Kinetic energy (KE) is the energy an object has because it is moving. A rolling cart has it. A falling book has it. A parked car has none, because it isn’t moving. It was introduced in Activity 4-1-1. Here is the full formula:
KE = ½ m v²
Mass m is in kilograms, speed v is in meters per second, and the answer is in joules (J), the same unit as work. That’s no accident: work is the way energy gets transferred, and kinetic energy is one place it can end up.
💡Check the units: kg × (m/s)² = kg·m²/s², and that is exactly what one joule is. Work (N × m) and kinetic energy come out in the same unit.
BMath
Where ½mv² Comes From
The formula isn’t a rule someone handed down. You can build it from things you already know. Imagine a constant net force F pushing an object of mass m from rest through a distance d. By the work-energy theorem, the work done becomes the object’s KE.
Deriving the kinetic energy formula
Given
Constant net force F on mass m, starting from rest
The force acts over distance d
Final speed v
Solve
W_net = F × d
a = F ÷ m (Newton’s second law)
v² = 2 a d (constant acceleration from rest, Unit 1)
d = v² ÷ (2a) = m v² ÷ (2F)
W_net = F × (m v² ÷ 2F) = ½ m v²
W_net = ΔKE and KE_start = 0, so KE = ½ m v²
✓ The ½ and the v² come straight from F = ma and constant acceleration.
🔑Notice the d in the third line. To reach twice the speed, the same force has to push over four times the distance (d ∝ v²). More distance means more work, so the energy grows with v², not v.
CConcept
Mass vs. Speed: The Scaling Rules
Mass appears once in the formula. Speed appears squared. So the two have very different effects when you change them:
Change
KE becomes
Why
Mass ×2, same speed
×2
Twice the matter moving at the same speed
Mass ×3, same speed
×3
KE is directly proportional to mass
Speed ×2, same mass
×4
2² = 4
Speed ×3, same mass
×9
3² = 9
Speed ×½, same mass
×¼
(½)² = ¼
Mass ×3, speed ×½
×¾
3 × ¼ = ¾
🔑Speed is the more powerful lever. Doubling an object’s speed has twice the effect of doubling its mass. That is why a baseball at 40 m/s can match the energy of a bowling ball that weighs 40 times as much.
DExample
KE Explorer
Drag the speed up and watch the dot climb. The curve gets steeper and steeper, which is what “squared” looks like. Then change the mass and notice that it only stretches the curve up, without changing its shape.
The curve shows kinetic energy for every speed from 0 to 10 m/s. Slide the mass and the speed. The orange dot sits at double your speed (when it fits on the graph).
KE = ½ × 2 × 3² = ½ × 2 × 9.00
Kinetic energy9.00 J
At double the speed (6 m/s): 36.0 J, exactly 4 times as much.
EWatch Out
Three Common Mix-Ups
⚠️“Heavier means more kinetic energy.” Only if the speeds are equal. A light object that is fast enough can beat a heavy one. Always use both numbers.
⚠️“Kinetic energy has a direction, or can be negative.” It can’t. KE uses v², so it is always zero or positive and has no direction. Two carts moving in opposite directions don’t cancel their KE. What can be negative is the change in KE (ΔKE), which tells you the object slowed down.
⚠️“Any units will work.” The joule only comes out if mass is in kilograms and speed is in meters per second. Convert grams (÷ 1000) and km/h (÷ 3.6) first, or the answer will be wrong by a large factor.
FMath
Worked Examples
Example 1: find the KE
Given
A lab cart with mass m = 0.50 kg
Speed v = 2.0 m/s
Solve
KE = ½ m v²
KE = ½ (0.50 kg)(2.0 m/s)²
KE = ½ (0.50)(4.0)
KE = 1.0 J
✓ The cart carries 1.0 J of kinetic energy.
Example 2: find the speed from the KE
Given
A 1.5 kg object has KE = 12 J
Find its speed
Solve
KE = ½ m v² so v² = 2 × KE ÷ m
v² = 2 × 12 ÷ 1.5 = 16
v = √16
v = 4.0 m/s
✓ The object moves at 4.0 m/s.
Example 3: a change in KE and the net work behind it
Given
A 0.40 kg ball speeds up from 3.0 m/s to 6.0 m/s
Solve
KE_start = ½ (0.40)(3.0)² = 1.8 J
KE_final = ½ (0.40)(6.0)² = 7.2 J
ΔKE = 7.2 − 1.8 = 5.4 J
W_net = ΔKE = 5.4 J
✓ The speed doubled, so KE went up 4 times (1.8 J to 7.2 J). Net work of +5.4 J did it.
GConcept
Kinetic Energy and Stopping
In Activity 4-1-1, the work-energy theorem said net work equals the change in KE. For a braking car, the brakes do negative work that must remove all of the car’s KE. With a constant braking force F, that gives:
F × d = ½ m v² → d = KE ÷ F
Here are the stopping distances for a 1,000 kg car with 8,000 N of braking force at different speeds:
Starting speed
KE to remove
Braking distance
10 m/s (about 22 mph)
50,000 J
6.3 m
20 m/s (about 45 mph)
200,000 J
25 m
30 m/s (about 67 mph)
450,000 J
56 m
40 m/s (about 89 mph)
800,000 J
100 m
⚠️These distances are braking only. A driver also travels a long way during reaction time, before the brakes are even touched. Real stopping distances are longer than this table.
Notice the pattern: double the speed, four times the distance. Triple it, nine times. That is the speed-squared rule showing up on the road, and it is why speed limits matter so much near schools.
HExample
KE Duel
Test the scaling rules yourself. Build a tie, then try to beat a heavy, slow object with something much lighter.
Object A is heavy and slow. Object B is light and fast. Change anything you like. Can you make them tie? Can you make the slow one win?
Object A
KE = 16.0 J
Object B
KE = 20.3 J
Object B has 1.3 times as much KE.
IConcept
Quick Check
Four questions. They are not graded. Answer them to find out what still needs work.
Q1A 2.0 kg cart rolls at 3.0 m/s. What is its kinetic energy?
Q2A cyclist speeds up from 5 m/s to 10 m/s. Her mass doesn’t change. Her KE becomes…
Q3Object A is 4 kg moving at 3 m/s. Object B is 1 kg moving at 6 m/s. Which has more KE?
Q4A 0.50 kg ball moving right at 4.0 m/s bounces off a wall and moves left at 4.0 m/s. What is the change in its KE?
Score: 0 of 4
JConcept
Looking Ahead
You can now put a number on energy of motion. Next you’ll learn where that energy comes from when something falls, and then you’ll use both ideas in the builds that follow.