Physics · Unit 4: Work, Energy & Power ·  Activity 4-1-1

Deep Dive: Work & the Work-Energy Theorem

🔬 Deep Dive
This is your textbook for this topic. Read it alongside the activity.
AConcept

What Physics Means by “Work”

In everyday life, work means effort. In physics, work is a force moving an object over a distance. When a force pushes or pulls something along its direction of motion, the force transfers energy to the object.

W = F × d

Force is measured in newtons (N) and distance in meters (m), so work is measured in newton-meters. We call one newton-meter a joule (J).

🔑1 joule is about the work of lifting a small apple (roughly 1 N of weight) up 1 meter. Lift it 3 m and you’ve done about 3 J.
BWatch Out

Effort Is Not Work

For a force to do work on an object, three things have to be true:

⚠️Pushing on a wall, or holding a heavy box still, does zero work on the wall or the box because nothing moves. Your muscles still burn energy, but that’s your body using energy to stay tense. It isn’t energy being transferred to the box.

The same goes for carrying a backpack across a room at a steady height. Your arms push up, the backpack moves sideways, and a force at 90° to the motion does no work.

CMath

When the Force Is at an Angle

Often a force doesn’t point exactly along the motion. If you pull a sled with a rope angled upward, only part of your pull drags the sled forward. The rest lifts it slightly. We handle that with the angle θ between the force and the direction of motion:

W = F × d × cosθ

Think of cosθ as the fraction of the force that points along the motion.

Angle θcosθWhat it means
0°1.00Force points straight along the motion
30°0.87Most of the force helps
45°0.71About 70% of the force helps
60°0.50Half of the force helps
90°0.00Force is straight across the motion: none of it helps
180°−1.00Force points straight against the motion
Worked example: pulling a sled
Given
  • Pull force F = 40 N
  • Rope angle θ = 30° above the ground (and the sled moves along the ground)
  • Distance d = 10 m
Solve
W = F × d × cosθ
W = (40 N)(10 m)(cos 30°)
W = (40)(10)(0.87)
W = 348 J
✓ Your pull does about 348 J of work on the sled.
DConcept

Positive, Negative & Zero Work

The cosine goes negative when the angle is bigger than 90°. That’s how the math tells you a force is working against the motion. Work is a quantity with a sign, and the sign tells you what happens to the object’s speed.

SituationAngleWorkEffect
Force points with the motion0°Positive (+)Speeds the object up
Force points against the motion180°Negative (−)Slows the object down
Force is straight across the motion90°Zero (0)Speed doesn’t change (direction can)
Object doesn’t moveanyZero (0)Nothing to change
💡Friction almost always does negative work on a moving object, because it points opposite the motion. That’s why friction slows things down.
EExample

Work Explorer

Play with the three variables. Set the angle to 90° and see what happens. Then push it past 90° and watch the sign flip.

The block always moves to the right. Change the force, the distance, and the angle of the force, and watch the sign of the work.

direction of motion40 Nθ = 30°
W = 40 × 10 × cos(30°) = 40 × 10 × 0.87
Work done by the force346.4 J
Positive work: the force has a part pointing along the motion, so the block speeds up.
FConcept

Kinetic Energy: The Energy of Motion

Kinetic energy (KE) is the energy an object has because it is moving. It depends on how much mass is moving and, much more strongly, on how fast:

KE = ½ m v²
ObjectMassSpeedKE
Lab cart2 kg3 m/s9 J
Person jogging60 kg3 m/s270 J
Same person, twice as fast60 kg6 m/s1,080 J
Car on a road1,000 kg20 m/s200,000 J
🔑Speed is squared. Double the speed and the kinetic energy becomes four times as big. Look at the jogger: 3 m/s gives 270 J, and 6 m/s gives 1,080 J.

You’ll go deeper on kinetic energy in Activity 4-1-2. For now, all you need is that a moving object carries energy, and the formula above tells you how much.

GMath

The Work-Energy Theorem

Here is the big idea. Add up the work done by every force on an object, and you get the change in the object’s kinetic energy:

W_net = ΔKE = KE_final − KE_start

In plain words: net positive work speeds an object up, net negative work slows it down, and net zero work leaves its speed unchanged. The theorem lets you skip the messy details of the motion. You just count the work.

Count every force.For the block in the Slide & Stop lab, here are the forces and what each one does:

ForceAngle to motionWork
Gravity (pulls down)90° to the slide0
Floor’s normal force (pushes up)90° to the slide0
Friction (points backward)180° to the slideNegative

Only friction does work, so friction’s work is the net work. That’s why measuring friction and stopping distance tells you the block’s starting energy.

Worked example 1: how fast did the block start?
Given
  • Mass m = 0.80 kg
  • Friction force F_f = 1.5 N
  • Stopping distance d = 1.2 m
  • Final speed = 0 (it stops)
Solve
W_f = −F_f × d = −(1.5 N)(1.2 m) = −1.8 J
W_net = ΔKE: −1.8 J = 0 − KE_start
KE_start = 1.8 J
v = √(2 × KE ÷ m) = √(2 × 1.8 ÷ 0.80) = √4.5
v ≈ 2.1 m/s
✓ The block started at about 2.1 m/s.
Worked example 2: speeding up
Given
  • A 2.0 kg cart starts from rest
  • A 5.0 N force pushes it 3.0 m along a track (ignore friction)
Solve
W_net = F × d = (5.0 N)(3.0 m) = 15 J
W_net = KE_final − 0, so KE_final = 15 J
v = √(2 × 15 ÷ 2.0) = √15
v ≈ 3.9 m/s
✓ Positive work of 15 J gives the cart 15 J of kinetic energy, about 3.9 m/s.
HExample

Stopping Explorer

The theorem gives a stopping-distance rule: d = ½ m v² ÷ F_f. Change the sliders and watch how strongly speed controls the distance.

A block slides to a stop with only friction acting. Use the theorem to find how far it goes: friction’s work must erase all the starting kinetic energy, so F_f × d = ½ m v².

Starting KE = 2.0 Jd = KE ÷ F_f
At 2.0 m/s1.00 m
At double the speed (4.0 m/s)4.00 m
Double the speed, four times the stopping distance. Try changing the mass or friction and see if that ever stops being true.
🔑This is why speed limits are about more than reaction time. It is also why the exit speed of a slide matters so much in Problem 4-3-1.
IConcept

Quick Check

Four questions. They are not graded. Answer them to find out what still needs work.

Q1A student holds a 5 kg box still at arm’s length for 30 seconds. How much work does her arm’s force do on the box?
Q2You pull a wagon with a 50 N force at 60° to the ground for 4.0 m. How much work do you do? (cos 60° = 0.50)
Q3A sliding crate has 90 J of kinetic energy and then stops. How much net work did friction do on it?
Q4A car goes twice as fast as before and brakes with the same force. Its stopping distance is…
Score: 0 of 4
JConcept

Looking Ahead

Work and the work-energy theorem show up in both of the big builds that this activity sets up:

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