D2.3 Solve problems using conservation of energy

Ontario Grade 11 Physics

Energy and Society

SPH3U D2.3 | Tracking energy changes in a physical system

Energy is a scalar: it has a size but no direction. Motion quantities such as velocity are vectors: they have both size and direction. This distinction matters in energy problems. A velocity can point upward or downward, but kinetic energy is never negative. Conservation of energy says that the total energy of a defined system stays constant when energy is not transferred into or out of that system. Energy can change form within the system. This lesson uses that idea to solve problems about motion, height, and friction.

What you will learn

Identify a physical system and choose a clear reference level for gravitational potential energy.
Use conservation of energy to relate an object's initial and final states.
Calculate kinetic energy and gravitational potential energy using SI units.
Include energy transferred to thermal energy when friction acts within the chosen system.
Check the units and reasonableness of a calculated result.

1. Prerequisite bridge: system, direction, and energy

A system is the object or group of objects you choose to study. State what belongs to it before writing an energy equation. For example, a falling ball alone is a different system from a ball together with Earth. If Earth is included, gravitational potential energy can be counted as energy stored in the system.

A reference frame is the viewpoint used to describe position and motion. In these examples, use a frame fixed to the ground. Choose a vertical coordinate direction before using height. We will take upward as positive, so an object's vertical velocity is positive while it moves upward and negative while it moves downward. Energy is scalar, so its value does not become negative just because motion is downward.

The SI unit of energy is the joule, symbol J. Mass is measured in kilograms, height in metres, and speed in metres per second. Speed is the size of velocity, so it is never negative. Use speed in the kinetic-energy equation.

1\,\mathrm{J}=1\,\mathrm{kg\,m^2/s^2}

State the system and reference frame.
Choose and label a positive direction for position and velocity.
Energy is scalar; velocity is a vector.

2. The energy model

Kinetic energy is energy associated with motion. An object's kinetic energy depends on its mass and speed. Gravitational potential energy is energy associated with the position of objects in a gravitational field. Near Earth's surface, its value depends on mass and height above a chosen reference level.

The reference level is where you define gravitational potential energy to be zero. It can be chosen for convenience, but use the same level throughout one problem. A height below that level can give a negative gravitational potential-energy value. This is a result of the chosen reference, not a negative amount of total energy.

For a system with no energy transferred into or out of it, the total energy before a change equals the total energy after it. When only kinetic and gravitational potential energy are relevant, add those forms at each state. If friction converts some mechanical energy to thermal energy within the system, include that increase as well. Thermal energy is associated with the random motion of particles in the materials. This accounting does not mean energy disappears.

A practical diagram can show the chosen height reference and direction. For a falling object, draw a vertical line labelled upward positive, mark the reference height, and label the initial and final positions. The diagram helps keep height signs consistent. The energy equation itself uses scalar energies.

E_k=\frac{1}{2}mv^2,\qquad E_g=mgh

Use mass in kg, height in m, and speed in m/s.
Choose one zero-height reference and keep it fixed.
Include relevant energy forms on both sides of the conservation equation.

3. Set up and solve an energy problem

First list what is known and what is unknown. Identify the system and its initial and final states. Then select the energy forms that matter in those states. If the object starts from rest, its initial kinetic energy is zero. If it stops, its final kinetic energy is zero. These statements simplify the equation, but they do not change the conservation rule.

Use the same units before substituting. If a question gives centimetres, convert to metres; if it gives grams, convert to kilograms. Keep units in the substitution so that a unit error is easier to notice. Carry extra digits through the calculation, then round the final result to a sensible number of significant figures based on the given values.

After calculating, check whether the answer makes physical sense. A falling object that loses height should gain speed if no other energy form takes energy away from motion. If friction is present, the gain in kinetic energy can be smaller because some energy becomes thermal energy. A final answer should include a unit and, when relevant, a direction for the velocity or motion.

E_{\mathrm{initial}}=E_{\mathrm{final}}

Compare clearly identified initial and final states.
Do not treat a velocity direction as a negative kinetic energy.
Round only after calculation and check units and physical sense.

4. Friction and energy accounting

Friction can change the form of energy in a system. If the object and the surface are both part of the system, energy associated with their motion can become thermal energy in those materials. The total energy remains accounted for, even though the object's kinetic and gravitational potential energies alone may not add to the same value at the start and finish.

Use the energy terms relevant to the stated situation. For example, if an object slides and slows on a level surface, its gravitational potential energy stays the same. The decrease in kinetic energy can equal the increase in thermal energy for the object-and-surface system, provided no energy is transferred out of that system. Define the system carefully so that the energy terms match it.

Avoid adding an energy form just because it exists in principle. Include it when the problem gives information about it or when the situation requires it to account for an energy change. This keeps the model clear and the calculation focused.

E_{k,i}+E_{g,i}=E_{k,f}+E_{g,f}+\Delta E_{\mathrm{th}}

Energy can change form without being lost.
Friction can increase thermal energy in an object-and-surface system.
Choose the system so that your energy accounting is consistent.

Worked example

A ball falls from a ledge

A ball is released from rest at a height of 5.0 m above the ground. Ignore air resistance. Find its speed just before it reaches the ground. Use

g=9.8\,\mathrm{m/s^2}

.

Define the system and direction
Take the ball and Earth as the system and use a ground-fixed reference frame. Let upward be positive. Choose the ground as zero height. The ball's final velocity points downward, but the requested speed is a positive scalar.
Choose the energy relationship
There is no energy transfer out of the system in this model. Initially the ball is at rest, and finally it is at ground level, so initial gravitational potential energy becomes final kinetic energy.
$mgh_i=\frac{1}{2}mv_f^2$
Substitute and solve
The mass cancels because it appears in both energy terms. Substitute the height and gravitational field strength, then solve for the positive speed.
$v_f=\sqrt{2gh_i}=\sqrt{2(9.8\,\mathrm{m/s^2})(5.0\,\mathrm{m})}=9.9\,\mathrm{m/s}$

Answer: The ball's speed just before impact is

9.9\,\mathrm{m/s}

. Its velocity is downward, or

-9.9\,\mathrm{m/s}

with upward chosen as positive.

Check: The units inside the square root are

\mathrm{m^2/s^2}

, so the result has units of

\mathrm{m/s}

. A fall of 5.0 m gives a speed below 10 m/s, which is reasonable. The negative velocity indicates downward motion; the speed remains positive.

Worked example

A ball rises after a throw

A ball leaves a person's hand moving upward at

12\,\mathrm{m/s}

. The release point is 1.5 m above the ground. Ignore air resistance. Find the maximum height above the ground. Use

g=9.8\,\mathrm{m/s^2}

.

Define the system and direction
Use the ball and Earth as the system and a ground-fixed frame. Upward is positive, and the ground is the zero-height reference. At the highest point, the ball's instantaneous speed is zero.
Relate the two states
The ball starts with kinetic energy and gravitational potential energy at 1.5 m. At the highest point, its kinetic energy is zero and its gravitational potential energy is greater.
$\frac{1}{2}mv_i^2+mgh_i=mgh_f$
Calculate the final height
Cancel the mass and rearrange to find the height gained. Then add that gain to the release height to get height above the ground.
$h_f=h_i+\frac{v_i^2}{2g}=1.5\,\mathrm{m}+\frac{(12\,\mathrm{m/s})^2}{2(9.8\,\mathrm{m/s^2})}=8.8\,\mathrm{m}$

Answer: The ball reaches a maximum height of

8.8\,\mathrm{m}

above the ground.

Check: The squared-speed term divided by gravitational field strength has units of metres. The final height is above the release point, as expected for an upward throw. At the top, the speed is zero, not negative.

Worked example

A sliding object slows because of friction

A

2.0\,\mathrm{kg}

object slides at

6.0\,\mathrm{m/s}

on a level surface and comes to rest. The object and surface are the system. Assume no energy leaves this system. Find the increase in thermal energy.

Define the system and direction
Use the object and the surface as the system in a ground-fixed frame. Choose the object's initial direction of motion as positive. The surface is level, so its gravitational potential energy does not change. Speed is positive initially and zero at the end.
Account for the energy change
The object's initial kinetic energy becomes thermal energy in the object and surface. The final kinetic energy is zero, and the thermal-energy increase is positive.
$\frac{1}{2}mv_i^2=\Delta E_{\mathrm{th}}$
Substitute the given values
Use mass in kilograms and speed in metres per second. The calculated energy is the increase in thermal energy, not a direction-dependent quantity.
$\Delta E_{\mathrm{th}}=\frac{1}{2}(2.0\,\mathrm{kg})(6.0\,\mathrm{m/s})^2=36\,\mathrm{J}$

Answer: The thermal energy of the object-and-surface system increases by

36\,\mathrm{J}

.

Check: The units reduce to joules. The initial kinetic energy is

36\,\mathrm{J}

and the object ends at rest, so assigning that energy increase to thermal energy is consistent with the stated system and assumptions.

Common mistakes and how to avoid them

Using a negative value for kinetic energy because the object moves downward.

Correction: Kinetic energy depends on speed squared and is non-negative. Use direction for velocity, not for kinetic energy.

Changing the zero-height reference halfway through a problem.

Correction: Choose one reference level at the start and use it for both states.

Leaving thermal energy out when friction slows an object.

Correction: If the chosen system includes the object and surface, include the thermal-energy increase when it accounts for the decrease in mechanical energy.

Reporting a numerical answer without units or a reasonableness check.

Correction: Include the SI unit and check whether the result fits the direction and energy changes in the situation.

Lesson summary

Define the system, reference frame, positive direction, and height reference.
Use kinetic energy for motion and gravitational potential energy for height near Earth's surface.
For an isolated system, the initial total energy equals the final total energy.
Include thermal energy when friction changes energy form within the chosen system.
Check units, significant figures, direction where relevant, and physical reasonableness.

Check your understanding

Question 1

A cart moves down a ramp. Which statement about its kinetic energy is correct?

It is negative because the cart moves downward.
It is a scalar and cannot be negative.
It has the same direction as the cart's velocity.
correctIndex: 1

Show answer and explanation

It is a scalar and cannot be negative.

Kinetic energy is based on speed squared, so it is a scalar and is non-negative. Velocity, unlike energy, includes a direction.

Question 2

A

1.0\,\mathrm{kg}

object starts from rest and falls

2.0\,\mathrm{m}

. Ignore air resistance. Approximately how fast is it moving just before reaching the lower point?

$2.0\,\mathrm{m/s}$
$6.3\,\mathrm{m/s}$
$19.6\,\mathrm{m/s}$
correctIndex: 1

Show answer and explanation

6.3\,\mathrm{m/s}

Conservation gives

mgh=\frac{1}{2}mv^2

, so

v=\sqrt{2gh}=\sqrt{2(9.8)(2.0)}\approx6.3\,\mathrm{m/s}

. The units are metres per second.

Question 3

A block slides on a level surface and stops. If the block and surface are the system and no energy leaves, where does the block's initial kinetic energy go?

It is destroyed by friction.
It becomes gravitational potential energy because the block stopped.
It increases thermal energy in the system.
correctIndex: 2

Show answer and explanation

It increases thermal energy in the system.

Energy is not destroyed. For the stated system, friction changes some of the initial kinetic energy into thermal energy.

Key terms

System: The object or group of objects chosen for an energy analysis.
Reference frame: The viewpoint used to describe an object's position and motion.
Scalar: A quantity with magnitude but no direction.
Vector: A quantity with magnitude and direction.
Kinetic energy: Energy associated with an object's motion.
Gravitational potential energy: Energy associated with position in a gravitational field, measured relative to a chosen height.
Thermal energy: Energy associated with the random motion of particles in materials.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation D2.3. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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D2.3 · Solve problems using conservation of energy

Ontario Grade 11 Physics

SPH3U D2.3 | Tracking energy changes in a physical system

What you will learn

1. Prerequisite bridge: system, direction, and energy

2. The energy model

3. Set up and solve an energy problem

4. Friction and energy accounting

A ball falls from a ledge

A ball rises after a throw

A sliding object slows because of friction

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH3U

About this lesson and its review