When only gravity does work, an object's total mechanical energy, kinetic plus potential, stays exactly the same. Energy just shifts between the two forms. This single law lets you connect the speed at one point of a motion to the height at another, without ever touching forces or time. It is one of the most powerful problem-solving tools in all of physics.
A pendulum is released from the side and swings down and back up to the other side.
Before reading: ignoring air resistance, does the pendulum rise higher than, lower than, or exactly to the same height it was released from? Write your prediction and one sentence of reasoning.
When only conservative forces act, $KE + U$ stays constant
When the only force doing work is gravity, the total of kinetic and potential energy never changes, it just shifts back and forth between the two forms.
The total mechanical energy is $E = KE + U$. The law of conservation of mechanical energy states that, when only conservative forces (such as gravity) do work, this total stays constant:
$$KE_i + U_i = KE_f + U_f \qquad (\text{no non-conservative forces})$$
The key condition is that no non-conservative force (like friction, air resistance, or an applied push) does any net work. If that holds, energy lost from one form (e.g. potential) appears exactly as gain in the other (kinetic). This is why a frictionless pendulum returns to the same height, its energy simply cycles between potential and kinetic.
Total mechanical energy $E = KE + U$ is conserved when only conservative forces (e.g. gravity) do work: $KE_i + U_i = KE_f + U_f$. Energy lost as potential energy reappears as kinetic energy and vice versa, so the sum stays constant.
Pause, copy the highlighted law and condition into your book before moving on.
Did you get this? True or false: mechanical energy is conserved only when no non-conservative force does work on the object.
Connect a height at one point to a speed at another
We just stated the law $KE_i + U_i = KE_f + U_f$. That raises a question: how does it actually solve a motion problem? This card answers it → equate the energy at the start and end, and solve for the unknown speed or height.
Energy conservation lets you jump straight from a starting height to a finishing speed, without working through forces or time.
To solve a conservation problem: write the total energy at the start ($KE_i + U_i$), set it equal to the total at the end ($KE_f + U_f$), and solve for the unknown. Often potential energy converts entirely into kinetic (or back), so the equation simplifies neatly.
A $30\ \text{kg}$ child starts from rest at the top of a smooth (frictionless) slide $2.5\ \text{m}$ high. Using $g = 9.8\ \text{m s}^{-2}$, find the child's speed at the bottom.
To solve with conservation: $KE_i + U_i = KE_f + U_f$. For a frictionless slide from rest, $mgh = \tfrac{1}{2}mv^2$, so $v = \sqrt{2gh}$, independent of mass. Here $v = \sqrt{2 \times 9.8 \times 2.5} = 7.0\ \text{m s}^{-1}$.
Add the highlighted method to your notes before the check below.
Quick check: A ball is dropped from $5.0\ \text{m}$ on a frictionless path. Using $g = 9.8\ \text{m s}^{-2}$ and energy conservation, its speed at the bottom is closest to…
Pattern B, Calculate then explain
A $2.0\ \text{kg}$ trolley is released from rest at the top of a $3.0\ \text{m}$-high ramp. Take $g = 9.8\ \text{m s}^{-2}$. In your book: (1) calculate the speed at the bottom assuming the ramp is frictionless; (2) suppose it actually arrives at $6.0\ \text{m s}^{-1}$, calculate how much mechanical energy was "lost" and identify which force did this work; (3) explain in one sentence where that energy went and why total energy is still conserved.
Odd one out: three of these are conditions under which total mechanical energy is conserved. Which one is not?
Dot point: State the conditions under which total mechanical energy is conserved
Mechanical energy falls, but total energy is conserved
We just solved frictionless problems. That raises a question: what happens when friction is present and the speed comes out lower? This card answers it → the missing mechanical energy has become heat, and which force did that work tells the story.
When a real object ends up slower than energy conservation predicts, the missing energy isn't gone, friction has quietly turned it into heat.
When a non-conservative force such as friction acts, the mechanical energy ($KE + U$) at the end is less than at the start. The difference is the work done by friction, which has been transformed into heat and sound. The total energy of everything (including the warmed surfaces) is still conserved, only the mechanical energy has decreased. To analyse such a case, identify which force does the work: gravity converts $U$ into $KE$ (conservative), while friction removes mechanical energy as heat (non-conservative). The energy balance becomes:
$$(KE_i + U_i) = (KE_f + U_f) + (\text{energy lost to friction})$$
This is why the skateboarder returns to almost the starting height: a small amount of energy is dissipated by friction and air resistance each cycle.
When friction acts, mechanical energy ($KE + U$) decreases, but total energy is conserved, the missing energy becomes heat and sound. Energy balance: $(KE_i + U_i) = (KE_f + U_f) + \text{energy lost to friction}$. Gravity (conservative) converts $U$ to $KE$; friction (non-conservative) dissipates energy.
Pause, copy the highlighted balance into your book before the check below.
Last check: Decide if the statement is true or false.
When friction slows a sliding block, the energy it removes is permanently destroyed.
A ball is thrown up and falls back. Match each stage with the energy transformation:
Dot point: Analyse energy transformations and calculate the work done when an object moves within a uniform gravitational field, including changes in gravitational potential energy due to height and identifying the forces responsible for doing work
Watch kinetic and gravitational potential energy trade as an object descends, with and without friction
Use the simulation. With friction off, set $h = 30\ \text{m}$ and $m = 2.0\ \text{kg}$, then move the height slider to $h = 15\ \text{m}$. What is the kinetic energy there?
Use the simulation. Switch on friction. True or false: the dashed total mechanical energy line stays flat, because energy is conserved.
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
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AnalyseBand 6(5 marks) 1. A $0.20\ \text{kg}$ ball is released from rest at the top of a curved ramp $1.5\ \text{m}$ high. It reaches the bottom moving at $4.0\ \text{m s}^{-1}$. Using $g = 9.8\ \text{m s}^{-2}$: (a) calculate the speed expected if the ramp were frictionless; (b) calculate the mechanical energy dissipated by friction; (c) state what happened to that energy.
1 mark: $v_{ideal} = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.5} \approx 5.4\ \text{m s}^{-1}$ · 1 mark: $U_i = mgh = 0.20 \times 9.8 \times 1.5 = 2.94\ \text{J}$ · 1 mark: actual $KE_f = \tfrac{1}{2}(0.20)(4.0)^2 = 1.6\ \text{J}$ · 1 mark: energy lost $= 2.94 - 1.6 = 1.3\ \text{J}$ · 1 mark: states it became heat/sound (friction), total energy still conserved
ApplyBand 6(4 marks) 2. A $0.25\ \text{kg}$ ball is released from rest at the top of a smooth $1.8\ \text{m}$ ramp. Taking $g = 9.8\ \text{m s}^{-2}$, use conservation of mechanical energy to find its speed at the bottom, and state why the ramp angle does not appear in the calculation.
1 mark: states the condition, only conservative forces act · 1 mark: sets $mg\Delta h = \tfrac{1}{2}mv^2$ · 1 mark: $v = 5.9\ \text{m s}^{-1}$ · 1 mark: explains gravity is conservative, so only the height change matters, not the path
Dot point: Apply the law of conservation of mechanical energy and use KE and U to solve problems
AnalyseBand 6(3 marks) 3. A $2.0\ \text{kg}$ box slides down a rough ramp of vertical height $3.0\ \text{m}$ and reaches the bottom at $6.0\ \text{m s}^{-1}$. Taking $g = 9.8\ \text{m s}^{-2}$, determine the energy dissipated by friction and identify the force responsible.
1 mark: $\Delta U = 58.8\ \text{J}$ · 1 mark: $KE = 36\ \text{J}$ · 1 mark: energy dissipated $= 22.8\ \text{J}$, by friction, a non-conservative force
Dot point: Analyse energy transformations and calculate the work done when an object moves within a uniform gravitational field, including changes in gravitational potential energy due to height and identifying the forces responsible for doing work
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 (5 marks): (a) If the ramp were frictionless, all potential energy would become kinetic: $v_{ideal} = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.5} = \sqrt{29.4} \approx 5.4\ \text{m s}^{-1}$. (b) The initial potential energy is $U_i = mgh = 0.20 \times 9.8 \times 1.5 = 2.94\ \text{J}$, and the actual kinetic energy at the bottom is $KE_f = \tfrac{1}{2}(0.20)(4.0)^2 = 1.6\ \text{J}$. The mechanical energy dissipated is $2.94 - 1.6 = 1.3\ \text{J}$. (c) That $1.3\ \text{J}$ was transformed by friction (and a little air resistance) into heat and sound. The mechanical energy decreased, but total energy is still conserved.
Q2 (4 marks): The ramp is smooth, so no non-conservative force does work and the total mechanical energy is conserved. All the gravitational potential energy lost becomes kinetic energy: $$mg\Delta h = \tfrac{1}{2}mv^2$$ The mass cancels, leaving $$v = \sqrt{2g\Delta h} = \sqrt{2 \times 9.8 \times 1.8} = \sqrt{35.28} = 5.9\ \text{m s}^{-1}$$ The ramp angle does not appear because gravity is a conservative force: the work it does depends only on the change in height, not on the path taken to achieve it. A steep ramp and a gentle one of the same vertical drop deliver the same final speed, though the gentle one takes longer. Introducing friction would break this, since the work done against friction does depend on the path length.
Q3 (3 marks): The gravitational potential energy released is $$\Delta U = mg\Delta h = 2.0 \times 9.8 \times 3.0 = 58.8\ \text{J}$$ The kinetic energy actually gained is $$KE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(2.0)(6.0)^2 = 36\ \text{J}$$ The shortfall is $$58.8 - 36 = 22.8\ \text{J}$$ This energy has not been destroyed; it has been transferred out of the mechanical store by friction between the box and the ramp, appearing as heat in both surfaces and a little sound. Friction is a non-conservative force, so the work it does depends on the path length and none of that energy returns to the box. Total energy is still conserved; it is total MECHANICAL energy that is not.
You can now state when mechanical energy is conserved, apply $KE_i + U_i = KE_f + U_f$, and account for energy "lost" to friction. Next you turn to momentum, the other great conserved quantity of mechanics.
↩ Back to focus-area hubAt the start you predicted how high the pendulum swings back. The answer: ignoring air resistance, it returns to exactly the same height, because mechanical energy is conserved, all the kinetic energy at the bottom converts back into potential energy. In reality it rises slightly lower because friction and air resistance dissipate a little energy as heat. If you said it rises higher, revisit Card 1.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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