth appear in the bars. Skater alone: open. Gravity does positive outside work and friction does negative outside work; there is no Ug bar. Both stories give the same speed.

With numbers: the skater

m = 50 kg, starts at rest 4.0 m above the bottom. System: skater + Earth + ramp (closed). Zero of Ug at the bottom.

Show all steps as text
  1. Start energy: Ug = m g h = 50 × 9.8 × 4.0 = 1960 J, K = 0. Total 1960 J. At rest, all the energy is stored as Ug.
  2. No friction: at the bottom Ug = 0, so K = 1960 J. Closed system, nothing turns into Eth.
  3. Speed: ½ (50) v² = 1960, v² = 78.4, v = 8.9 m/s. (Shortcut: v = √(2 g h); the mass cancels.) Every kilogram gets the same energy per metre dropped.
  4. With friction turning 400 J into thermal energy: K = 1960 − 400 = 1560 J. Start total = K + Ug + Eth at the end.
  5. Speed with friction: v² = 2 × 1560 / 50 = 62.4, v = 7.9 m/s. Less K, less speed.
  6. Highest point on the other side (if friction takes another 400 J on the way up): Ug = 1960 − 800 = 1160 J, so h = 1160 / (50 × 9.8) = 2.4 m. This is why each pass is lower.

Play: the skate park and the spring launcher

Release the skater from rest. The bars show kinetic energy K, gravitational potential energy Ug, thermal energy Eth and the total, all live. With friction 0 she returns to the dashed start height forever. Turn friction up and watch Eth grow while the total bar stays exactly the same. Can she get over the bump if you start her low?

Simple model: friction force = μ m g cos θ along the track (it ignores the extra push in the curved bottom). Ug = 0 at the ground line.

A spring launches a block up a ramp

Predict first. With no friction, you double the spring's compression. How far up the ramp does the block go now?

Now check it: note the "Max up the ramp" value, double the compression and play again.

Worked examples

Dropped ball basic

A 0.20 kg ball is dropped from 10 m. Ignoring air resistance, how fast is it moving just before it lands?

Show solution
Show all steps as text
  1. System ball + Earth is closed: m g h = ½ m v². Ug turns into K.
  2. v = √(2 g h) = √(2 × 9.8 × 10) = √196 = 14 m/s. The mass cancels; 0.20 kg is not needed.

Pendulum medium

A pendulum bob is pulled aside so that it is 0.45 m above its lowest point, then released. How fast is it moving at the bottom?

Show solution
Show all steps as text
  1. The string tension is always perpendicular to the motion, so it does no work. θ = 90°.
  2. So bob + Earth is closed for energy: ½ v² = g h, v = √(2 × 9.8 × 0.45) = √8.82 = 3.0 m/s. Only the height drop matters, not the curved path.

Spring launcher medium

A toy launcher has k = 500 N/m and is compressed 0.10 m. It fires a 0.050 kg ball straight up. How high does the ball rise above the point where it was released?

Show solution
Show all steps as text
  1. System ball + spring + Earth, closed. Start: Us = ½ × 500 × 0.10² = 2.5 J, K = 0. Zero Ug at the release point.
  2. Top: K = 0, Us = 0, all the energy is Ug: 0.050 × 9.8 × h = 2.5. At the top the ball stops for an instant.
  3. h = 2.5 / 0.49 = 5.1 m. Energy bars: 2.5 J of Us becomes 2.5 J of Ug.

Sliding to a stop AP

A 2.0 kg block slides on a level floor at 4.0 m/s. μk = 0.30. How far does it slide? Describe the energy bars for block + floor.

Show solution
Show all steps as text
  1. Start: K = ½ × 2.0 × 4.0² = 16 J. End: K = 0. It stops.
  2. Block + floor is closed, so all 16 J becomes Eth: fk d = 16 J. Bars: K 16 J at the start, Eth 16 J at the end.
  3. fk = μ m g = 0.30 × 2.0 × 9.8 = 5.88 N, so d = 16 / 5.88 = 2.7 m. ΔEth = fk d.

Roller coaster with friction AP

A 500 kg coaster car starts from rest at the top of a 30 m hill and reaches the top of a 20 m hill at 12 m/s. How much energy became thermal energy?

Show solution
Show all steps as text
  1. Ug drop: m g Δh = 500 × 9.8 × 10 = 49 000 J. Only the 10 m difference matters.
  2. K gained: ½ × 500 × 12² = 36 000 J. Starts from rest.
  3. ΔEth = 49 000 − 36 000 = 13 000 J. Closed system: what Ug lost but K did not gain went into internal energy.
  4. Without friction the car would have v = √(2 × 9.8 × 10) = 14 m/s at the second hill. A quick check that 12 m/s is reasonable.

Practice

  1. A ball is released from rest at height h on a frictionless track. What is its speed when it is at height h/2?

    Worked answer

    It has dropped h/2: ½mv² = mg(h/2), so v² = gh and v = √(gh).

  2. A ball falls with no air resistance. Which choice of system is closed (no outside work), so its total energy stays constant?

    Worked answer

    With Earth inside the system, gravity is internal and does no outside work. The ball alone is open: gravity does outside work on it.

  3. A block slides from rest down a rough ramp. For the block + ramp + Earth system, which bar chart is right (start → bottom)?

    Worked answer

    Closed system: the total stays 5 blocks. Friction turns some into Eth, so K is less than 5 and K + Eth = 5. "K 3" alone loses 2 blocks; "K 5, Eth 2" creates 2.

  4. Two identical balls are thrown from the same cliff with the same speed, one straight up and one horizontally. Ignore air. Which hits the ground faster?

    Worked answer

    Both start with the same K and the same Ug, and both end at the same height. Energy is a scalar, so direction does not matter: same final K, same speed (but different directions and times).

  5. A cart rolls to a stop on a rough floor. For the cart + floor system, what happens to the total energy?

    Worked answer

    Both rubbing surfaces are inside the system, so no outside work is done. The cart's K turns into Eth of cart and floor. Total unchanged.

  6. A spring launcher fires a ball straight up to a height of 2.0 m. The compression is increased from 5 cm to 10 cm. The new height is about

    Worked answer

    ½kx² = mgh, so h ∝ x². Doubling x gives 4 × 2.0 = 8.0 m.

  7. Short answer. A 1.0 kg block starts from rest 2.0 m high on a rough ramp and reaches the bottom at 5.0 m/s. How much energy became thermal energy? Show the energy bars in words.

    Worked answer

    Start: Ug = 1.0 × 9.8 × 2.0 = 19.6 J. End: K = ½ × 1.0 × 5.0² = 12.5 J.

    ΔEth = 19.6 − 12.5 = 7.1 J. Bars: start Ug 19.6 J; end K 12.5 J and Eth 7.1 J; no outside work (block + ramp + Earth).

  8. Short answer. A ball is thrown straight up. Describe the energy bars for the ball + Earth system at the moment it leaves the hand and at its highest point. Then describe them for the ball alone.

    Worked answer

    Ball + Earth: leaving the hand, all K (with Ug = 0 there); at the top, all Ug, same total, no outside work.

    Ball alone: start K only; top K = 0. The bars do not balance by themselves: Earth's gravity does negative outside work equal to the starting K. ΔE = Wext.

AP question types for this topic

These four free response questions match the four question types on the AP Physics 1 exam, each sized for conservation of energy. Work each part on paper first, then open the worked answer to check it and see how points are given.

1. Mathematical Routines

Derive with symbols first, then put in numbers. Tomas, mass m, starts from rest at the top of a playground slide of height h.

  1. Derive his speed at the bottom if there were no friction.
  2. He actually reaches the bottom at speed vb. Derive the energy turned into internal (thermal) energy of the Tomas-slide-Earth system.
  3. Take m = 30 kg, h = 3.0 m, vb = 6.0 m/s. Calculate the frictionless speed and the internal energy, and the fraction of the starting energy it is.
Worked answer and scoring

(a) mgh = ½mv², so v = √(2gh); the mass cancels.

(b) mgh = ½mvb² + ΔEint, so ΔEint = mgh − ½mvb².

(c) v = √(2 × 9.8 × 3.0) = 7.7 m/s. mgh = 882 J, ½mvb² = 540 J, so ΔEint = 342 J, about 39% of 882 J.

Scoring (6 points)

  • 1 point for mgh = ½mv²
  • 1 point for √(2gh)
  • 1 point for including ΔEint
  • 1 point for 7.7 m/s
  • 1 point for 342 J
  • 1 point for about 39%

2. Translation Between Representations

Move between graphs, pictures, equations and words, and justify. Sana releases a 0.50 kg cart from rest at the top of a track 0.80 m high. The system is cart + Earth; first assume no friction.

Four energy bar charts (left of the dashed line: at the top; right: at the bottom):

KUKUA
Chart A: before, all U (4 blocks); after, all K (4 blocks).
KUKUB
Chart B: before, all U (4 blocks); after, 2 blocks K and 2 blocks U.
KUKUC
Chart C: before, 4 blocks U; after, 4 blocks K and still 4 blocks U.
KUKUD
Chart D: before, all K (4 blocks); after, all U (4 blocks).
  1. Which chart is correct for the frictionless track (zero of U at the bottom)? Explain why each other chart is wrong.
  2. Write the energy equation that matches your chart and calculate the speed at the bottom.
  3. With friction, the cart reaches the bottom at 3.5 m/s. Describe how the chart changes and calculate the energy that becomes internal energy.
Worked answer and scoring

(a) Chart A. All U at the top becomes K at the bottom. B leaves U at the bottom, but U = 0 there. C has more energy after than before, which an isolated system cannot do. D has the cart moving at the top, but it starts at rest.

(b) mgh = ½mv²: v = √(2 × 9.8 × 0.80) = 4.0 m/s.

(c) Add an internal-energy bar after: the K bar is shorter and K + Eint equals the starting U. Utop = 0.50 × 9.8 × 0.80 = 3.92 J; K = ½ × 0.50 × 3.5² = 3.06 J; ΔEint = 0.86 J.

Scoring (5 points)

  • 1 point for chart A
  • 1 point for rejecting B, C and D with reasons
  • 1 point for mgh = ½mv²
  • 1 point for 4.0 m/s
  • 1 point for an Eint bar and 0.86 J

3. Experimental Design and Analysis

Plan a measurement, make a straight-line graph and read its slope. Mei wants to find the spring constant of a launcher using energy. She has a 0.50 kg cart, a spring launcher with a compression scale, a level low-friction track and a photogate with a 5.0 cm flag.

Her data:

compression x (m)0.010.020.030.04
speed v (m/s)0.210.390.610.79
  1. Describe her procedure.
  2. Derive what to plot for a straight line and what the slope means.
  3. Use the data to find k.
Worked answer and scoring

(a) Compress the launcher to a marked x, release the cart on the level track, and time the flag through the photogate just after launch: v = 0.050 m / time. Repeat three times for each of several compressions.

(b) ½kx² = ½mv², so v = √(k/m) × x. Plot v against x: slope = √(k/m), so k = m × slope².

(c) Best-fit slope ≈ 0.79/0.04 ≈ 20 s-1. k = 0.50 × 20² = 200 N/m (about 199 N/m with the exact fit).

Scoring (6 points)

  • 1 point for marked compressions
  • 1 point for speed from flag / gate time, with repeats
  • 1 point for ½kx² = ½mv²
  • 1 point for plotting v against x
  • 1 point for slope = √(k/m)
  • 1 point for k ≈ 200 N/m

4. Qualitative/Quantitative Translation

Explain a claim in words, back it with a derivation, and connect the two. Ade stands on a cliff 15 m high and throws a stone at 10 m/s. He says: Whether I throw it straight up, straight down or sideways, it hits the ground below at the same speed (ignore air resistance).

  1. In words, explain whether Ade is right.
  2. Derive the landing speed and calculate it.
  3. Connect your expression in (b) to your reasoning in (a).
Worked answer and scoring

(a) Yes. In every throw the stone-Earth system starts with the same K (same speed) and the same U (same height), and ends at the same height. Energy is conserved and is a scalar, so the direction does not matter: the final K, and so the speed, is the same.

(b) ½mv0² + mgh = ½mv², so v = √(v0² + 2gh) = √(100 + 294) = 19.8 m/s.

(c) The expression contains only v0, g and h: no angle and no mass. That is the maths form of (a): only the starting speed and the drop matter, not the direction.

Scoring (5 points)

  • 1 point for “yes”
  • 1 point for same initial K and U, same final U
  • 1 point for the energy equation
  • 1 point for 19.8 m/s
  • 1 point for noting the expression has no angle

Common mistakes

The mistake: "Energy is lost to friction" and then leaving it out of the total.

Why it is wrong: the energy is not destroyed; it becomes internal (thermal) energy of the surfaces.

How to spot it: your end bars add up to less than your start bars and you have no Eth bar and no outside work.

The mistake: counting gravity twice: putting Ug in the bars and also adding work done by gravity.

Why it is wrong: if Earth is in the system, gravity's effect is ΔUg. Work by gravity only appears when Earth is outside.

How to spot it: say your system out loud first, then list only the outside forces that do work.

The mistake: "Energy is conserved, so the kinetic energy stays the same."

Why it is wrong: the total stays the same in a closed system; K can grow or shrink as U changes.

How to spot it: a ball rolling downhill clearly speeds up.

The mistake: "A heavier skater reaches the bottom of a frictionless ramp faster."

Why it is wrong: m g h = ½ m v², the mass cancels: v = √(2gh) for every mass.

How to spot it: if mass is not given, the answer probably does not depend on it.

The mistake: using v instead of v² (v = g h) or forgetting the ½.

Why it is wrong: K = ½mv². Units warn you: g h has units m²/s², which is a speed squared.

How to spot it: check units at the end; take the square root last.