Lesson 1 of 4Energy and Momentum

Unit 2 · Lesson 1 of 4

Work, Energy and the Work-Energy Theorem

Work done by a force at an angle, the theorem that links net work to a change in kinetic energy, conservation of energy when friction turns some of it into thermal energy, and power.

4 min read11 practice questionsAsk AI about this →

Key things to remember

1Work by a constant force: $W = F Δ d cos θ$ . It is a scalar in joules and is negative when the force opposes the motion.
2A force perpendicular to the displacement ( $θ = 9 0^{\circ}$ ) does zero work.
3Work-energy theorem: $W_{n e t} = Δ E_{k} = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2} .$
4With friction: $E_{k 1} + E_{g 1} = E_{k 2} + E_{g 2} + E_{t h}$ , where $E_{t h} = F_{f} Δ d$ and $F_{f} = μ_{k} F_{N}$ .
5Power: $P = W /Δ t$ , and for a constant force along the motion $P = F v$ . $1 W = 1 J/s$ .

Newton's laws tell you what happens at every instant, which is powerful but often slow: to find how fast a roller coaster is going at the bottom of a curved hill you would need the force at every point on the track. Energy methods skip all of that. They compare only the start and the end, and they work no matter how twisted the path in between. The price of admission is one idea you met in Grade 11, work, now handled carefully with angles, signs and friction.

Work done by a force at an angle

W = F Δ d cos θ

F is the size of the force, Delta d the size of the displacement, theta the angle between them.

Fig 1Only the component of the force along the displacement does work.

Work is a scalar, measured in joules ( $1 J = 1 N \cdot m$ ), but it can be negative. The factor $cos θ$ picks out the part of the force that lies along the motion. A force that helps the motion adds energy to the object (positive work); a force that opposes it removes energy (negative work); a force perpendicular to the motion does no work at all, however large it is.

Angle between force and displacement	$cos θ$	Work	Example
$0^{\circ}$	$1$	Positive, the maximum $F Δ d$	Pushing a cart straight ahead
Between $0^{\circ}$ and $9 0^{\circ}$	Positive	Positive, less than $F Δ d$	Pulling a wagon by a handle angled upward
$9 0^{\circ}$	$0$	Zero	Normal force on a box sliding along the floor; gravity on a car on a level road
Between $9 0^{\circ}$ and $18 0^{\circ}$	Negative	Negative	Gravity on a ball on its way up a ramp
$18 0^{\circ}$	$- 1$	Negative, the most negative	Kinetic friction on a sliding puck

The work-energy theorem

W_{n e t} = Δ E_{k} = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}

The total work done by all forces equals the change in kinetic energy.

Here is why it is true. For a constant net force along a straight line, $F_{n e t} = ma$ and $v_{2}^{2} = v_{1}^{2} + 2 a Δ d$ . Multiply the second equation by $\frac{1}{2} m$ and rearrange: $ma Δ d = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}$ . The left side is $F_{n e t} Δ d$ , the net work. So net work is exactly the change in kinetic energy, which is why kinetic energy has the formula it does. Use it when the forces are known and you want a speed, or the speeds are known and you want a force or distance.

Conservation of energy when friction acts

E_{k 1} + E_{g 1} = E_{k 2} + E_{g 2} + E_{t h} E_{t h} = F_{f} Δ d

Energy is never lost. Kinetic friction converts mechanical energy into thermal energy.

Fig 2Bar charts at two moments: the bars change height but their total stays the same.

Choosing the right tool

Gravity does work that depends only on the change in height, so it is easiest to account for it as a change in

E_{g} = m g h

rather than as work. Friction's work depends on the path length, so it appears as

E_{t h} = F_{f} Δ d

, the thermal energy added to the surfaces. Put every other force in as work. Then the total energy at the start, plus any work added, equals the total at the end.

Why no perpetual motion machine can exist

Every real machine has some friction, and friction always turns mechanical energy into thermal energy that spreads into the surroundings. A machine that kept running forever with no energy input would have to create energy to replace that loss, which conservation of energy forbids. This is why a swinging playground swing always slows down unless someone keeps pushing.

Power: how fast work is done

P = \frac{W}{Δ t} = \frac{Δ E}{Δ t} P = F v (constant force along the motion)

Power is measured in watts, where one watt is one joule per second.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

25 kg

crate starts from rest and is pulled

12 m

across a level floor by a rope with tension

150 N

3 0^{\circ}

above the horizontal. The coefficient of kinetic friction is

0.20

. Find the work done by each force and the crate's final speed.

Example 2

450 kg

roller coaster car moves at

4.0 m/s

over the top of a hill

32 m

high. Friction converts

1.5 \times 1 0^{4} J

into thermal energy by the time the car reaches a point

12 m

high. How fast is it moving there?

Example 3

2.0 kg

block is released from rest at the top of a ramp

3.0 m

long, inclined at

3 5^{\circ}

. The coefficient of kinetic friction is

0.25

. Use energy to find its speed at the bottom.

Example 4

An elevator motor lifts an

800 kg

elevator car upward at a constant

1.5 m/s

. What power must the motor supply, ignoring friction?

Where marks get lost

Using the ramp length instead of the vertical height in $E_{g} = m g h$ . Gravitational energy depends only on height; the ramp length belongs in the friction term.
Taking the normal force as $m g$ when a rope pulls at an angle or the surface is tilted. Resolve forces first, then find $F_{f} = μ_{k} F_{N}$ .
Forgetting that work can be negative. Friction and any force opposing the motion do negative work and reduce kinetic energy.
Using the applied force alone in the work-energy theorem. It is the net work, from all the forces together, that equals the change in kinetic energy.
Counting gravity twice, once as work done by gravity and again as a change in $E_{g}$ . Choose one way to account for it.

Flashcards 8

Work by a constant force at an angle

When is work zero?

Work-energy theorem

Kinetic energy

+ 4 more

Practice

A child pulls a sled

8.0 m

along level snow with a force of

60 N

directed

4 0^{\circ}

above the horizontal. How much work does the child do, in joules? Give three significant figures.

A box slides

5 m

across a level floor. How much work does the normal force do on it?

1200 kg

car speeds up from

15 m/s

25 m/s

. How much net work was done on it, in kilojoules?

In which case does the named force do negative work?

A frictionless roller coaster car starts from rest at the top of a hill and drops a vertical distance of

20 m

. What is its speed at the bottom, in metres per second?

m/s

1500 kg

car travelling at

20 m/s

brakes on a level road with a constant friction force of

9000 N

. How far does it travel before stopping, in metres?

70 kg

student runs up a flight of stairs with a vertical rise of

4.5 m

6.0 s

. What average power do they develop against gravity, in watts?

When kinetic friction slows a sliding object, into what form of energy is the lost kinetic energy mainly converted?

10 N

force acts on an object that moves

2 m

. Order these angles between force and displacement from the most positive work to the most negative work.

1 $18 0^{\circ}$
2 $12 0^{\circ}$
3 $9 0^{\circ}$
4 $0^{\circ}$
5 $6 0^{\circ}$

Match each quantity to its expression.

A. $μ_{k} F_{N} Δ d$
B. $\frac{1}{2} m v^{2}$
C. $F Δ d cos θ$
D. $W /Δ t$

Work by a force at an angle

Kinetic energy

Thermal energy from friction

Power

Two identical balls leave the edge of the same cliff at the same speed. One is thrown horizontally and one is thrown straight up. Ignoring air resistance, which is moving faster when it reaches the ground?

Spot a mistake on this page? Let us know