Unit 3 · Topic 3.4 Beta

Conservation of Energy

Pick the system first.

Practice 1: Creating RepresentationsPractice 2: Mathematical RoutinesPractice 3: Scientific Questioning and Argumentation

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Question set for this topic

Part 1 · Hook

Why this matters

A roller coaster has no engine after the first hill, yet it races through loops and over smaller hills for a full minute. Nothing pushes it; energy simply changes form, from height to speed and back. Learning to track that energy, and to choose what is inside the system, solves problems that would take pages with forces.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Near Earth's surface, the change in gravitational potential energy of an object-Earth system is

  1. mgΔy
  2. ½mv²
  3. ½k(Δx)²
  4. mg/Δy
Show the answer

ΔU_g = mgΔy.

  • Correct: mgΔy:
  • ½mv²:
  • ½k(Δx)²:
  • mg/Δy:

2. The net work done on an object equals its

  1. change in kinetic energy
  2. kinetic energy
  3. change in potential energy
  4. speed
Show the answer

The work-energy theorem: W_net = ΔK.

  • Correct: change in kinetic energy:
  • kinetic energy:
  • change in potential energy:
  • speed:

3. A force exerted on the system by an object outside it is

  1. an external force
  2. an internal force
  3. a conservative force
  4. a normal force
Show the answer

External forces come from outside the chosen system.

  • Correct: an external force:
  • an internal force:
  • a conservative force:
  • a normal force:

Part 4 · See it

See it first

A block released from rest at A, 6.0 m above the lowest point B of a frictionless curved track, slides through B and up to C, 2.0 m above B. Bar charts at A, B and C show kinetic energy, gravitational potential energy and their constant total: 0 and 6 units at A, 6 and 0 at B, 4 and 2 at C.
Energy bar charts for a block on a frictionless track: the bars change, the total does not. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Choose a system and draw its boundaryevery energy is either inside it (K, U, thermal) or crossing the boundary as work done by external forces
  2. External forces do work on the systemthe system's energy changes by exactly that work: ΔE_system = W_external
  3. No external work, and no friction insidemechanical energy K + U is constant: K_i + U_i = K_f + U_f
  4. Kinetic friction acts inside the systemmechanical energy decreases and thermal energy increases by the same amount
  5. The energies are drawn as barsthe bars before plus any work in equal the bars after: the bar chart is the equation

Part 6 · Key ideas

Key ideas

  • A one-object system has only kinetic energy. Potential energy appears only when the interacting objects (Earth, a spring) are inside the system.
  • Mechanical energy is K + U. It is constant in a closed system with no external work and no internal nonconservative forces.
  • Conservation of energy: total energy never changes; it moves across the boundary as work or changes form inside.
  • Kinetic friction turns mechanical energy into thermal energy; only mechanical energy is calculated, but the loss equals f_k d.
  • An energy bar chart shows each energy at two moments; the bars must add up (with any work) and match the situation.

Part 7 · Misconception

A common mistake

The wrong idea: Mechanical energy is conserved because energy is always conserved.

What actually happens: Total energy is always conserved. Mechanical energy K + U stays constant only when no external force does work on the system and no friction acts inside it. A justification must state that condition.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

A block on a frictionless track

A 2.0 kg block on a frictionless track, released from rest at AUg = 0 at BABC2.0 m6.0 mv

The block has a mass of 2.0 kg. Use g = 9.8 m/s². Air resistance is negligible.

1. How fast is the block moving at B, in m/s?

Type a number in m/s.

Show the answer

System: block + track + Earth, no friction, so K + U_g is constant: mg(6.0 m) = ½mv_B². v_B = √(2 × 9.8 m/s² × 6.0 m) = 11 m/s.

  • Answer: 11 m/s

2. How fast is the block moving at C, in m/s?

Type a number in m/s.

Show the answer

mg(6.0 m) = mg(2.0 m) + ½mv_C², so v_C = √(2g(4.0 m)) = √(78.4) m/s = 8.9 m/s.

  • Answer: 8.9 m/s

3. Which describes an energy bar chart for the block-track-Earth system at C, with the zero of U_g at B?

  1. K is twice U_g, and K + U_g equals the U_g bar at A
  2. K and U_g are equal, each one half of the total energy
  3. U_g is twice K, because C is high on the track
  4. K is twice U_g, and the total is less than at A
Show the answer

Total = mg(6.0 m). At C: U_g = mg(2.0 m) = ⅓ of the total and K = ⅔ of it. The bars must add to the same total at every point.

  • Correct: K is twice U_g, and K + U_g equals the U_g bar at A: Right: at C, U_g = mg(2.0 m) and K = mg(4.0 m), adding to mg(6.0 m).
  • K and U_g are equal, each one half of the total energy: Bars are not split evenly by default; the heights decide. C is at one third of A's height.
  • U_g is twice K, because C is high on the track: C is 2.0 m up out of 6.0 m, so U_g is one third of the total, not two thirds.
  • K is twice U_g, and the total is less than at A: With no friction the total stays the same as at A.

4. Now take the block alone as the system. Which statement is correct as it slides from A to B?

  1. The block alone has gravitational potential energy that turns into kinetic energy
  2. The block's energy is constant, because the track is frictionless and nothing slows it
  3. It has just kinetic energy, and Earth's pull, now external, does positive work on it
  4. The normal force from the track does the work that speeds the block up
Show the answer

Choosing the system decides the bookkeeping. Block alone: only K, and W_grav = +mg(6.0 m) = ΔK. Block + Earth: U_g changes into K with no external work. Both give the same speed.

  • The block alone has gravitational potential energy that turns into kinetic energy: Potential energy belongs to a system of interacting objects. With Earth outside, there is no U_g.
  • The block's energy is constant, because the track is frictionless and nothing slows it: Earth is outside the system and does positive work, so the block's energy (all kinetic) increases.
  • Correct: It has just kinetic energy, and Earth's pull, now external, does positive work on it: Right: a one-object system has no potential energy, so gravity's work is an energy transfer in.
  • The normal force from the track does the work that speeds the block up: The normal force is perpendicular to the track, so it does no work.

5. A block slides down a frictionless ramp. Which is the best justification that the mechanical energy of the block-Earth system stays constant?

  1. Mechanical energy is constant because energy is conserved
  2. The block's speed stays constant as it slides down the ramp
  3. Gravity does no work on the block, since the ramp supports it
  4. No friction acts, and the normal force is perpendicular to the motion, so it does no work
Show the answer

K + U_g stays constant when no external force does net work on the system and no nonconservative force acts inside it. State that condition, not just the conservation law.

  • Mechanical energy is constant because energy is conserved: This is circular. Total energy is always conserved; mechanical energy is constant only under a condition, which must be stated.
  • The block's speed stays constant as it slides down the ramp: The block speeds up; its kinetic energy grows as U_g falls.
  • Gravity does no work on the block, since the ramp supports it: Gravity does positive work. Inside the block-Earth system that appears as a drop in U_g.
  • Correct: No friction acts, and the normal force is perpendicular to the motion, so it does no work: Right: it names the condition (no external work, no internal nonconservative force).

6. A roller coaster car passes over a 30.0 m high hill at 4.00 m/s. Friction is negligible. How fast is it moving when it is 10.0 m above the ground? Use g = 9.8 m/s².

Type a number in m/s.

Show the answer

½mv_i² + mg(30.0 m) = ½mv_f² + mg(10.0 m), so v_f = √(v_i² + 2g(20.0 m)) = √(16.0 + 392) m/s = 20.2 m/s.

  • Answer: 20.2 m/s

Part 9 · Summary

Summary

Pick the system first. Inside it energy changes form; across its boundary it moves as work by external forces. With no external work and no friction, K + U is constant. Friction turns mechanical energy into thermal energy, and energy bar charts must add up to match the equation.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections