Define kinetic energy as the energy of motion (KE = 1/2 mv^2) and gravitational potential energy as the energy of position (PE = mgh), and calculate each (MA STE Introductory Physics, Energy, HS-PS3-1, HS-PS3-2).

What this topic is asking

Energy comes in two mechanical forms the Massachusetts Introductory Physics MCAS tests directly: the energy of motion and the energy of position. You must define kinetic energy as the energy an object has because it is moving, $KE = \tfrac{1}{2}mv^2$, define gravitational potential energy as the energy an object has because of its height, $PE = mgh$, and calculate both. Both formulas are on the reference sheet. The crosscutting idea is that energy at the everyday scale is the sum of energy from motion and energy from position, which is exactly these two forms.

Kinetic energy

A moving object can do work on whatever it hits, so it carries energy. The reference-sheet formula is

where $KE$ is the kinetic energy (J), $m$ is the mass (kg), and $v$ is the speed (m/s). The single most important feature, and the one the MCAS tests most, is that the speed is squared:

• Doubling the speed multiplies the kinetic energy by four ($2^2$).
• Tripling the speed multiplies it by nine ($3^2$).

This is why a car at highway speed has far more kinetic energy than the same car in town, and why stopping distances grow so fast with speed. Mass matters too, but only in proportion: doubling the mass doubles the kinetic energy.

Gravitational potential energy

Lifting an object takes work against gravity, and that work is stored as potential energy. The reference-sheet formula is

where $PE$ is the gravitational potential energy (J), $m$ is the mass (kg), $g$ is the acceleration due to gravity (taken as $10$ m/s squared), and $h$ is the height above a chosen reference level (m). The "reference level" is just where you call $h = 0$, usually the ground or the floor. Only changes in height matter, so it does not matter where you set zero as long as you are consistent.

The two forms convert

This is why the two forms are taught together. A dropped object loses height (losing potential energy) and gains speed (gaining kinetic energy); the totals trade off. The MCAS often asks you to find the speed at the bottom of a drop by setting the potential energy lost equal to the kinetic energy gained, which is the subject of conservation of energy.

Worked example

Reference-sheet note

The reference sheet prints kinetic energy as $KE = \tfrac{1}{2}mv^2$ and gravitational potential energy as $PE = mgh$, and gives $g = 10$ m/s squared. What you recall is that the speed is squared in the kinetic energy (so doubling the speed quadruples it), that only changes in height matter for potential energy, and that the two forms convert into each other.

Try this

Q1. A $4.0$ kg object moves at $5.0$ m/s. Calculate its kinetic energy. [2]

• Cue. $KE = \tfrac{1}{2}(4.0)(5.0)^2 = \tfrac{1}{2}(4.0)(25) = 50$ J.

Q2. A $10$ kg box is lifted $2.0$ m. Calculate the gravitational potential energy gained. Use $g = 10$ m/s squared. [2]

• Cue. $PE = mgh = (10)(10)(2.0) = 200$ J.