Work and conservation of energy AP Physics 1 · Lesson 7 of 7

Q: Does it matter where I put the zero of height?

No, only changes in Ug matter. Choose the lowest point in the problem to keep everything positive, and keep the same zero throughout.

Written by STEM-Tutor's tutors · Updated October 2026 · 7 topics, 18 practice problems, 3 free-response questions

The start of Unit 3. Energy methods solve many force problems in a few lines. You will learn what work is, which forces are conservative, how to use conservation of energy on roller coasters, slides and springs, how friction turns mechanical energy into thermal energy, and how to draw the energy bar charts and graphs the AP exam asks for.

On this page

In this lesson
Work and the work-energy theorem
Conservative forces and conservation of energy
Elastic potential energy
Thermal energy vs work done by friction
Energy bar charts
Graphing energy: free fall vs springs
Atwood machine with energy
Key equations
Practice problems
Free-response practice
Common mistakes
FAQs
Keep going

In this lesson

This lesson covers work (conservative vs non-conservative forces), conservation of energy, mechanical energy on a slide, spheres on different ramps, elastic potential energy, thermal energy vs work done by friction, energy bar charts, graphing energy, and the Atwood machine with energy:

Work and the work-energy theorem
Conservative forces and conservation of energy
Elastic potential energy
Thermal energy and friction
Energy bar charts
Graphing energy: free fall vs springs
Atwood machine with energy

All numbers use g = 9.8 m/s², as on the AP Physics 1 equation sheet. Every formula used here is on that sheet: AP Physics 1 equation sheet.

Work and the work-energy theorem

Work is how a force transfers energy into or out of a system by acting through a displacement. For a constant force:

W = F_∥d = Fd cosθ

θ is the angle between the force and the displacement. Work is a scalar measured in joules (1 J = 1 N·m).

A force F at angle θ to the displacement d. Only the component along the displacement, F cosθ, does work: W = Fd cosθ.

Angle between F and d	Work	Example
0° (same direction)	Positive: energy into the object	A push in the direction of motion
90° (perpendicular)	Zero	The normal force on a sliding box; tension in uniform circular motion
180° (opposite)	Negative: energy out of the object	Kinetic friction on a sliding box

For a force that changes with position, work is the area under a force-position graph. For a spring, that area is a triangle, ½kx².

Force against stretch for a spring with k = 40 N/m. The work done stretching it from 0 to 0.20 m is the area of the triangle: ½(0.20)(8.0) = 0.80 J, which is the same as ½kx².

The work-energy theorem links work to kinetic energy, K = ½mv²: the net work on an object equals its change in kinetic energy.

W_net = ΔK = ½mv_f² − ½mv_i²

Worked example 1: work at an angle

A 12 kg sled is pulled 15 m across level snow by a rope with tension 40 N at 30° above the horizontal. Kinetic friction is 20 N. Find the work done by each force and the net work.

Tension: W = 40 × 15 × cos 30° = 520 J.
Friction: W = 20 × 15 × cos 180° = −300 J.
Gravity and the normal force are perpendicular to the motion: 0 J each.
Net work = 520 − 300 = 220 J, which is the sled's gain in kinetic energy.

Worked example 2: work-energy theorem

If the sled in example 1 started at 1.0 m/s, how fast is it moving after 15 m?

ΔK = 220 J, and K_i = ½(12)(1.0)² = 6 J.
K_f = 226 J, so v = √(2 × 226/12) = 6.1 m/s.

Conservative forces and conservation of energy

A force is conservative if the work it does on an object moving between two points does not depend on the path taken. Gravity and spring forces are conservative. For these forces we can define a potential energy, energy stored in the arrangement of a system, and the work they do equals minus the change in potential energy: W_cons = −ΔU.

Friction is non-conservative: drag a box around a long path and friction takes out more energy than on a short path. Friction has no potential energy; it turns mechanical energy into thermal energy (internal energy of the surfaces). Applied pushes and pulls by people and motors are also non-conservative.

Force	Type	Energy it is linked to
Gravity (near Earth)	Conservative	U_g = mgh, in the object-Earth system
Ideal spring	Conservative	U_s = ½kx²
Kinetic friction	Non-conservative	Converts mechanical energy to thermal energy, ΔE_th = F_fd
Applied force, tension from outside	Non-conservative	Adds or removes energy as work from outside the system

Key idea

Potential energy belongs to a system of interacting objects, not to one object. Gravitational potential energy is energy of the object-Earth system. If your system is the object alone, gravity is an external force that does work on it, and there is no U_g. AP questions often say which system to use: read it carefully.

Conservation of energy. The total energy of a system changes only by the work done on it by external forces. If no external forces do work and no friction acts inside, the mechanical energy is constant:

K_i + U_i = K_f + U_f ½mv_i² + mgh_i = ½mv_f² + mgh_f

A roller coaster released from rest at A. Without friction, the total mechanical energy K + U_g is the same at A, B, C and D, so the speed at each point depends only on how far it has dropped below A.

Worked example 1: roller coaster

A car starts from rest at A (20 m high) in the diagram. Ignoring friction, find its speed at B (h = 0), C (12 m) and D (2 m).

mgh_A = ½mv² + mgh, so v = √(2g(h_A − h)). The mass cancels.
B: v = √(2 × 9.8 × 20) = 19.8 m/s. C: v = √(2 × 9.8 × 8) = 12.5 m/s. D: v = √(2 × 9.8 × 18) = 18.8 m/s.

Worked example 2: mechanical energy on a slide

A 30 kg child slides down a 3.5 m tall slide, starting at 0.50 m/s. Ignoring friction, how fast is the child going at the bottom?

½v_i² + gh = ½v_f².
v_f = √(0.50² + 2 × 9.8 × 3.5) = √68.85 = 8.3 m/s.

Spheres released from the same height on different ramps. With no energy lost to friction or air, both reach the bottom with the same speed, √(2gh), whatever the ramp's shape. The curved ramp gets its sphere there first, because it speeds up sooner.

The spheres on different ramps demo shows the power of energy methods: the speed at the bottom depends only on the drop in height, not on the shape of the ramp, as long as no energy is lost. Newton's laws would need a different calculation for every ramp shape. (For rolling balls, part of the energy goes into spinning, so they are slower than sliding blocks, but they still all reach the bottom with the same speed as each other. Rolling is Unit 6.)

Practice derivation: conservation of energy

Choose the system (for gravity, include Earth) and the initial and final states.
Write each energy that is present at each state: K, U_g, U_s.
Write E_i + W_ext = E_f + ΔE_th, crossing out zero terms (starting from rest, zero height, and so on).
Solve in symbols, then check the units and limiting cases (for example, more height should give more speed).

Elastic potential energy

A stretched or compressed spring stores elastic potential energy, equal to the work done to deform it (the area under the F-x graph):

U_s = ½kx²

x is the stretch or compression from the natural length. Because x is squared, doubling the compression stores four times the energy.

Worked example 1: spring launcher

A spring with k = 400 N/m is compressed 0.10 m and launches a 0.050 kg ball horizontally on a frictionless surface. Find the ball's launch speed.

½kx² = ½mv².
v = x√(k/m) = 0.10 × √(400/0.050) = 0.10 × 89.4 = 8.9 m/s.

Worked example 2: dropped onto a spring

A 2.0 kg block is dropped from rest 0.50 m above the top of a vertical spring with k = 600 N/m. What is the maximum compression?

At maximum compression the block is momentarily at rest. It has fallen 0.50 + x in total.
mg(0.50 + x) = ½kx²: 19.6(0.50 + x) = 300x², so 300x² − 19.6x − 9.8 = 0.
x = [19.6 + √(19.6² + 4 × 300 × 9.8)]/600 = 0.22 m.

Common trap

A common exam trap: when a block is dropped onto a spring, its gravitational potential energy keeps falling during the compression too. Measure the drop all the way to the lowest point (h + x).

Thermal energy vs work done by friction

When kinetic friction acts between two surfaces, mechanical energy is converted into thermal energy, which warms the surfaces. For an object sliding a distance d:

ΔE_th = F_fd = μ_kF_Nd

There are two equally good ways to account for it, depending on the system:

System	How friction appears	Energy equation
Block alone	Friction is an external force doing negative work, −F_fd	K_i + W_other − F_fd = K_f
Block + surface (+ Earth)	Friction is internal; mechanical energy becomes thermal energy inside the system	K_i + U_i = K_f + U_f + ΔE_th

Either way, the numbers come out the same. Do not count friction twice, as both negative work and thermal energy, in the same equation.

A block held against a compressed spring on a level floor. When released, the spring's elastic energy becomes kinetic energy, which friction then turns into thermal energy as the block slides to a stop.

Worked example 1: spring release with friction

In the diagram, k = 250 N/m, the spring is compressed 0.20 m and the 1.5 kg block is released. After leaving the spring it slides across a rough section with μ_k = 0.30. How far does it slide on the rough section?

Energy stored: ½(250)(0.20)² = 5.0 J. This all becomes kinetic energy before the rough section (the floor under the spring is smooth).
Friction: F_f = 0.30 × 1.5 × 9.8 = 4.41 N.
5.0 = 4.41d, so d = 1.13 m.

Worked example 2: slide with friction

The 30 kg child from earlier actually reaches the bottom of the 3.5 m slide at 6.0 m/s, starting from rest. How much thermal energy was produced?

mgh = 30 × 9.8 × 3.5 = 1029 J; K_f = ½(30)(36) = 540 J.
ΔE_th = 1029 − 540 = 489 J, nearly half the energy.

Energy bar charts

An energy bar chart (an LOL diagram) shows how energy moves within and into a system between two moments. Draw a bar for each kind of energy at the start, a bar for any work done by external forces, and a bar for each kind of energy at the end. The total height on the left plus the work must equal the total height on the right.

Energy bar chart for a child sliding down a slide with friction, system = child + slide + Earth. 1 unit of kinetic energy and 4 units of gravitational potential energy at the top become 3 units of kinetic energy and 2 units of thermal energy at the bottom. No external work.

How to draw an energy bar chart

State the system (it decides whether gravity's energy is U_g or work by an external force).
Pick a zero for height, usually the lowest point, so U_g is never negative.
Draw the initial bars, then the work bar (positive if energy goes in, negative if it comes out), then the final bars.
Check: total initial + work = total final, counting blocks.

Key idea

Bar charts are a favourite Translation Between Representations (TBR) question. The AP exam often asks you to draw the bar chart first and then write the matching energy equation: every bar becomes one term.

Graphing energy: free fall vs springs

Graphs of energy against position show the same conservation idea visually. With no friction, the total mechanical energy is a horizontal line, and K and U trade places beneath it.

Free fall from rest: as the object falls, U_g (pink) falls in a straight line and K (blue) rises in a straight line. Their total (green, dashed) stays constant.

A mass on a horizontal spring oscillating between −A and +A: U_s = ½kx² is a parabola, largest at the turning points, and K is the upside-down parabola, largest at x = 0. The total is constant.

	Free fall (from rest)	Mass on a horizontal spring
Potential energy	U_g = mgh: straight line, falls as the object drops	U_s = ½kx²: parabola, zero at x = 0
Kinetic energy	K = mg × (distance fallen): straight line	K = ½k(A² − x²): upside-down parabola
Where speed is greatest	At the lowest point	At x = 0, the equilibrium position
Against time instead	K rises as t² (curved), since the distance fallen is ½gt²	K and U oscillate, each going through two cycles per period

Worked example: reading a spring graph

A 0.50 kg mass on a horizontal spring (k = 200 N/m) oscillates between x = −0.10 m and x = +0.10 m. Find its speed at x = 0 and at x = 0.05 m.

Total energy = ½kA² = ½(200)(0.10)² = 1.0 J.
At x = 0: K = 1.0 J, so v = √(2 × 1.0/0.50) = 2.0 m/s.
At x = 0.05 m: U_s = ½(200)(0.05)² = 0.25 J, K = 0.75 J, v = √(2 × 0.75/0.50) = 1.7 m/s.

Atwood machine with energy

Energy gives a quick route to the speed in connected systems. For an Atwood machine, take the system as both blocks, the string and Earth. The string's tension does +W on one block and −W on the other, so it does no net work on the system.

An Atwood machine released from rest. As the 3m block falls a height h, the m block rises h. Treating both blocks and Earth as one system, the tension does no net work and energy is conserved.

Worked example: Atwood machine with energy

Blocks of mass m and 3m (in the diagram) are released from rest. Find their speed after the heavier block has fallen h = 0.60 m.

Energy lost: the 3m block loses 3mgh, the m block gains mgh. Net loss of U_g = 2mgh.
Both blocks gain kinetic energy at the same speed: ½(m + 3m)v² = 2mv².
2mv² = 2mgh, so v = √(gh) = √(9.8 × 0.60) = 2.4 m/s.
Check with Lesson 2: a = (3m − m)g/(4m) = g/2, and v² = 2(g/2)h = gh. It agrees.

Key idea

Energy gives speeds, not times or accelerations directly. If a question asks “how long”, use Newton's second law and kinematics; if it asks “how fast after falling h” or “how far before stopping”, energy is usually quicker.

Key equations

Work by a constant force	W = Fd cosθ
Kinetic energy	K = ½mv²
Work-energy theorem	W_net = ΔK
Gravitational potential energy (near Earth)	ΔU_g = mgΔh
Elastic potential energy	U_s = ½kx²
Conservation of energy	K_i + U_i + W_ext = K_f + U_f + ΔE_th
Thermal energy from friction	ΔE_th = F_fd

Practice problems

18 problems, from warm-up to exam level. Answer each one before you look: use the hint if you are stuck, then reveal the worked solution one step at a time. Number answers are marked within 2.5%, so g = 10 m/s² still counts.

Q1Warm-up

A box is pulled 8.0 m across a floor by a 50 N force at 25° above the horizontal. How much work does the force do?

Hint

Only the component along the displacement does work.

Worked solution

W = Fd cosθ = 50 × 8.0 × cos 25° = 363 J.

Q2Warm-up

A ball on a string moves in a horizontal circle at constant speed. How much work does the tension do in one full revolution?

Hint

What is the angle between the tension and the velocity?

Worked solution

The tension points to the centre, perpendicular to the velocity at every instant.
cos 90° = 0, so no work, consistent with the constant speed. The answer is C.

Q3Warm-up

A 4.0 kg cart starts from rest. The net work done on it is 120 J. What is its final speed?

Hint

Work-energy theorem: W_net = ΔK.

Worked solution

120 = ½(4.0)v², so v = √60 = 7.75 m/s.

Q4Warm-up

Which force is non-conservative?

Hint

Does the work depend on the path?

Worked solution

Friction always opposes motion, so a longer path means more energy converted to thermal energy.
Its work depends on the path, so it is non-conservative. The answer is C.

Q5Warm-up

A roller coaster starts from rest at the top of a 25 m drop. Ignoring friction, how fast is it moving at the bottom?

Hint

mgh = ½mv².

Worked solution

v = √(2gh) = √(2 × 9.8 × 25) = 22.1 m/s.

Q6Exam-style

A coaster car passes a 15 m high point at 3.0 m/s. Ignoring friction, how fast is it moving at a later point 6.0 m high?

Hint

½v₁² + gh₁ = ½v₂² + gh₂.

Worked solution

v₂ = √(3.0² + 2 × 9.8 × 9.0) = 13.6 m/s.

Q7Warm-up

Two identical balls are released from rest at the same height, one down a straight ramp and one down a steeper curved ramp. Friction is negligible. Which statement is correct?

Hint

Speed at the bottom depends only on the drop. Time depends on how quickly each speeds up.

Worked solution

Same height drop, so the same speed at the bottom: v = √(2gh).
The curved ramp drops steeply first, so its ball gains speed sooner and arrives first. The answer is A.

Q8Warm-up

A spring with k = 300 N/m is compressed by 0.12 m. How much elastic potential energy does it store?

Hint

U_s = ½kx².

Worked solution

U_s = ½(300)(0.12)² = 2.16 J.

Q9Warm-up

A spring is compressed by x and stores energy E. It is then compressed by 3x. How much energy does it store?

Hint

U_s ∝ x².

Worked solution

Tripling x multiplies x² by 9.
So the energy is 9E. The answer is C.

Q10Warm-up

A spring (k = 500 N/m) compressed 0.080 m launches a 0.20 kg block across a frictionless floor. What is the block's speed?

Hint

All the elastic energy becomes kinetic energy.

Worked solution

½kx² = ½mv², so v = x√(k/m) = 0.080 × √2500 = 4.00 m/s.

Q11Exam-style

A spring (k = 250 N/m) compressed 0.20 m launches a 1.5 kg block onto a rough floor with μ_k = 0.30 (the floor under the spring is smooth). How far does the block slide on the rough floor?

Hint

Elastic energy = thermal energy produced by friction.

Worked solution

½kx² = 5.0 J.
5.0 = μ_kmgd = 4.41d, so d = 1.13 m.

Q12Exam-style

A 25 kg child slides from rest down a slide 4.0 m high and reaches the bottom at 7.0 m/s. How much thermal energy was produced?

Hint

Compare the gravitational energy lost with the kinetic energy gained.

Worked solution

mgh = 980 J; K = ½(25)(49) = 612 J.
ΔE_th = 368 J.

Q13Exam-style

A block slides down a rough ramp. The system is the block, the ramp and Earth. Which energy bar chart change is correct from top to bottom?

Hint

With the ramp and Earth inside the system, are gravity and friction internal or external?

Worked solution

Both gravity and friction are internal, so there is no external work.
U_g becomes K and thermal energy. The answer is A.

Q14Challenge

A mass on a horizontal spring oscillates between x = −A and x = +A with no friction. At which position is its kinetic energy equal to its elastic potential energy?

Hint

Equal energies means each is half the total: ½kx² = ½(½kA²).

Worked solution

x² = A²/2.
x = A/√2 ≈ 0.71A. The answer is B.

Q15Exam-style

A 0.50 kg mass on a horizontal spring (k = 200 N/m) oscillates with amplitude 0.10 m. What is its speed when x = 0.060 m?

Hint

Total energy = ½kA²; subtract the elastic energy at x.

Worked solution

Total = ½(200)(0.10)² = 1.0 J; U_s at 0.060 m = ½(200)(0.060)² = 0.36 J.
K = 0.64 J, so v = √(2 × 0.64/0.50) = 1.60 m/s.

Q16Exam-style

An Atwood machine has blocks of 1.0 kg and 3.0 kg, released from rest. Using energy, find their speed after the heavier block has fallen 0.80 m.

Hint

The heavy block loses m₂gh, the light one gains m₁gh, and both gain kinetic energy.

Worked solution

Net loss of U_g = (3.0 − 1.0)(9.8)(0.80) = 15.7 J.
½(4.0)v² = 15.7, so v = 2.80 m/s.

Q17Challenge

A 2.0 kg block is dropped from rest 0.50 m above the top of a vertical spring (k = 600 N/m). What is the maximum compression of the spring?

Hint

The block falls 0.50 m plus the compression x before stopping.

Worked solution

mg(0.50 + x) = ½kx², so 300x² − 19.6x − 9.8 = 0.
x = [19.6 + √(19.6² + 11760)]/600 = 0.216 m.

Q18Exam-style

A graph shows the kinetic energy of a falling object (dropped from rest, no air resistance) against the distance it has fallen. What is the shape, and what is its gradient?

Hint

K = mg × (distance fallen).

Worked solution

The gravitational energy lost, mgd, becomes kinetic energy.
So K = (mg)d: a straight line through the origin, gradient mg (the weight). The answer is A.

Free-response practice

Original questions in the style of the AP exam's free-response section. Write a full answer on paper, with words as well as equations, then open the scoring guide and tick each point you earned.

Mathematical Routines (MR) style · 8 points · about 15 minutes

A roller coaster car of mass M is released from rest at point A, at height H above the ground. It rolls down to the lowest point B (at ground level), then up over a hill at C, at height h_C. The top of hill C is part of a circle of radius R. Friction is negligible until part (d).

(a) Derive an expression for the speed of the car at C in terms of H, h_C and physical constants.
(b) Show that the car stays in contact with the track at the top of hill C only if H − h_C ≤ R/2.
(c) For H = 30.0 m and h_C = 18.0 m, calculate the speed at C and the smallest radius R that keeps the car on the track.
(d) In reality, the 500 kg car reaches C at 12 m/s. Calculate the thermal energy produced between A and C, and state where this energy has gone.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Energy of the car-Earth system: MgH = ½Mv_C² + Mgh_C, so v_C = √(2g(H − h_C)).

1 point for: energy conservation with U_g at both points and K = 0 at A
1 point for: correct expression for v_C

Part (b)

At the top of hill C, the centre is below: Mg − F_N = Mv_C²/R. The car stays on if F_N ≥ 0, so v_C² ≤ gR. Substituting: 2g(H − h_C) ≤ gR, so H − h_C ≤ R/2.

1 point for: Newton's second law at the top of the hill with the centre below
1 point for: condition F_N ≥ 0 leading to v² ≤ gR
1 point for: combines with (a) to reach the result

Part (c)

v_C = √(2 × 9.8 × 12) = 15.3 m/s. R ≥ 2(H − h_C) = 24 m.

1 point for: v_C ≈ 15 m/s and R_min = 24 m

Part (d)

Expected: MgH = 147000 J. Actual at C: ½(500)(144) + 500(9.8)(18) = 36,000 + 88,200 = 124,200 J. ΔE_th = 147,000 − 124,200 = 22,800 J. It has become thermal energy in the wheels, track and air (by friction and air resistance).

1 point for: correct energy bookkeeping, about 2.3 × 10⁴ J
1 point for: identifies the energy as thermal (internal) energy of the surroundings/track/air

Your mark: 0 of 8

Translation Between Representations (TBR) style · 7 points · about 14 minutes

A block of mass m is pushed against a horizontal spring of spring constant k, compressing it by x₀, and released from rest. The floor under the spring is frictionless; after the block leaves the spring it crosses a rough section with coefficient of kinetic friction μ and comes to rest.

(a) Draw energy bar charts for the system of block, spring, floor and Earth: (i) just before release, (ii) just as the block leaves the spring, (iii) when the block has stopped.
(b) Using your bar charts, derive an expression for the distance d the block slides on the rough section.
(c) Calculate d for k = 400 N/m, x₀ = 0.15 m, m = 0.8 kg and μ = 0.25.
(d) The compression is doubled. A student predicts the block will slide twice as far. Explain whether the student is correct.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

(i) All elastic energy U_s, for example 4 blocks. (ii) K = 4 blocks, U_s = 0. (iii) E_th = 4 blocks, K = 0. U_g stays zero throughout (level floor) and there is no external work.

1 point for: U_s becomes K, with equal totals
1 point for: K becomes E_th at the end, with no external work bar

Part (b)

½kx₀² = μmgd, so d = kx₀²/(2μmg).

1 point for: equation matching the bar charts: U_s = ΔE_th
1 point for: correct expression for d

Part (c)

½(400)(0.15)² = 4.50 J; friction = 0.25 × 0.80 × 9.8 = 1.96 N; d = 4.50/1.96 = 2.30 m.

1 point for: d ≈ 2.3 m

Part (d)

Incorrect. The stored energy is proportional to x₀², so doubling the compression stores 4 times the energy. The friction force is unchanged, so the block slides 4 times as far (d ∝ x₀²).

1 point for: 4 times as far
1 point for: reason: U_s ∝ x² while the friction force is unchanged

Your mark: 0 of 7

Qualitative/Quantitative Translation (QQT) style · 7 points · about 12 minutes

Blocks of mass m and 3m hang from the ends of a light string over an ideal pulley and are released from rest. The heavier block falls a distance h.

An Atwood machine released from rest. As the 3m block falls a height h, the m block rises h. Treating both blocks and Earth as one system, the tension does no net work and energy is conserved.

(a) Without calculating, explain how the speed of the blocks after the heavier block falls h compares with the speed of an object dropped from rest through the same height h.
(b) Use conservation of energy to derive an expression for the speed v in terms of g and h.
(c) Show that your answer to (b) is consistent with (a), and with the acceleration found from Newton's second law, a = (3m − m)g/(4m).

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Slower. In free fall, all the gravitational energy lost goes into the falling object's own kinetic energy. Here the heavy block's energy loss is partly cancelled by the light block gaining gravitational energy, and the remaining energy is shared between both blocks' kinetic energy.

1 point for: slower
1 point for: the light block gains gravitational energy and both blocks share the kinetic energy

Part (b)

Loss of U_g: 3mgh − mgh = 2mgh. Gain of K: ½(4m)v² = 2mv². So 2mv² = 2mgh and v = √(gh).

1 point for: net change in gravitational energy 2mgh
1 point for: kinetic energy of both blocks, ½(4m)v²
1 point for: v = √(gh)

Part (c)

Free fall gives √(2gh), and √(gh) is smaller, consistent with (a). With a = g/2, kinematics gives v² = 2ah = gh, so v = √(gh), the same as the energy result.

1 point for: compares with √(2gh)
1 point for: shows kinematics with a = g/2 gives the same result

Your mark: 0 of 7

Common mistakes

Counting gravity twice: as work done by gravity and as a change in U_g, in the same equation. Pick the system first.
Forgetting that work by a force perpendicular to the motion (the normal force on level ground, tension in uniform circular motion) is zero.
Using the spring's length instead of its stretch or compression from natural length in U_s = ½kx².
Assuming doubling the compression doubles the energy. Energy goes as x², so it quadruples.
Using energy to find a time. Energy gives speeds and distances; use kinematics or Newton's laws for times.
Measuring height from different zero levels at the start and end. Choose one zero and stick to it.
Ignoring the extra drop x when a block falls onto a spring.

Frequently asked questions

When should I use energy instead of Newton's laws?

Use energy when the question links speeds to heights or distances, especially on curved paths (roller coasters, ramps, pendulums) or with springs, where the force changes. Use Newton's laws when you need accelerations, forces or times.

Can kinetic energy be negative?

No. K = ½mv², and both m and v² are positive. Potential energy can be negative depending on where you put the zero, and work can be negative.

Does it matter where I put the zero of height?

No, only changes in U_g matter. Choose the lowest point in the problem to keep everything positive, and keep the same zero throughout.

Is the normal force ever doing work?

Not on an object sliding over a fixed surface, because it is perpendicular to the motion. It can do work when the surface itself moves, for example the floor of an elevator pushing you upward as the elevator rises.

What is the difference between thermal energy and heat?

Thermal energy is the internal energy of the random motion of particles in an object. Heat is energy transferred because of a temperature difference. In AP Physics 1, friction converts mechanical energy into thermal energy of the surfaces; you will not need to calculate temperature changes.

Is power part of this lesson?

Power, P = W/Δt (the rate of doing work), is the last topic in Unit 3. It usually follows straight after this lesson. See the AP Physics 1 energy guide for power.

Why does a ball on a curved ramp reach the bottom first if the speeds are equal?

Speed at the bottom depends only on the height dropped. Time depends on the whole journey: a ramp that drops steeply at the start gets the ball moving fast early, so its average speed is higher, even if the path is a little longer.

Do I need to know rolling energy for this lesson?

Not yet. Rolling objects also have rotational kinetic energy, which is in Unit 6. In this lesson, treat objects as sliding without friction, or as points.

Keep going

Related guides and practice:

All lessons in the AP Physics 1 forces and energy course · AP Physics 1 past papers