D. Energy and Society · Work, Energy and Power

Conservation of Mechanical Energy

D2.3 D3.1 D3.3About 75 minutes

Learning goals

I can state when mechanical energy is conserved.
I can write and use Ek₁ + Eg₁ = Ek₂ + Eg₂ to find speeds and heights.
I can explain why the speed of a falling object does not depend on its mass when air resistance is ignored.
I can describe the energy changes of a pendulum and a roller coaster.

Words to know

mechanical energy (Em)
The total of kinetic and gravitational potential energy: Em = Ek + Eg.

conservation of mechanical energy
When only gravity does work (no friction or air resistance), Em stays constant: Ek₁ + Eg₁ = Ek₂ + Eg₂.

isolated system
A system with no energy entering or leaving it. Its total energy is constant.

pendulum
A mass (the bob) hanging from a fixed point, free to swing back and forth.

turning point
The highest point of a swing or throw, where the object stops for an instant: all its mechanical energy is Eg.

The ride that runs on height

On a classic roller coaster, a chain drags the train up the first hill. After that, there is no motor. The whole ride, every drop, loop and hill, runs on the gravitational potential energy stored on that first climb. In this lesson you will predict the speed of the train at any point using one powerful idea: if friction is small enough to ignore, mechanical energy is conserved.

When is mechanical energy conserved?

Mechanical energy is Em = Ek + Eg. As an object rises and falls, energy changes between Ek and Eg. If gravity is the only force doing work, no energy leaves the kinetic–potential "accounts", and the total stays constant:

Ek₁ + Eg₁ = Ek₂ + Eg₂

½mv₁² + mgh₁ = ½mv₂² + mgh₂

This holds for free fall, a frictionless roller coaster, a pendulum (the string's pull is always perpendicular to the bob's motion, so it does no work) and a thrown ball, as long as friction and air resistance are small enough to ignore. When they are not, some mechanical energy becomes thermal energy; that is the next lesson.

Mass cancels

Every term has an m in it, so divide it out:

½v₁² + gh₁ = ½v₂² + gh₂

The speed at each height does not depend on the mass. A full train and an empty train (ignoring friction) reach the bottom at the same speed, just as Galileo argued that heavy and light objects fall together.

Worked example

Example 1: Free fall

A ball is dropped from rest 20 m above the ground. Ignoring air resistance, how fast is it moving just before it lands?

Choose the ground as the reference level. Point 1: top (v₁ = 0, h₁ = 20 m). Point 2: just before landing (h₂ = 0).
Ek₁ + Eg₁ = Ek₂ + Eg₂ becomes 0 + mgh₁ = ½mv₂² + 0.
Cancel m: v₂² = 2gh₁ = 2(9.8)(20) = 392.
v₂ = √392 ≈ 19.8 m/s

About 20 m/s (19.8 m/s), whatever the ball's mass. This matches v² = v₁² + 2aΔd from the Kinematics unit.

A 500 kg cart passes A at 3.0 m/s. With no friction, its mechanical energy is the same at A, B and C. It is fastest at the lowest point B. (Heights drawn to scale; horizontal distances are not.)

Worked example

Example 2: Roller coaster

A 500 kg roller coaster cart passes point A, 40 m above the ground, at 3.0 m/s. Ignoring friction, find its speed at B (10 m high) and at C (25 m high).

Total mechanical energy at A: Ek = ½(500)(3.0)² = 2250 J; Eg = (500)(9.8)(40) = 196 000 J. Em = 198 250 J.
At B: Eg = (500)(9.8)(10) = 49 000 J, so Ek = 198 250 − 49 000 = 149 250 J.
v_B = √(2 × 149 250 ÷ 500) = √597 ≈ 24 m/s (24.4 m/s).
At C, using the mass-free form: ½v_C² = ½(3.0)² + 9.8(40 − 25) = 4.5 + 147 = 151.5, so v_C² = 303.
v_C = √303 ≈ 17 m/s (17.4 m/s).

About 24 m/s at B and 17 m/s at C.

A swinging pendulum trades Eg for Ek and back. With no air resistance it would rise to the same height h on each side forever.

Worked example

Example 3: Pendulum

A pendulum bob is pulled aside so that it is 0.20 m higher than its lowest point, then released from rest. How fast is it moving at the bottom of its swing?

Reference level: the lowest point. Point 1: release (v₁ = 0, h₁ = 0.20 m). Point 2: bottom (h₂ = 0).
gh₁ = ½v₂², so v₂ = √(2gh₁) = √(2 × 9.8 × 0.20) = √3.92.
v₂ ≈ 2.0 m/s (1.98 m/s)

About 2.0 m/s. The length of the string and the mass of the bob do not matter; only the drop in height does.

Worked example

Example 4: How high does it go?

A ball is thrown straight up at 12 m/s. Ignoring air resistance, how high does it rise above the release point?

At the top the ball stops for an instant, so Ek₂ = 0.
½v₁² + 0 = 0 + gh₂, so h₂ = v₁² ÷ (2g) = 144 ÷ 19.6.
h₂ ≈ 7.3 m

About 7.3 m above the release point.

Solving conservation problems

Check the condition: can friction and air resistance be ignored? (The problem usually says so.)
Choose a reference level (often the lowest point) and label point 1 and point 2.
Write Ek₁ + Eg₁ = Ek₂ + Eg₂, crossing out any term that is zero (at rest → Ek = 0; at the reference → Eg = 0).
Cancel the mass if you are finding a speed or height.
Solve, then ask: is the answer reasonable? Lower point → faster; higher point → slower.

Just for fun

A coaster with a single lift hill and no extra boosts can't have a later hill noticeably taller than its first one: apart from the small Ek the train has as it crawls over the first crest, there just isn't enough stored Eg. Friction makes later hills need to be a bit shorter still. Physics: the original ride operator.

Hands-on activity

Galileo's interrupted pendulum

You need: about 1 m of string, a metal nut, washer or key ring as a bob, tape, a pencil, a ruler and a doorway or a hook to hang the string from.

Tape or tie the string to the top of a door frame (or a hook) so the bob hangs about 10 cm above the floor.
Pull the bob to one side and hold it beside a ruler taped to a wall or chair. Note its height above the lowest point. Release it from rest (do not push).
Watch the other side. How high does the bob rise? Compare it with the release height.
Now have a helper hold a pencil horizontally against the door frame, halfway down the string, so the string catches on the pencil as the bob swings through the bottom.
Release from the same height. The bob now swings on a shorter string on the far side. How high does it rise now?

What to look for: in both cases the bob rises to nearly the same height it was released from, even though its path changes when the string catches. That is exactly what conservation of mechanical energy predicts: the bob needs all of its kinetic energy at the bottom to climb back to the same Eg. It falls slightly short because air resistance and friction at the support turn a little energy into thermal energy. Galileo described this experiment in 1638.

Your turn

An apple falls from rest from a branch 3.2 m high. How fast is it moving just before it hits the ground? Ignore air resistance.

A 0.30 kg ball is thrown straight up at 8.0 m/s. How high does it rise?

A skier starts from rest at the top of a 45 m high frictionless hill. How fast is she going at the bottom?

A cart moving at 10 m/s on a frictionless track reaches the bottom of a hill. How high a hill can it climb?

A pendulum bob moves at 1.5 m/s at the bottom of its swing. How far above the bottom will it rise?

At what point in a pendulum's swing is its kinetic energy greatest? Its gravitational potential energy?

Why does a pendulum's string do no work on the bob?

More practice: warm-up

Write the equation for conservation of mechanical energy.

A 2.0 kg rock has 50 J of Eg and 30 J of Ek. What is its mechanical energy?

When can we use conservation of mechanical energy?

If a ball's Eg decreases by 40 J as it falls (no air resistance), what happens to its Ek?

Does a heavier ball dropped from the same height land faster? (Ignore air resistance.)

More practice: on your own

A 60 kg skateboarder rolls at 4.0 m/s at the top of a 2.5 m ramp. Ignoring friction, how fast is she going at the bottom?

A coaster car moves at 25 m/s at ground level. Ignoring friction, what is its speed at the top of a 20 m hill?

A 0.15 kg ball is thrown from a 1.8 m high balcony at 6.0 m/s (in any direction). How fast is it moving just before it hits the ground? Why doesn't the direction matter?

A roller coaster starts from rest at the top of a 30 m hill. Can it get over a second hill that is 32 m high? Explain with energy.

A 2.0 kg ball falls from 10 m. Find Eg, Ek and v at 10 m, 5.0 m and 0 m.

Challenge

A pendulum is 1.2 m long. It is released from rest with the string at 30° from vertical. Find the speed at the bottom. (Hint: the bob's height above the lowest point is L − L cos θ.)

A ball dropped from height H hits the ground at speed v. From what height must it be dropped to hit at 2v? Explain.

At what height, as a fraction of the starting height h, does a ball dropped from rest have equal Ek and Eg (reference = ground)? Find its speed there in terms of g and h.

Check your understanding

Quiz

Mechanical energy is conserved when:

A ball is dropped from 5.0 m. Ignoring air resistance, its speed just before landing is about:

Where is a pendulum bob's speed greatest?

Two balls, 1 kg and 5 kg, are dropped from the same height with no air resistance. At the ground:

A cart has 400 J of Ek and 600 J of Eg at the top of a hill. With no friction, at the bottom (Eg = 0) its Ek is:

A ball is thrown up at 9.8 m/s. It rises (ignoring air resistance):

For parents and teachers

Big idea: When only gravity does work, Ek + Eg stays constant (D2.3, D3.1). This one equation replaces many kinematics steps and works for curved paths like coasters and pendulums.

How to help: Ask your teen to point to the places in a problem where the object is fastest and slowest before calculating. Watching a playground swing works too: highest point, zero speed; lowest point, maximum speed.

Common difficulties:

Forgetting the Ek₁ term when the object does not start at rest.
Using inconsistent reference levels for h₁ and h₂.
Not cancelling the mass and then getting stuck because the mass is not given.
Using the conservation equation when the problem includes friction (that is next lesson).

Safety: the pendulum should hang clear of faces; release it gently and keep it away from glass.