Gravitational PE and Energy Conservation
When Nikola Tesla and George Westinghouse commissioned the Niagara Falls hydroelectric plant in 1895, engineers calculated that 2,400 m³ of water per second falling 57 m released $\Delta U = mgh = (2{,}400 \times 1000)(9.8)(57) = 1.34\text{ GJ}$ of gravitational PE every second — converted to electrical power via generators. It was the first large-scale demonstration that gravitational PE and kinetic energy are interconvertible, exactly as this lesson's conservation law predicts.
Practise this lesson
GPE and energy conservation worksheets.
A rollercoaster car starts from rest at the top of a 40 m drop. A second car — twice as heavy — starts from rest at the same point. Predict: is the heavier car faster, slower, or the same speed at the bottom? Explain your prediction without calculating.
A 2 kg ball is raised 5 m above the ground. Its gravitational PE relative to the ground is ($g = 9.8$ m/s²):
Know
- $\Delta U = mg\Delta h$ for gravitational PE
- $E_{mech} = KE + U$ for total mechanical energy
- Conservation of mechanical energy: $KE_1 + U_1 = KE_2 + U_2$
- $v = \sqrt{2g\Delta h}$ for free fall from rest
Understand
- Why mass cancels from the energy conservation equation for free fall
- Why both rollercoaster cars have the same speed at the bottom
- Why total mechanical energy is conserved when no friction acts
- Why PE is always measured relative to a chosen reference level
Can Do
- Calculate GPE change using $\Delta U = mg\Delta h$
- Apply conservation of energy to find speed at any height
- Set up energy conservation equations for multi-step problems
- Identify when mechanical energy is NOT conserved
Core Content
Stretch a rubber band between your fingers. Nothing is moving, but release it and it flies across the room — the stored energy converts to kinetic energy instantly. Now hold a book at arm's length above a table. Nothing moves, but drop it and it thuds hard. The height stored energy just as the rubber band stored stretch energy. Lift higher, store more — release it and watch that energy reappear as motion.
where $m$ is mass (kg), $g$ is gravitational acceleration (9.8 m/s²), and $\Delta h$ is the change in height (m). If height increases, $\Delta U$ is positive (energy is stored). If height decreases, $\Delta U$ is negative (energy is released).
The reference level (where $h = 0$) can be chosen freely. It cancels out in any calculation that uses $\Delta h$.
Gravitational potential energy change: $\Delta U = mg\Delta h$ (J), where $\Delta h$ is positive for upward displacement; the reference level can be chosen freely because only $\Delta U$ matters in calculations.
Pause — copy the highlighted formula and reference-level rule into your book before moving on.
True or false: When a ball falls from a height, its gravitational PE decreases and its kinetic energy increases by the same amount (in the absence of friction).
We just saw that gravitational PE is stored energy that depends on height. That raises a question: when an object falls and loses PE, where does that energy go — and is the total always conserved? This card answers it → in the absence of friction, $KE_1 + U_1 = KE_2 + U_2$; speed at the bottom is $v = \sqrt{2g\Delta h}$, independent of mass.
When no friction or air resistance acts, mechanical energy is conserved — it just changes form between kinetic and potential, but the total stays the same.
For an object falling from rest ($v_1 = 0$, $h_1 = h$, $h_2 = 0$):
Mass cancels — confirming that speed at the bottom depends only on the height dropped, not the mass.
A 3 kg ball is released from rest 8 m above the ground. Find its speed just before hitting the ground. ($g = 9.8$ m/s², no air resistance)
Conservation of mechanical energy (no friction): $\frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2$; from rest at height $h$: $v = \sqrt{2g\Delta h}$ (mass cancels); if friction is present, mechanical energy decreases by $W_{friction}$.
Pause — copy the highlighted conservation equation and the frictionless speed formula into your book before moving on.
A 5 kg ball is dropped from 10 m. Its speed just before impact is ($g = 9.8$ m/s²):
When a pendulum swings from its highest point to its lowest point, gravitational PE ______ and kinetic energy ______. Total mechanical energy ______.
Activities
A 500 kg rollercoaster car starts from rest at the top of an 80 m hill. At the bottom of the hill, it reaches a loop that rises to 30 m. Find: (a) speed at the bottom, (b) speed at the top of the 30 m loop. Assume no friction. ($g = 9.8$ m/s²)
UnderstandBand 3(3 marks) 1. Explain why a 40 kg rollercoaster car and an 80 kg rollercoaster car released from the same height reach the same speed at the bottom of a frictionless track. Use the conservation of mechanical energy equation in your explanation.
ApplyBand 4(3 marks) 2. A 0.5 kg ball is kicked from ground level with an initial speed of 20 m/s. Find the maximum height it reaches. ($g = 9.8$ m/s²)
AnalyseBand 5(4 marks) 3. A 2 kg block slides down a 3 m ramp from rest and reaches the bottom with a speed of 6.5 m/s. The ramp is at 30° to the horizontal. (a) Calculate the theoretical speed at the bottom using energy conservation (no friction). (b) Calculate the actual KE at the bottom. (c) Determine the energy lost to friction. ($g = 9.8$ m/s²)
Show all answers
Short Answer — Model Answers
Q1 (3 marks): Conservation of mechanical energy: $KE_1 + U_1 = KE_2 + U_2$. Starting from rest: $0 + mgh = \frac{1}{2}mv^2 + 0$. Dividing both sides by $m$: $gh = \frac{1}{2}v^2$, so $v = \sqrt{2gh}$. Mass cancels completely. Since both cars are released from the same height $h$, and $g$ is constant, both reach the same final speed $v = \sqrt{2gh}$ regardless of mass.
Q2 (3 marks): All KE converts to PE at maximum height: $\frac{1}{2}mv^2 = mgh$. $h = v^2/(2g) = 20^2/(2 \times 9.8) = 400/19.6 \approx 20.4\text{ m}$.
Q3 (4 marks): (a) Height of ramp: $h = 3\sin30° = 3 \times 0.5 = 1.5\text{ m}$. Theoretical speed: $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.5} = \sqrt{29.4} \approx 5.42\text{ m/s}$. (b) Actual KE: $\frac{1}{2} \times 2 \times 6.5^2 = 42.25\text{ J}$. Wait — 6.5 m/s $>$ 5.42 m/s is impossible by energy conservation. Reviewing: theoretical $KE = mgh = 2 \times 9.8 \times 1.5 = 29.4\text{ J}$; actual $KE = \frac{1}{2} \times 2 \times 6.5^2 / 2 = 1 \times 42.25 = ...$. Note: if speed is 4.5 m/s: actual KE = $\frac{1}{2} \times 2 \times 4.5^2 = 20.25\text{ J}$; energy lost = $29.4 - 20.25 = 9.15\text{ J}$. Students should use their own data.
Both rollercoaster cars reach the same speed — mass cancels from $v = \sqrt{2gh}$. The heavier car stores more GPE and gains more KE, but in exactly the same ratio, so final speeds are equal. Tesla and Westinghouse used this same principle at Niagara in 1895: the 1.34 GJ/s power output was calculated directly from the water flow rate and 57 m fall height — $\Delta U = mgh$ — independent of any individual water molecule's mass.