Grade by Grade
D. Energy and Society · Work, Energy and Power

Kinetic Energy and the Work–Energy Theorem

D3.3 D2.3 D3.2 D2.1About 70 minutes
Learning goals
  • I can calculate kinetic energy with Ek = ½mv² and solve for mass or speed.
  • I can explain why doubling the speed of an object multiplies its kinetic energy by four.
  • I can derive the work–energy theorem, W_net = ΔEk, from Newton's second law and a kinematics equation.
  • I can use W_net = ΔEk to find forces, distances and speeds.

Words to know

kinetic energy (Ek)
The energy of a moving object: Ek = ½mv², with m in kilograms, v in metres per second and Ek in joules.
work–energy theorem
The net work done on an object equals the change in its kinetic energy: W_net = ΔEk = Ek₂ − Ek₁.
speed (v)
How fast an object moves, in m/s. Kinetic energy depends on speed, not direction, so it is a scalar.
scalar
A quantity with size only (no direction), such as energy, mass and speed.

Twice as fast, four times the trouble

Driving instructors in Canada warn that you need much more room to stop on a snowy road when you go faster. Going from 50 km/h to 100 km/h does not just double the braking distance; with the same braking force it quadruples it. The reason is kinetic energy: the energy of motion grows with the square of the speed. In this lesson you will see where the formula comes from and use it with work.

The kinetic energy formula

An object of mass m moving at speed v has kinetic energy

Ek = ½mv²

Units: kg × (m/s)² = kg·m²/s² = J. (In the last lesson you showed 1 J = 1 kg·m²/s².)

Two things to notice:

  • Ek is proportional to mass: twice the mass at the same speed means twice the kinetic energy.
  • Ek is proportional to speed squared: twice the speed means 2² = 4 times the kinetic energy; three times the speed means 9 times.

Kinetic energy is never negative (v² is never negative), and it does not depend on direction. A puck moving left at 10 m/s has the same Ek as one moving right at 10 m/s.

52 kJ (×1)30 km/h208 kJ (×4)60 km/h469 kJ (×9)90 km/ha 1500 kg car at three speeds
Doubling the speed of a car from 30 km/h to 60 km/h makes its kinetic energy four times as large. Tripling it makes it nine times as large.
Worked example

Example 1: Kinetic energy of a puck

A 0.17 kg hockey puck moves at 40 m/s. Find its kinetic energy.

  1. Ek = ½mv²
  2. Ek = ½ (0.17 kg)(40 m/s)²
  3. Ek = ½ (0.17)(1600) = 136 J ≈ 1.4 × 10² J

The puck has about 1.4 × 10² J of kinetic energy.

Worked example

Example 2: Finding the speed

A 70 kg skier has 7.0 × 10³ J of kinetic energy. How fast is she moving?

  1. Rearrange: v² = 2Ek ÷ m, so v = √(2Ek ÷ m).
  2. v = √(2 × 7000 J ÷ 70 kg) = √(200 m²/s²)
  3. v ≈ 14 m/s (about 51 km/h)

About 14 m/s.

Where does ½mv² come from? The work–energy theorem

Suppose a constant net force F_net pushes an object of mass m through a displacement Δd along a straight line, changing its speed from v₁ to v₂.

  1. The net work is W_net = F_net Δd.
  2. Newton's second law (from the Forces unit): F_net = ma, so W_net = maΔd.
  3. From the Kinematics unit: v₂² = v₁² + 2aΔd, so aΔd = (v₂² − v₁²) ÷ 2.
  4. Substitute: W_net = m(v₂² − v₁²) ÷ 2 = ½mv₂² − ½mv₁².

So the net work done on an object equals the change in the quantity ½mv². That quantity is the kinetic energy. This is the work–energy theorem:

W_net = ΔEk = Ek₂ − Ek₁

Positive net work speeds an object up; negative net work slows it down; zero net work leaves its speed unchanged. The theorem also shows why kinetic energy is measured in joules: it is exactly the work needed to get an object from rest up to speed v.

v₁ = 20 m/sv₂ = 0braking forceΔd = 40 mW_net = −(6000 N)(40 m) = −240 000 J = ΔEk
A 1200 kg car braking from 20 m/s to rest. Its kinetic energy drops from 240 000 J to 0, so the braking force must do −240 000 J of work. Over 40 m that takes a 6000 N force.
Worked example

Example 3: Braking on a winter road

A 1200 kg car moving at 20 m/s brakes to a stop in 40 m. (a) Find the work done by the braking (friction) force. (b) Find the average braking force. (c) With the same braking force, how far would it take to stop from 40 m/s?

  1. (a) Ek₁ = ½(1200)(20)² = 240 000 J; Ek₂ = 0. W_net = ΔEk = 0 − 240 000 J = −2.4 × 10⁵ J.
  2. (b) Friction is the only horizontal force, so W = −FΔd: −240 000 J = −F(40 m), F = 6.0 × 10³ N.
  3. (c) At 40 m/s: Ek₁ = ½(1200)(40)² = 960 000 J, four times as much.
  4. Δd = 960 000 J ÷ 6000 N = 160 m: four times the distance.

(a) −2.4 × 10⁵ J; (b) 6.0 × 10³ N; (c) 160 m. Doubling the speed quadruples the braking distance.

Worked example

Example 4: Speeding up a cart

A student pushes a 2.0 kg cart from rest. The net work done on it is 9.0 J. How fast is it moving afterward?

  1. W_net = Ek₂ − Ek₁, and Ek₁ = 0 (starts at rest).
  2. 9.0 J = ½(2.0 kg)v₂²
  3. v₂² = 9.0 m²/s², so v₂ = 3.0 m/s.

3.0 m/s.

Watch the units and the square

  • Convert km/h to m/s first: divide by 3.6 (90 km/h = 25 m/s).
  • Square only the speed: ½ × m × (v × v). A common slip is to calculate (½mv)².
  • Use the net work in W_net = ΔEk, not the work done by one force (unless it is the only force doing work).
Just for fun

A 4 kg Canada goose flying at 15 m/s has about 450 J of kinetic energy, more than three times as much as a hockey puck shot at 40 m/s (136 J). Good advice for both: duck.

Hands-on activity

Your own kinetic energy

You need: a bathroom scale, a tape measure, a stopwatch (a phone works) and a helper. Use a flat, clear area outside, such as a quiet park path or a schoolyard.

Grown-up note: choose a safe, dry surface away from traffic, and jog only at a pace that is comfortable.

  1. Find your mass on the scale (in kg).
  2. Mark a 10.0 m course. Get moving before the start line so you cross it at a steady speed.
  3. Have your helper time you over the 10.0 m while you walk, then while you jog, then while you run (at a safe pace).
  4. Calculate each speed (v = 10.0 m ÷ time) and each kinetic energy (Ek = ½mv²).
  5. Compare: when your speed roughly doubled, what happened to your kinetic energy?

What to look for: a 60 kg student walking 10.0 m in 7.1 s (1.4 m/s) has Ek ≈ 59 J. Jogging it in 3.6 s (2.8 m/s) gives Ek ≈ 230 J: about twice the speed, four times the energy. Your values will differ, but the pattern should match.

Your turn

1

Find the kinetic energy of a 55 kg cyclist moving at 8.0 m/s.

2

A 0.045 kg golf ball has 50 J of kinetic energy. Find its speed.

3

A truck has 4.0 × 10⁵ J of kinetic energy at 20 m/s. What is its mass?

4

If a car's speed triples, by what factor does its kinetic energy change?

5

A 3.0 kg box at rest has 24 J of net work done on it. Find its final speed.

6

Convert 72 km/h to m/s and find the Ek of a 1000 kg car at that speed.

7

A 0.5 kg ball moving at 6.0 m/s is caught and stopped. How much work did the catcher's hand do on the ball?

More practice: warm-up

1

Write the formula for kinetic energy.

2

Find the Ek of a 2.0 kg object moving at 3.0 m/s.

3

Can kinetic energy be negative? Why or why not?

4

State the work–energy theorem.

5

Two pucks have the same speed; one has twice the mass. Compare their kinetic energies.

More practice: on your own

1

A 1500 kg car speeds up from 10 m/s to 25 m/s. How much net work was done on it?

2

A 60 kg skater glides to a stop over 30 m on ice. She started at 4.0 m/s. Find the average friction force on her.

3

A 0.20 kg ball is thrown at 15 m/s. Another ball, 0.40 kg, is thrown at 7.5 m/s. Which has more kinetic energy?

4

A net force of 50 N acts on a 4.0 kg cart at rest over 2.0 m. Use the work–energy theorem to find its final speed. Check with kinematics.

5

Explain, using kinetic energy, why speed limits are lower near schools.

Challenge

1

A car's stopping distance (braking only, same braking force) from 50 km/h is 14 m. Estimate the stopping distance from 80 km/h.

2

A 0.010 kg arrow leaves a bow at 60 m/s. The bowstring pushed it over 0.50 m. Find the average force on the arrow from the string.

3

An object's kinetic energy increases by 44%. By what percent did its speed increase?

Check your understanding

Quiz

1.

The kinetic energy of a 4.0 kg object moving at 5.0 m/s is:

2.

If a car's speed is doubled, its kinetic energy is:

3.

The work–energy theorem says that:

4.

A 2.0 kg cart at rest receives 16 J of net work. Its final speed is:

5.

Which has the most kinetic energy?

6.

A sled slows from 6.0 m/s to 2.0 m/s. The net work done on it is:

For parents and teachers

Big idea: Kinetic energy, Ek = ½mv², and the work–energy theorem, W_net = ΔEk, link forces and motion to energy (D3.3, D2.3). The derivation ties this unit to the Kinematics and Forces units.

How to help: The "twice the speed, four times the energy" idea matters for new drivers. Talk about why braking distances on snowy roads grow so quickly with speed.

Common difficulties:

  • Forgetting to square the speed, or squaring the ½ and the mass too.
  • Not converting km/h to m/s (divide by 3.6).
  • Using one force's work instead of the net work in the theorem.
  • Rearranging for v: remember to take the square root at the end.

Safety: for the activity, choose a flat, dry space away from traffic.