Quantify kinetic and gravitational potential energy and use conservation of energy to find speeds and heights

Learning goal: Calculate kinetic energy from mass and speed and gravitational potential energy from mass, gravitational field strength and height above a chosen reference; explain how the potential energy of an object depends on where the reference point is placed and why only changes in energy can be measured; and use conservation of energy to find an unknown speed, height or mass, including for a falling object and a pendulum.

From words to numbers. Middle school established that energy is conserved and that kinetic energy (energy of motion) and potential energy (stored energy of position) can change into one another. This course makes those ideas quantitative. Energy is a scalar: it has a size but no direction, so energies add as plain numbers. Its unit is the joule (J). One joule is a small amount: lifting a small apple about a meter stores roughly one joule. Two equations carry the whole lesson, and each has conditions attached.

Kinetic energy. A moving object of mass m (in kilograms) and speed v (in meters per second) has kinetic energy Eₖ = ½ m v². The speed is squared, so doubling the speed quadruples the kinetic energy while doubling the mass only doubles it; that asymmetry is why a car at 60 km/h has four times the kinetic energy it has at 30 km/h, not twice. Units check: kg × (m/s)² = kg·m²/s², which is the definition of a joule. Speed is measured, not guessed. For an object moving at constant speed, divide the distance covered by the time taken; a meter stick and a stopwatch, or a video stepped frame by frame with a known frame rate, are enough. Model: a 0.50 kg puck slides 3.0 m across smooth ice in 1.5 s at constant speed, so v = 3.0 ÷ 1.5 = 2.0 m/s and Eₖ = ½ × 0.50 × 2.0² = ½ × 0.50 × 4.0 = 1.0 J. (A sliding puck is chosen on purpose: a rolling ball also has kinetic energy in its spin, which ½ m v² does not count.)

Gravitational potential energy. An object of mass m at height h above a reference level has gravitational potential energy E_g = m g h, where g is the gravitational field strength, 9.8 N/kg at Earth's surface (the same number as the free-fall acceleration 9.8 m/s²). Units: kg × N/kg × m = N·m = J. A 2.0 kg bag on a shelf 1.5 m above the floor has E_g = 2.0 × 9.8 × 1.5 = 29.4 J relative to the floor. Strictly, this energy belongs to the object and Earth together, because it comes from the gravitational pull between them; "the object's potential energy" is the usual shorthand, and this course uses it. The equation holds for heights small compared with Earth's radius, which covers every classroom and roller coaster.

The reference point is a choice. Potential energy is measured relative to a level chosen to be zero, and nothing physical fixes that level. The bag on the shelf has 29.4 J relative to the floor, 0 J relative to the shelf itself, and, if the reference is a cupboard top 0.50 m higher than the shelf, a negative value, 2.0 × 9.8 × (−0.50) = −9.8 J, because it is below the reference. None of these numbers is "the" energy of the bag. What is the same in every case is the change in potential energy between two positions: lowering the bag 1.5 m to the floor releases 29.4 J whichever reference was chosen, because the drop is 1.5 m in all of them. Only changes in energy can be measured absolutely, and a good choice of reference, usually the lowest point in the problem, makes the arithmetic simplest.

Conservation of energy as a tool. In a system where friction and air resistance are negligible and nothing else (such as a spring, a motor or a push) adds or removes energy, the total of kinetic plus gravitational potential energy stays constant as an object moves. Written for two moments, Eₖ(start) + E_g(start) = Eₖ(end) + E_g(end). Choosing the reference at the lowest point removes one term; releasing from rest removes another. For an object dropped from rest at height h and landing at the reference level: m g h = ½ m v², and the mass cancels, so v² = 2 g h. Conversely, a measured impact speed gives the drop height, h = v² ÷ (2 g). The cancellation of mass is a real prediction: with air resistance negligible, a heavy and a light ball dropped from the same height reach the same speed. The equations hold as long as no energy leaves the system as thermal energy; when it does, the measured speed is lower than the calculated one, and the difference is the energy transferred out, which is the next lesson.

A fully worked model. A 3.0 kg object is released from rest 4.0 m above the ground. Take the ground as reference. Start: Eₖ = 0, E_g = 3.0 × 9.8 × 4.0 = 117.6 J, total 117.6 J. At the ground: E_g = 0, so Eₖ = 117.6 J. Then ½ × 3.0 × v² = 117.6, v² = 117.6 × 2 ÷ 3.0 = 78.4, v = √78.4 ≈ 8.9 m/s. Halfway down, at 2.0 m: E_g = 3.0 × 9.8 × 2.0 = 58.8 J, so Eₖ = 117.6 − 58.8 = 58.8 J and v = √(2 × 58.8 ÷ 3.0) = √39.2 ≈ 6.3 m/s, not half of 8.9, because kinetic energy depends on speed squared. The diagram plots both energies against height for a 2.0 kg object released from 4.0 m: E_g falls in a straight line from 78.4 J to 0 while Eₖ rises from 0 to 78.4 J, and the two always sum to 78.4 J.

Investigation: does a pendulum keep its energy? Materials: a 0.20 kg mass on a 1.0 m string hung from a clamp, a meter stick, a photogate or a phone camera with a known frame rate, a card of known width taped to the bob, goggles. Steps: with the bob hanging at rest, mark the height of its bottom edge as the reference; pull it aside until the bob is 0.10 m higher than the reference and measure that height with the stick; release from rest and measure the speed at the bottom of the swing (the photogate gives card width ÷ blocked time; a video gives distance moved between frames ÷ frame interval); repeat for release heights of 0.05 m and 0.15 m, three trials each. Record: release height, predicted speed from √(2 g h) and measured speed for each trial. For 0.10 m the prediction is √(2 × 9.8 × 0.10) = √1.96 = 1.4 m/s, and E_g at release is 0.20 × 9.8 × 0.10 = 0.196 J. A result that supports the claim: the measured speeds are close to the predicted ones and rise with release height in the predicted way, showing that E_g at release became Eₖ at the bottom with little loss. Safety: goggles; keep the swing path clear; clamp the stand or hold its base. What the model cannot show: a measured speed a little below the prediction is expected, because some energy goes to the air and the string; the experiment cannot separate those losses, and it says nothing about energy forms other than the two it tracks.

Worked example 1 — kinetic energy from measured speed. A 1.5 kg cart travels 3.0 m along a level track in 1.5 s at constant speed. Speed v = 3.0 m ÷ 1.5 s = 2.0 m/s. Eₖ = ½ × 1.5 kg × (2.0 m/s)² = ½ × 1.5 × 4.0 = 3.0 J. If the same cart moved at 4.0 m/s, Eₖ would be ½ × 1.5 × 16 = 12 J, four times as much for twice the speed.

Worked example 2 — potential energy and the reference point. A 1.2 kg book rests on a table 0.80 m above the floor. Relative to the floor, E_g = 1.2 × 9.8 × 0.80 = 9.408 J, about 9.4 J. Relative to the table top, h = 0 and E_g = 0 J. Relative to a shelf 0.50 m above the table, h = −0.50 m and E_g = 1.2 × 9.8 × (−0.50) = −5.88 J, about −5.9 J. Moving the book from the table to the floor changes E_g by −9.4 J in every one of these frames, because the change in height is 0.80 m downward in all of them.

Worked example 3 — drop height from impact speed. A ball hits the ground at 7.0 m/s after being released from rest, with air resistance negligible. Conservation: m g h = ½ m v², so h = v² ÷ (2 g) = 7.0² ÷ (2 × 9.8) = 49 ÷ 19.6 = 2.5 m. The mass was not needed and would have canceled; the result holds for any mass under the stated condition.

Worked example 4 — an unknown mass. A falling object arrives at the reference level with 117.6 J of kinetic energy after a 4.0 m drop from rest. Conservation gives m g h = 117.6, so m = 117.6 ÷ (9.8 × 4.0) = 117.6 ÷ 39.2 = 3.0 kg. The same equation can be solved for any one of mass, height or speed when the others are known, provided the energy stayed in the system.

Representation task: an energy ledger. For a 2.0 kg object released from rest at 4.0 m above the ground, make a table with rows for heights 4.0, 3.0, 2.0, 1.0 and 0 m and columns for E_g, Eₖ, total energy and speed. Fill every cell with a calculation shown (E_g = 2.0 × 9.8 × h; Eₖ = 78.4 − E_g; v = √(2 Eₖ ÷ 2.0)). Beneath the table, sketch the two energy lines against height and write two sentences: one explaining why the total column is constant and one explaining why the speed at 2.0 m is not half the speed at 0 m.

Independent performance: predict, measure, compare. Roll a low-friction cart with light wheels (not a ball: a rolling ball keeps part of its kinetic energy as spin, so it moves more slowly at the bottom even if nothing is dissipated) down a ramp from three different release heights and measure its speed at the bottom with a photogate or frame-by-frame video. Success criteria: the reference level is stated and the release heights are measured from it; a predicted speed from √(2 g h) is shown for each height with the arithmetic; the measured speed is computed from a distance and a time you recorded, with units; a table compares predicted and measured speeds and states the difference as a percentage; a paragraph explains why the measured speed is expected to be a little lower and what the missing energy became; and a final sentence says what a change of reference level would and would not change in your calculations.