Kinetic energy
A car at 100 km/h carries four times the kinetic energy it has at 50 km/h.
The energy of motion
Anything that moves can do work on whatever it hits. A bowling ball knocks pins over, wind turns a turbine, a hammer drives a nail. This energy of motion is called kinetic energy, and for an object of mass m moving at speed v it is KE = ½ m v². The mass is in kilograms, the speed in metres per second, and the answer comes out in joules.
OpenStax gives a quiet example: a 30.0 kg package riding a roller conveyor at 0.500 m/s. Its kinetic energy is ½ × 30.0 × 0.500² = 3.75 J. That is a heavy box, yet the energy is tiny, because the speed is low and the speed is squared.
Why the square matters
Mass enters the formula plainly: twice the mass at the same speed means twice the kinetic energy. Speed enters squared. Double the speed and the energy goes up four times; triple it and the energy goes up nine times. That is why a car at 100 km/h has four times the kinetic energy it has at 50 km/h, and why high-speed crashes are so much worse than slow ones.
The square also explains braking distances. To stop, the brakes must do negative work equal to the car's kinetic energy. With the same braking force, the work is force times distance, so the stopping distance grows like v². Going from 60 km/h to 90 km/h is 1.5 times the speed, but (1.5)² = 2.25 times the braking distance, before you even count the time it takes to react.
The work-energy theorem
Unit 2 said that work moves energy. The work-energy theorem says exactly where it goes when the object speeds up or slows down: the net work done on an object equals the change in its kinetic energy, W_net = ½ m v² − ½ m v₀². Net work means the work done by all the forces together, or equally the work done by the net force.
Go back to the 30.0 kg package. Someone pushes it with 120 N for 0.800 m while friction pulls back with 5.00 N. Gravity and the support force are at right angles to the motion and do no work. The person does 120 × 0.800 = 96.0 J and friction does −5.00 × 0.800 = −4.00 J, so the net work is 92.0 J. The package started with 3.75 J, so it ends with 95.75 J. Solving ½ × 30.0 × v² = 95.75 gives v = 2.53 m/s.
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You can run the formula backwards to learn something. OpenStax lists a tennis ball at 100 km/h as having about 22 J of kinetic energy. Change the speed to metres per second, 100 ÷ 3.6 = 27.8 m/s, and solve ½ m v² = 22 for the mass: m = 2 × 22 ÷ 27.8² = 0.057 kg, or 57 g, which is the mass of a real tennis ball. A mosquito, at about a hundredth of a gram flying at 0.5 m/s, has 1.3 × 10⁻⁶ J. The two differ by a factor of about 17 million.
| Kinetic energy | KE = ½ m v² |
| Double the speed | Four times the kinetic energy |
| Work-energy theorem | Net work = change in kinetic energy |
| Braking distance | Grows with the square of speed (same braking force) |
- KE = ½ m v², in joules.
- Twice the mass gives twice the KE; twice the speed gives four times.
- Net work equals the change in kinetic energy.
- Stopping distance grows like v² for the same braking force.
- Kinetic energy of a 30.0 kg package at 0.500 m/s, in joules?
- A 2 kg ball moves at 3 m/s. Kinetic energy in joules?
- The same ball at 6 m/s. Kinetic energy in joules?
- A truck has twice the mass of a car and both go at the same speed. The truck's KE is:
- Twice the car's
- Four times the car's
- The same
- Half
- Speed goes from 50 km/h to 100 km/h. Kinetic energy becomes:
- 4 times as large
- 2 times as large
- The same
- 8 times as large
- A car speeds up from 60 km/h to 90 km/h. By what factor does the braking distance grow (same braking force)?
- Net work on the package: 120 N push and 5.00 N friction over 0.800 m. In joules?
- Final speed of that package (starting at 0.500 m/s), in m/s? (two decimals)
- Kinetic energy of a 1,200 kg car at 20 m/s, in kJ?
- In the work-energy theorem, which work counts?
- The net work of all forces
- Only the work you do
- Only friction's work
- Only gravity's work
Sources for this course (8)
- OpenStax College Physics 2e (openstax.org/books/college-physics-2e): 7.1 Work: the scientific definition (W = F d cos θ, briefcase cases, 1 J = 1 N·m, 1 J lifts a 100 g apple about 1 m, lawn mower 75.0 N at 35° for 25.0 m = 1.54 × 10³ J, 1 kcal = 4186 J, under 10% of food energy goes into work)
- 7.2 Kinetic energy and the work-energy theorem (KE = ½ m v², 30.0 kg package at 0.500 m/s = 3.75 J, 120 N push with 5.00 N friction over 0.800 m = 92.0 J net, final speed 2.53 m/s, a car at 100 km/h has four times the KE it has at 50 km/h)
- 7.3 Gravitational potential energy (ΔPE = m g h, reference level, path independence, 0.500 kg cuckoo clock weight raised 1.00 m = 4.90 J, 60.0 kg person landing stiffly from 3.00 m, roller coaster 20.0 m hill: 19.8 m/s, and 20.4 m/s with a 5.00 m/s start, mass cancels)
- 7.4 Conservative forces and potential energy (spring PE = ½ k x², 0.100 kg toy car, k = 250 N/m, 4.00 cm, 0.180 m rise)
- 7.5 Nonconservative forces (friction turns mechanical energy into thermal energy)
- 7.6 Conservation of energy (statement of the law, forms of energy, Table 7.1: 1000 kg car at 90 km/h 3.1 × 10⁵ J, tennis ball at 100 km/h 22 J, mosquito 1.3 × 10⁻⁶ J, daily food intake 1.2 × 10⁷ J, 1 gallon gasoline 1.2 × 10⁸ J, 1 g fat 3.9 × 10⁴ J; efficiency, Table 7.2: cycling and climbing 20%, swimming at the surface 2%, gasoline engine 30%, diesel 35%, coal power plant 42%, electric motor 98%, solar cell 10%)
- 7.7 Power (P = W ÷ t, 1 W = 1 J/s, 1 hp = 746 W, 60.0 kg woman up 3.00 m stairs in 3.50 s reaching 2.00 m/s = 538 W, Table 7.3: Sun 4 × 10²⁶ W, car 8 × 10⁴ W, person at rest 100 W, heart at rest 8 W, clothes dryer 4 × 10³ W, a 60 W incandescent bulb gives 5 W of light, power plants turn 35 to 40% of fuel energy into electricity, E = P t, 0.200 kW computer 6.00 h a day for 30.0 days = 36.0 kWh, $4.32 at $0.120 per kWh)
- g = 9.8 m/s² as in OpenStax examples (NIST standard gravity 9.80665 m/s², physics.nist.gov/cgi-bin/cuu/Value?gn).
Checked: 2026-10-17