C. Energy and Momentum · Work, Energy and Simple Harmonic Motion

Kinetic Energy and the Work–Energy Theorem in Two Dimensions

C2.2 C2.1About 70 minutes

Learning goals

I can apply the work–energy theorem, W_net = ΔEk, to motion in one and two dimensions.
I can find the work done by each force on an incline and use it to find a speed or a distance.
I can use the work–energy theorem with a force that changes (from a graph).
I can decide when an energy method is better than a force-and-kinematics method.

Words to know

kinetic energy
The energy of motion, Ek = ½mv². It depends on speed, not direction.

work–energy theorem
The net work done on an object equals its change in kinetic energy: W_net = Ek₂ − Ek₁.

energy bar chart
A diagram that shows initial energy, the work done by each force, and final energy as bars, to keep track of energy transfers.

incline
A sloped surface; on it, gravity has a component mg sin θ along the slope and mg cos θ into the slope.

Why bother with energy?

A snowboarder slides down a curved half-pipe. The slope changes at every point, so the acceleration changes too, and Newton's laws with the kinematics equations get messy fast. Yet a single energy equation can give her speed at the bottom in one line. The tool that makes this possible is the work–energy theorem, which you met in Grade 11. In Grade 12 you will use it where it really shines: forces at angles, inclines, curved paths and changing forces.

Remember from Grade 11: kinetic energy is Ek = ½mv², and the work–energy theorem says that the net work done on an object equals its change in kinetic energy: W_net = ΔEk = ½mv₂² − ½mv₁². SPH3U proved it for a constant force along a straight line using v₂² = v₁² + 2aΔd.

The theorem in two dimensions

In two dimensions the theorem still holds, and it is still a scalar equation:

W_net = Ek₂ − Ek₁

Find the work done by each force with W = FΔd cos θ (or as an area under a graph), then add them. Each force's work can be positive, negative or zero.
Kinetic energy depends only on speed, not direction. A puck moving north at 5 m/s and one moving east at 5 m/s have the same Ek.
The theorem tells you the speed after a motion, but not the direction or the time taken. For those you still need Newton's laws and kinematics (as in the sibling unit Dynamics: Forces and Motion in a Plane).

This is why energy methods are so powerful: you can skip the acceleration completely.

An energy bar chart for the sled in Example 2 of Work Done by a Constant Force in One and Two Dimensions. Start with Ek₁, add the work of each force (positive works raise the total, negative works lower it), and you get Ek₂.

Worked example

Example 1: Final speed of the sled

The sled from Example 2 of Work Done by a Constant Force in One and Two Dimensions (15 kg) starts at rest. The rope does +1087.6 J of work and friction does −243.3 J over 20 m. What is the sled's speed at the end?

W_net = 1087.6 − 243.3 + 0 + 0 = 844.3 J.
W_net = Ek₂ − Ek₁, and Ek₁ = 0 (rest): 844.3 J = ½(15 kg)v₂².
v₂² = 2(844.3) ÷ 15 = 112.57 m²/s², so v₂ = 10.6 m/s.

About 10.6 m/s (38 km/h). No acceleration or time was needed.

A block sliding up a rough 30° incline. The component of gravity along the slope and the friction force both point down the slope, so both do negative work while the block rises.

Worked example

Example 2: Sliding up a rough incline

A 2.0 kg block is launched up a 30° incline at 6.0 m/s. μ_k = 0.20. How far up the incline does it slide before stopping?

Let Δd be the distance along the slope. Forces: gravity, normal force, friction (down the slope, because the block moves up).
Normal force: F_N = mg cos 30° = (2.0)(9.8)(0.866) = 16.97 N, so F_f = 0.20 × 16.97 = 3.395 N.
Gravity: the angle between mg (down) and Δd (up the slope) is 90° + 30° = 120°, so W_g = mgΔd cos 120° = −mgΔd sin 30° = −9.8Δd.
Friction: W_f = −3.395Δd. Normal force: 0.
W_net = −(9.8 + 3.395)Δd = −13.195Δd.
ΔEk = 0 − ½(2.0)(6.0)² = −36 J. So −13.195Δd = −36, and Δd = 2.73 m.

The block slides about 2.7 m up the incline. Notice the mass cancels if you keep it as a symbol: Δd = v₁² ÷ [2g(sin θ + μ_k cos θ)].

Worked example

Example 3: Using a force graph

A 0.50 kg cart starts at rest on a smooth track. A horizontal force on it rises steadily from 0 to 8.0 N over the first 1.0 m. What is its speed at 1.0 m?

Work = area of the triangle under the graph = ½(1.0 m)(8.0 N) = 4.0 J.
That is the only force doing work, so W_net = 4.0 J = ½(0.50)v².
v² = 16, so v = 4.0 m/s.

4.0 m/s. The work–energy theorem works just as well for a changing force.

Worked example

Example 4: A throw at any angle

A student throws a ball from 2.0 m above level ground at 15 m/s. Ignoring air resistance, how fast is it moving when it hits the ground? Does the launch angle matter?

Only gravity does work. W_g = mgΔh, where Δh = 2.0 m is the drop, whatever the path.
mgΔh = ½mv₂² − ½mv₁². Divide by m: v₂² = v₁² + 2gΔh.
v₂² = 15² + 2(9.8)(2.0) = 225 + 39.2 = 264.2, so v₂ = 16.3 m/s.

16 m/s (16.3 m/s), for any launch angle. The angle changes the path and the time in the air, but not the landing speed. Solving this with projectile components (as in Dynamics: Forces and Motion in a Plane) gives the same answer with much more work.

When to choose energy, and when to choose forces

Use the work–energy theorem when the question asks for a speed or a distance, especially on curves, slopes, or with changing forces.
Use Newton's laws and kinematics when the question asks for time, acceleration, or a direction.
Many problems need both: forces to find the friction or normal force, then energy to find the speed.

Just for fun

Doubling your speed quadruples your kinetic energy. That is why a car moving at 100 km/h needs about four times the braking distance it needs at 50 km/h, and why every Canadian driving instructor sounds slightly nervous in January.

Hands-on activity

Measure kinetic energy from a ramp roll

You need: a toy car or a can of soup, a long board or a stiff piece of cardboard (about 1 m) propped on books, a tape measure, a phone with a stopwatch or slow-motion video, and a kitchen scale.

Weigh the car or can. Prop the board so its top end is about 10 cm high and measure the length of the board.
Release the car from rest at the top. Mark a 50 cm stretch of flat floor just after the ramp and time how long the car takes to cross it. Its speed is v = 0.50 m ÷ time.
Calculate Ek = ½mv² at the bottom and compare with mgh, the work gravity did on the way down.
Repeat with the board at 15 cm and 20 cm.

What to look for: Ek at the bottom should be less than mgh, because friction and air resistance do negative work and a rolling object also stores energy in its spinning wheels. The ratio Ek ÷ mgh is often 0.6 to 0.9. The higher the ramp, the larger both values.

Your turn

Find the kinetic energy of a 57 g tennis ball served at 50 m/s.

A 6.0 kg object moving at 4.0 m/s has 300 J of net work done on it. Find its new speed.

A 1500 kg car slows from 25 m/s to 15 m/s. How much net work was done on it?

A 0.17 kg puck slides on ice at 8.0 m/s with μ_k = 0.10. How far does it slide before stopping?

By what factor does Ek change if speed doubles? If speed is cut in half?

The net work on an object is zero. What can you say about its speed? About its velocity?

A 60 kg skier starts from rest and slides 50 m down a 15° slope with μ_k = 0.080. Find her speed at the bottom.

More practice: warm-up

Write the formula for kinetic energy.

Find Ek for a 2.0 kg ball moving at 3.0 m/s.

State the work–energy theorem in words.

50 J of net work is done on a 1.0 kg object at rest. Find its speed.

Can kinetic energy be negative? Why or why not?

More practice: on your own

A 2000 kg truck moving at 20 m/s brakes to a stop in 50 m. Find the average braking force.

A 10 kg box starts at rest. It is pulled 8.0 m across a level floor by a 50 N force at 30° above the horizontal. μ_k = 0.20. Find its final speed.

A bow does 60 J of work on a 25 g arrow. If all of it becomes the arrow's kinetic energy, how fast does the arrow leave the bow?

A stone is thrown straight down at 5.0 m/s from a 20 m cliff. Ignoring air resistance, how fast does it hit the water below?

A 3.0 kg cart moving at 2.0 m/s meets a constant opposing force of 6.0 N. How far does it travel before stopping?

Challenge

The 2.0 kg block in Example 2 slides 2.73 m up the 30° incline (μ_k = 0.20), stops, and slides back down. How fast is it moving when it returns to its starting point?

Show that, for a car braking on a level road with friction coefficient μ, the stopping distance is d = v² ÷ (2μg). Then find d for μ = 0.70 at 50 km/h and at 100 km/h.

A 1200 kg car coasts in neutral up a 5.0° hill, starting at 25 m/s. Air drag and rolling resistance together are a constant 500 N. How far up the hill does it coast?

Check your understanding

Quiz

The work–energy theorem states that:

A 4.0 kg object moving at 3.0 m/s has 32 J of net work done on it. Its new speed is:

Two identical balls are thrown from the same height at the same speed, one straight up and one horizontally. Ignoring air resistance, how do their speeds compare when they hit the ground?

A block slides up a rough incline. Which forces do negative work on it?

What can the work–energy theorem NOT tell you directly?

A car's speed triples. Its kinetic energy becomes:

For parents and teachers

Big idea (C2.2): the work–energy theorem, W_net = ΔEk, lets students find speeds and distances without finding the acceleration. In Grade 12 they apply it in two dimensions: on slopes, with angled forces and with changing forces.

How to help: ask your teen to explain why a ball thrown up and one thrown sideways from the same balcony land at the same speed. Talk about braking distances in winter: doubling speed means four times the braking distance.

Common difficulties:

Using the work of a single force instead of the net work.
Squaring the wrong thing: Ek = ½mv² squares only v.
Forgetting that the theorem gives speed, not direction or time.
Sign errors on inclines. A sketch with the displacement arrow drawn first helps.