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Work and Kinetic Energy

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Work and Kinetic Energy - Complete Interactive Lesson

Part 1: Definition of Work

⚙️ Work: $W = Fd\cos\theta$

Part 1 of 7 — Work and Kinetic Energy

In everyday language, "work" means effort. In physics, work has a precise definition: it's the transfer of energy to an object by a force acting over a displacement. Understanding work is the gateway to all of energy physics.

In this lesson, we'll define work mathematically and explore when work is positive, negative, or zero.

The Definition of Work

The work done by a constant force $\vec{F}$ on an object that undergoes a displacement $\vec{d}$ is:

Positive, Negative, and Zero Work

The sign of work is determined entirely by $\cos\theta$ :

Angle $\theta$	$\cos\theta$	Work	Meaning
$0°$

Special Cases

Force Along the Displacement ( $\theta = 0°$ )

$W = Fd\cos(0°) = Fd$

Maximum work — the entire force contributes.

Force Opposite to Displacement ( $θ$ )

Work Concepts Quiz 🎯

Work Calculations 🧮

Use $g = 10$ m/s² where needed.

A force of 80 N pulls a sled at $\theta = 60°$ above the horizontal for a distance of 10 m. What is the work done (in J)?
A 5 kg block slides 3 m across a floor. The kinetic friction force is 20 N. What is the work done by friction (in J)?
A 2 kg object is lifted straight up by 4 m at constant velocity. What is the work done by the lifting force (in J)?

Classify the Work 🔍

Exit Quiz — Work ✅

Part 2: Work by Angled Forces

⚡ The Work-Energy Theorem

Part 2 of 7 — Work and Kinetic Energy

The Work-Energy Theorem is one of the most powerful principles in mechanics. It directly connects the net work done on an object to its change in kinetic energy. This theorem often lets you bypass complex force analysis and jump straight to the answer.

Statement of the Work-Energy Theorem

$W_{\text{net}} = \Delta KE = KE_f - KE_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

Part 3: Kinetic Energy

🏃 Kinetic Energy: $KE = \frac{1}{2}mv^2$

Part 3 of 7 — Work and Kinetic Energy

Kinetic energy is the energy of motion. Every moving object — from a tiny electron to a massive truck — possesses kinetic energy. In this lesson, we'll explore the properties of kinetic energy and how it depends on mass and speed.

Part 4: Work-Energy Theorem

🔧 Work by Multiple Forces

Part 4 of 7 — Work and Kinetic Energy

Real-world problems rarely involve a single force. Objects are typically acted on by gravity, normal forces, applied forces, and friction simultaneously. In this lesson, we'll practice computing the work done by each force and finding the net work.

Strategy for Multiple Forces

Step-by-Step

Draw a free-body diagram — identify every force
Find the displacement — direction and magnitude
Calculate work for each force: $W = Fd\cos\theta$
Sum all works to get $W_{\text{net}}$

Part 5: Power

⚡ Power: $P = W/t = Fv$

Part 5 of 7 — Work and Kinetic Energy

Work tells us how much energy is transferred. Power tells us how fast that energy is transferred. A powerful engine doesn't necessarily do more work — it does the same work in less time.

Defining Power

Power is the rate at which work is done (or energy is transferred):

$P = \frac{W}{t}$

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

Part 6 of 7 — Work and Kinetic Energy

This lesson is a hands-on workshop where we'll work through challenging work and energy problems step by step. These problems combine multiple concepts: work by various forces, the Work-Energy Theorem, and power.

Problem-Solving Strategy Review

When to Use Work-Energy vs. Newton's Laws

Use Work-Energy When...	Use Newton's Laws When...
You need final speed from forces and distance	You need acceleration
You want to avoid finding acceleration	You need time information
Forces act over a displacement	Forces act over a time interval
Problem involves multiple forces at different angles	Free-body diagrams suffice

Master Formula

$W_{\text{net}} = \Delta KE$

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Part 7 of 7 — Work and Kinetic Energy

This final lesson brings together everything from the unit: the definition of work, the Work-Energy Theorem, kinetic energy, power, and multi-force problems. These AP-style questions test your conceptual understanding and problem-solving skills.

Key Equations Summary

Concept	Equation	Notes
Work	$W = Fd\cos\theta$	$\theta$ = angle between and

$F$ = magnitude of the force (N)
$d$ = magnitude of the displacement (m)
$\theta$ = angle between the force and displacement vectors
$W$ = work (measured in Joules, J = N·m)

Work depends on three things:

How strong the force is ( $F$ )
How far the object moves ( $d$ )
The angle between force and displacement ( $\theta$ )

If any of these is zero, the work is zero!

Positive work: You push a box forward, and it moves forward ( $\theta = 0°$ )
Negative work: Friction acts backward on a sliding box ( $\theta = 180°$ )
Zero work: A waiter carries a tray horizontally — gravity pulls down, motion is horizontal ( $\theta = 90°$ )
Zero work: The normal force on a box sliding on a flat surface ( $\theta = 90°$ )

Only the component of force parallel to displacement does work. The perpendicular component changes the direction of motion but does no work.

$W = Fd\cos(180°) = -Fd$

Maximum negative work — the force removes energy from the object.

When you pull a sled at angle $\theta$ above horizontal with force $F$ :

Only the horizontal component $F\cos\theta$ does work
The vertical component $F\sin\theta$ partially supports the weight but does no work (perpendicular to motion)

In words: The net work done on an object equals the change in its kinetic energy.

$W_{\text{net}}$ is the total work done by all forces acting on the object
If $W_{\text{net}} > 0$ : the object speeds up (gains KE)
If $W_{\text{net}} < 0$ : the object slows down (loses KE)
If $W_{\text{net}} = 0$ : speed is unchanged (constant KE)

Starting from Newton's 2nd Law: $F_{\text{net}} = ma$

Multiply both sides by displacement $d$ :

$F_{\text{net}} \cdot d = ma \cdot d$

Using kinematics ( $v_f^2 = v_i^2 + 2ad \Rightarrow ad = \frac{v_f^2 - v_i^2}{2}$ ):

$W_{\text{net}} = m \cdot \frac{v_f^2 - v_i^2}{2} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

Calculating Net Work

There are two equivalent approaches:

Method 1: Sum of Individual Works

Calculate the work done by each force, then add them:

$W_{\text{net}} = W_1 + W_2 + W_3 + \cdots$

Method 2: Net Force Method

Find the net force first, then calculate work:

$W_{\text{net}} = F_{\text{net}} \cdot d \cdot \cos\theta$

Example

A 5 kg box is pulled with 40 N at $0°$ across a surface with 10 N friction for 6 m.

Method 1:

$W_{\text{pull}} = 40 \times 6 = 240$ J
$W_{\text{friction}} = -10 \times 6 = -60$ J

Method 2:

$F_{\text{net}} = 40 - 10 = 30$ N (horizontal)
$W_{\text{net}} = 30 \times 6 = 180$ J ✓

Applying the Work-Energy Theorem

Finding Final Speed

If a 5 kg box starts from rest and $W_{\text{net}} = 180$ J:

$180 = \frac{1}{2}(5)v_f^2 - 0$ $v_f = \sqrt{\frac{2 \times 180}{5}} = \sqrt{72} \approx 8.5 \text{ m/s}$

Finding Stopping Distance

A 2 kg ball moving at 10 m/s is stopped by friction ( $f_k = 8$ N):

$W_{\text{net}} = \Delta KE$ $-8d = 0 - \frac{1}{2}(2)(10)^2$

The Work-Energy Theorem is especially powerful when you don't need to find acceleration or time.

Work-Energy Theorem Concepts 🎯

Work-Energy Theorem Calculations 🧮

A 4 kg object starts from rest and has 200 J of net work done on it. What is its final speed (in m/s)?
A 3 kg ball moving at 8 m/s is brought to rest by a net force over 6 m. What is the magnitude of the net force (in N)?
A 1500 kg car accelerates from 10 m/s to 30 m/s. What is the net work done on the car (in kJ)?

Work-Energy Theorem Analysis 🔍

Exit Quiz — Work-Energy Theorem ✅

Defining Kinetic Energy

The kinetic energy of an object with mass $m$ moving at speed $v$ is:

$KE = \frac{1}{2}mv^2$

Properties

Property	Detail
Units	Joules (J) = kg·m²/s²
Sign	Always $\geq 0$ (mass and $v^2$ are both non-negative)
Scalar	Not a vector — depends on speed, not direction
Zero when	$v = 0$ (object is at rest)

The $v^2$ Dependence

Kinetic energy depends on the square of the speed:

Speed	KE
$v$	$\frac{1}{2}mv^2$

Doubling speed quadruples KE. This is why highway accidents at 120 km/h are four times more destructive than at 60 km/h.

Mass vs. Speed: Which Matters More?

KE depends linearly on mass but quadratically on speed:

$KE = \frac{1}{2}mv^2$

Comparison

Doubling mass → KE doubles
Doubling speed → KE quadruples

This means speed has a greater effect on kinetic energy than mass.

Real-World Examples

Object	Mass (kg)	Speed (m/s)	KE (J)
Walking person	70	1.5	79
Running person	70	8	2,240
Car on highway	1,500	30	675,000
Baseball pitch	0.145	40	116
Bullet	0.01	700	2,450

A tiny bullet can have more KE than a walking person because of its enormous speed!

KE and Reference Frames

Kinetic energy depends on the reference frame:

A passenger sitting on a train has $KE = 0$ relative to the train
The same passenger has $KE = \frac{1}{2}mv_{\text{train}}^2$ relative to the ground

In AP Physics 1, we typically use the ground as our reference frame unless stated otherwise.

KE Is Not Conserved

Unlike total energy, kinetic energy alone is not necessarily conserved. It can be:

Converted to potential energy (ball thrown upward)
Converted to thermal energy (friction)
Transferred between objects (collisions)

Kinetic Energy Concepts 🎯

Kinetic Energy Calculations 🧮

What is the kinetic energy of a 2 kg ball moving at 6 m/s (in J)?
A 1200 kg car has a kinetic energy of 150,000 J. What is its speed (in m/s, to the nearest whole number)?
A 0.5 kg object moving at 10 m/s has its speed tripled. What is the new kinetic energy (in J)?

Kinetic Energy Comparisons 🔍

Exit Quiz — Kinetic Energy ✅

Apply the Work-Energy Theorem:

W_{\text{net}} = \Delta KE

Forces That Often Do Zero Work

Force	Why Zero Work?
Normal force (flat surface)	Perpendicular to motion
Gravity (horizontal motion)	Perpendicular to motion
Centripetal force	Always perpendicular to velocity
Tension in a pendulum	Perpendicular to the arc of motion

Example: Box on a Horizontal Surface

A 10 kg box is pulled 8 m across a floor by a 60 N force at $30°$ above horizontal. The coefficient of kinetic friction is $\mu_k = 0.3$ . ( $g = 10$ m/s²)

Step 1: Identify Forces

Applied force $F = 60$ N at $30°$
Weight $W = mg = 100$ N (down)
Normal force $N$ (up)
Kinetic friction $f_k$ (backward)

Step 2: Find Normal Force

Vertical equilibrium: $N + F\sin(30°) = mg$ $N = 100 - 60(0.5) = 100 - 30 = 70 \text{ N}$

Step 3: Find Friction

$f_k = \mu_k N = 0.3 \times 70 = 21 \text{ N}$

Step 4: Calculate Each Work

$W_{\text{applied}} = 60 \times 8 \times \cos(30°) = 480 \times 0.866 \approx 416$ J

Step 5: Net Work

$W_{\text{net}} = 416 - 168 + 0 + 0 = 248 \text{ J}$

Work on an Incline

When an object moves along an incline at angle $\phi$ :

Gravity component along incline: $mg\sin\phi$ (down the incline)
Normal force: Perpendicular to incline (does zero work)

Moving Up the Incline (distance $d$ )

$W_{\text{gravity}} = -mgd\sin\phi$ (opposes motion)
Height gained: $h = d\sin\phi$
So $W_{\text{gravity}} = -mgh$ ✓

Moving Down the Incline (distance $d$ )

$W_{\text{gravity}} = +mgd\sin\phi$ (aids motion)
Height lost: $h = d\sin\phi$

Multiple Forces — Concepts 🎯

Multi-Force Work Problems 🧮

Use $g = 10$ m/s².

A 5 kg box is pushed 4 m across a floor by a horizontal force of 30 N. Kinetic friction is 10 N. What is the net work done (in J)?
A 2 kg block slides 5 m down a frictionless incline at $30°$ to the horizontal. What is the work done by gravity (in J)?
In problem 2, starting from rest, what is the block's speed at the bottom (in m/s, to 3 significant figures)?

Work Sign Analysis 🔍

Exit Quiz — Work by Multiple Forces ✅

$P$ = power (Watts, W)
$W$ = work done (Joules, J)
$t$ = time interval (seconds, s)

$1 \text{ Watt} = 1 \text{ J/s} = 1 \text{ kg·m}^2/\text{s}^3$

Unit	Value
1 Watt (W)	1 J/s
1 kilowatt (kW)	1,000 W
1 horsepower (hp)	746 W
1 kilowatt-hour (kWh)	$3.6 \times 10^6$ J (energy, not power!)

Note: A kilowatt-hour is a unit of energy (power × time), not power.

Instantaneous Power

For an object moving at velocity $v$ under a force $F$ :

$P = Fv\cos\theta$

When the force is in the direction of motion ( $\theta = 0°$ ):

$P = Fv$

Derivation

$P = \frac{W}{t} = \frac{Fd\cos\theta}{t} = F\left(\frac{d}{t}\right)\cos\theta = Fv\cos\theta$

Important Consequence

At constant power $P$ , if speed increases, force must decrease:

$F = \frac{P}{v}$

This is why cars have a maximum speed — as speed increases, the engine force decreases until it equals the drag force, and acceleration stops.

Power and Efficiency

Average vs. Instantaneous Power

Average power: $P_{\text{avg}} = W_{\text{total}} / t_{\text{total}}$
Instantaneous power: $P = Fv$ at a specific moment

Efficiency

In real systems, not all energy goes where we want. Efficiency measures this:

$\text{efficiency} = \frac{P_{\text{useful}}}{P_{\text{total}}} \times 100\%$

For example, if a motor uses 1000 W but only 800 W goes to lifting a load:

$\text{efficiency} = \frac{800}{1000} \times 100\% = 80\%$

Power Concepts 🎯

Power Calculations 🧮

Use $g = 10$ m/s².

A crane lifts a 500 kg load 20 m in 25 s. What is the average power output (in W)?
A car engine provides 5000 W. If the car moves at a constant 25 m/s against friction, what is the friction force (in N)?
An elevator motor has an efficiency of 80%. To lift a 600 kg load at 2 m/s, what total (input) power is needed (in W)?

Power Analysis 🔍

Exit Quiz — Power ✅

$\sum F_i d_i \cos\theta_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

Forgetting that $\cos(180°) = -1$ for opposing forces
Using the wrong angle (the angle is between force and displacement)
Confusing force magnitude with weight
Not including ALL forces when computing net work

Worked Example 1: Pulling at an Angle

A 12 kg box, initially at rest, is pulled 6 m across a rough horizontal floor by a rope at $37°$ above horizontal with tension $T = 100$ N. The coefficient of kinetic friction is $\mu_k = 0.25$ ( $g = 10$ m/s²).

Step 1: Normal force $N = mg - T\sin(37°) = 120 - 100(0.6) = 120 - 60 = 60 \text{ N}$

Step 2: Friction $f_k = \mu_k N = 0.25 \times 60 = 15 \text{ N}$

Step 3: Work by each force

$W_T = 100(6)\cos(37°) = 600(0.8) = 480$ J
J

Step 4: Final speed $W_{\text{net}} = 480 - 90 = 390 \text{ J}$ $390 = \frac{1}{2} (12) v_{f}^{2} \Rightarrow v_{f} = \sqrt{\frac{780}{12}} = \sqrt{65} \approx$

Workshop Problems 🎯

Workshop Calculations 🧮

Use $g = 10$ m/s². Use $\cos(37°) = 0.8$ , $\sin(37°) = 0.6$ .

A 3 kg object moving at 4 m/s has a 15 N net force applied in the direction of motion for 2 m. What is the final speed (in m/s, to 3 significant figures)?
A 50 kg skier starts from rest and descends a slope, dropping 20 m vertically. At the bottom, their speed is 15 m/s. How much energy was lost to friction (in J)?
A 1500 kg car engine produces 45 kW. What is the maximum speed the car can travel against a 1500 N drag force (in m/s)?

Problem Strategy Selection 🔍

Exit Quiz — Problem Solving ✅

Key Conceptual Points

Positive work → adds energy → object speeds up
Negative work → removes energy → object slows down
Zero work → perpendicular force or zero displacement
KE depends on $v^2$ → doubling speed quadruples KE
Power is about how fast work is done, not how much

AP-Style Conceptual Questions 🎯

AP-Style Calculations 🧮

Use $g = 10$ m/s².

A 0.2 kg ball is dropped from 20 m. What is its speed just before hitting the ground (in m/s)?
A 1500 kg car moving at 25 m/s brakes to 15 m/s over 40 m. What is the average braking force (in N)?
A motor lifts a 200 kg elevator at a constant 3 m/s. What minimum power does the motor need (in W)?

AP Review — True or False 🔍

Final AP Exit Quiz — Work & Kinetic Energy ✅

W_{\text{gravity}} = 0

J (perpendicular)

W_{\text{normal}} = 0

J (perpendicular)

W_{\text{net}} = 240 - 60 + 0 + 0 = 180

d = 12.5 \text{ m}

W_{\text{friction}} = 21 \times 8 \times \cos(180°) = -168

W_{\text{gravity}} = 100 \times 8 \times \cos(90°) = 0

W_{\text{normal}} = 70 \times 8 \times \cos(90°) = 0

W_{\text{gravity}} = +mgh

✓

W_{f} = - 15 (6) = - 90

W_g = 0

W_N = 0

390 = \frac{1}{2} (12) v_{f}^{2} \Rightarrow v_{f} = \frac{780}{12} = 65 \approx 8.1 m/s

Work and Kinetic Energy

Work and Kinetic Energy - Complete Interactive Lesson

Part 1: Definition of Work

⚙️ Work: W=Fdcos⁡θW = Fd\cos\thetaW=Fdcosθ

The Definition of Work

Positive, Negative, and Zero Work

Special Cases

Force Along the Displacement (θ=0°\theta = 0°θ=0°)

Force Opposite to Displacement (θ)

Part 2: Work by Angled Forces

⚡ The Work-Energy Theorem

Statement of the Work-Energy Theorem

Part 3: Kinetic Energy

🏃 Kinetic Energy: KE=12mv2KE = \frac{1}{2}mv^2KE=21​mv2

Part 4: Work-Energy Theorem

🔧 Work by Multiple Forces

Strategy for Multiple Forces

Step-by-Step

Part 5: Power

⚡ Power: P=W/t=FvP = W/t = FvP=W/t=Fv

Defining Power

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

Problem-Solving Strategy Review

When to Use Work-Energy vs. Newton's Laws

Master Formula

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Key Equations Summary

Key Insight

Examples

Crucial Point

Force at an Angle

Key Points

Where It Comes From

Calculating Net Work

Method 1: Sum of Individual Works

Method 2: Net Force Method

Example

Applying the Work-Energy Theorem

Finding Final Speed

Finding Stopping Distance

Defining Kinetic Energy

Properties

The v2v^2v2 Dependence

Mass vs. Speed: Which Matters More?

Comparison

Real-World Examples

KE and Reference Frames

KE Is Not Conserved

Forces That Often Do Zero Work

Example: Box on a Horizontal Surface

Step 1: Identify Forces

Step 2: Find Normal Force

Step 3: Find Friction

Step 4: Calculate Each Work

Step 5: Net Work

Work on an Incline

Moving Up the Incline (distance ddd)

Moving Down the Incline (distance ddd)

Units

Common Units

Instantaneous Power

Derivation

Important Consequence

Power and Efficiency

Average vs. Instantaneous Power

Efficiency

Common Pitfalls

Worked Example 1: Pulling at an Angle

Key Conceptual Points

⚙️ Work: $W = Fd\cos\theta$

Force Along the Displacement ( $\theta = 0°$ )

Force Opposite to Displacement ( $θ$ )

🏃 Kinetic Energy: $KE = \frac{1}{2}mv^2$

⚡ Power: $P = W/t = Fv$

The $v^2$ Dependence

Moving Up the Incline (distance $d$ )

Moving Down the Incline (distance $d$ )