Lift something up and you store energy in it, gravitational potential energy, just waiting to be released. Let it go and that stored energy becomes kinetic energy, the energy of motion. Two simple formulas, $\Delta U = mg\Delta h$ and $KE = \tfrac{1}{2}mv^2$, let you track energy as it flows between height and speed, the foundation for conservation of energy in the next lesson.
You lift a $2\ \text{kg}$ book from the floor onto a shelf $2\ \text{m}$ high. Then you lift it again from a table $1\ \text{m}$ high to the same shelf.
Before reading: in which lift does the book gain more gravitational potential energy, from the floor, from the table, or the same amount? Write your prediction and one sentence of reasoning.
Lifting stores energy; only the height change counts
Raising an object stores energy in it, the higher you lift, the more energy is banked, ready to be released if it falls.
When you raise an object of mass $m$ through a height $\Delta h$ in a uniform gravitational field, you do work against gravity, and that energy is stored as gravitational potential energy. The change is given on the NESA formula sheet by:
$$\Delta U = mg\Delta h$$
Only the vertical change in height $\Delta h$ matters, not the path taken or the choice of "ground level". Lifting through the same height always changes $U$ by the same amount, whether straight up or along a ramp.
Gravitational potential energy change is $\Delta U = mg\Delta h$, where $\Delta h$ is the vertical change in height and $g = 9.8\ \text{m s}^{-2}$ near Earth's surface. Only the change in height matters, the path and the chosen zero level do not affect $\Delta U$.
Pause, copy the highlighted formula into your book before moving on.
Did you get this? A $3.0\ \text{kg}$ mass is raised $4.0\ \text{m}$. Using $g = 9.8\ \text{m s}^{-2}$, its gain in gravitational potential energy is…
The energy of motion grows with the square of speed
We just saw how height stores potential energy. That raises a question: what happens to that energy when the object is released and starts to move? This card answers it → it becomes kinetic energy, $KE = \tfrac{1}{2}mv^2$.
A moving object carries energy, and because the speed is squared in the formula, going faster increases that energy dramatically.
Kinetic energy is the energy of motion, given on the NESA formula sheet by:
$$KE = \tfrac{1}{2}mv^2$$
The $v^2$ dependence is crucial: doubling the speed quadruples the kinetic energy. When potential energy is released, it can become kinetic energy, so a falling object speeds up as $U$ falls and $KE$ rises.
A $0.50\ \text{kg}$ ball is dropped from a height of $1.8\ \text{m}$. Ignoring air resistance, use energy to find its speed just before it hits the ground. Take $g = 9.8\ \text{m s}^{-2}$.
Kinetic energy is $KE = \tfrac{1}{2}mv^2$, growing with the square of speed. For a dropped object, lost PE becomes KE: $mg\Delta h = \tfrac{1}{2}mv^2$, so $v = \sqrt{2g\Delta h}$. Here $v = \sqrt{2 \times 9.8 \times 1.8} \approx 5.9\ \text{m s}^{-1}$, independent of mass.
Add the highlighted formula and method to your notes before the check below.
Quick check: A car's speed doubles from $15\ \text{m s}^{-1}$ to $30\ \text{m s}^{-1}$. Its kinetic energy becomes…
Pattern A, Calculate then reason
A $4.0\ \text{kg}$ rock falls from a cliff $20\ \text{m}$ high. Take $g = 9.8\ \text{m s}^{-2}$ and ignore air resistance. In your book: (1) calculate the gravitational potential energy lost; (2) assuming it all becomes kinetic energy, find the rock's speed at the bottom using $KE = \tfrac{1}{2}mv^2$; (3) explain in one sentence why your impact speed would be the same for a $40\ \text{kg}$ rock falling the same height.
Odd one out: three of these leave the kinetic energy UNCHANGED. Which one does not?
Dot point: Solve problems involving kinetic energy using KE = ½mv²
Why a falling object speeds up and a rising one slows down
We just calculated both kinds of energy. That raises a question: how do they relate as an object moves up and down? This card answers it → potential and kinetic energy trade places, with the total staying the same when only gravity acts.
Height and speed are two forms of the same energy account: as one goes down, the other goes up.
When an object falls, it loses height, so its gravitational potential energy decreases ($\Delta U$ is negative). With only gravity acting, that energy reappears as kinetic energy, the object speeds up. When an object rises, the reverse happens: kinetic energy is converted into potential energy and the object slows down. The bookkeeping is exact: the kinetic energy gained equals the potential energy lost. This is why the impact speed depends only on the drop height (the mass cancels), and it sets up the principle of conservation of mechanical energy in the next lesson.
Gravitational potential energy and kinetic energy convert into each other: a falling object loses $U$ and gains $KE$ (speeds up); a rising object loses $KE$ and gains $U$ (slows down). With only gravity acting, the kinetic energy gained equals the potential energy lost.
Pause, copy the highlighted idea into your book before the check below.
Last check: Decide if the statement is true or false.
As an object falls freely, its kinetic energy increases by the same amount that its gravitational potential energy decreases.
Match each change with its effect on the gravitational potential energy (constant $g$):
Dot point: Solve problems involving changes in gravitational potential energy in a uniform gravitational field using ΔU = mgΔh
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A $0.30\ \text{kg}$ ball is thrown so that it rises and momentarily stops at a height of $5.0\ \text{m}$. Using $g = 9.8\ \text{m s}^{-2}$ and ignoring air resistance, calculate (a) the gravitational potential energy gained at the top, and (b) the kinetic energy it had at launch.
1 mark: $\Delta U = mg\Delta h = 0.30 \times 9.8 \times 5.0$ · 1 mark: $\Delta U = 14.7\ \text{J}$ · 1 mark: states KE at launch equals PE gained (energy conserved) · 1 mark: $KE = 14.7\ \text{J} \approx 15\ \text{J}$
ApplyBand 5(4 marks) 2. A $0.60\ \text{kg}$ ball is raised $2.5\ \text{m}$ and then dropped. Taking $g = 9.8\ \text{m s}^{-2}$ and ignoring air resistance, calculate the gain in gravitational potential energy and the speed of the ball just before it lands.
1 mark: applies $\Delta U = mg\Delta h$ · 1 mark: $\Delta U = 14.7\ \text{J}$ · 1 mark: equates $\Delta U$ to $KE = \tfrac{1}{2}mv^2$ · 1 mark: $v = 7.0\ \text{m s}^{-1}$
Dot point: Solve problems involving changes in gravitational potential energy in a uniform gravitational field using ΔU = mgΔh
AnalyseBand 6(3 marks) 3. Explain why doubling the speed of a car quadruples its kinetic energy, and state one road-safety consequence of this relationship.
1 mark: $KE = \tfrac{1}{2}mv^2$ depends on the SQUARE of the speed · 1 mark: doubling $v$ multiplies $v^2$ by four · 1 mark: a valid consequence, such as braking distance roughly quadrupling
Dot point: Solve problems involving kinetic energy using KE = ½mv²
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): (a) The gravitational potential energy gained at the top is $\Delta U = mg\Delta h = 0.30 \times 9.8 \times 5.0 = 14.7\ \text{J}$. (b) Ignoring air resistance, all of the ball's initial kinetic energy is converted into potential energy by the time it momentarily stops at the top. Therefore the kinetic energy at launch equals the potential energy gained: $KE = 14.7\ \text{J} \approx 15\ \text{J}$.
Q2 (4 marks): Raising the ball increases its gravitational potential energy by $$\Delta U = mg\Delta h = 0.60 \times 9.8 \times 2.5 = 14.7\ \text{J}$$ Ignoring air resistance, all of this converts to kinetic energy by the time the ball reaches its original height, so $$\tfrac{1}{2}mv^2 = 14.7 ;\Rightarrow; v^2 = \frac{2 \times 14.7}{0.60} = 49$$ giving $v = 7.0\ \text{m s}^{-1}$. Note that the mass cancels if the working is done symbolically, $v = \sqrt{2g\Delta h}$, so a ball of any mass dropped from $2.5\ \text{m}$ lands at the same speed.
Q3 (3 marks): Kinetic energy is $KE = \tfrac{1}{2}mv^2$, so it depends on the square of the speed rather than on the speed itself. Doubling $v$ multiplies $v^2$ by four, and since the mass and the factor of one half are unchanged, the kinetic energy is four times larger. The road-safety consequence follows from the work-energy theorem: to stop, the brakes must do work equal to the whole kinetic energy, and if the braking force is roughly constant then $Fs = \tfrac{1}{2}mv^2$ means the stopping distance is proportional to $v^2$ as well. A car travelling at $80\ \text{km h}^{-1}$ therefore needs roughly four times the braking distance of one at $40\ \text{km h}^{-1}$, not twice, which is why small increases in speed produce disproportionately worse crashes.
You can now calculate gravitational potential energy changes with $\Delta U = mg\Delta h$ and kinetic energy with $KE = \tfrac{1}{2}mv^2$, and convert one into the other. Next lesson formalises this as conservation of mechanical energy.
↩ Back to focus-area hubAt the start you predicted which lift gives the book more potential energy. The answer: lifting from the floor ($2\ \text{m}$) gives more than from the table ($1\ \text{m}$), because $\Delta U = mg\Delta h$ depends on the change in height, not the final shelf height. If you said they were the same, revisit Card 1.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
Tell us what you would like help with. Register interest; support details will be confirmed before enrolment.
Register interest →