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Work

Lesson ~11 min read 8 MCQs

In simple terms: In simple terms, work is about how a force changes an object's energy when it moves that object over a distance. It's the physics way of measuring how "effective" a push or pull is.

Why this matters

Imagine you're helping your friend Liam move. You spend all afternoon pushing a heavy sofa across the living room floor. It's exhausting! Then, you spend another ten minutes pushing with all your might against a wall that just won't budge. You feel equally tired from both, but in the world of physics, you only did "work" on the sofa. Why?

Physics has a very specific definition of work that's different from our everyday use of the word. It’s not just about effort; it’s about accomplishment. It’s about a force successfully causing motion. Understanding this precise definition is the key to unlocking one of the most powerful ideas in physics: the relationship between forces and energy. We'll break down what work really is, how to calculate it, and how it explains the energy changes happening all around you.

Diagram

Work Done by a Force at an Angle A diagram showing a box on a flat surface being pulled by a force at an angle. The force vector is broken into its parallel and perpendicular components. The formula W = Fd cos(θ) is shown, relating the force, displacement, and angle to the work done. Displacement, d = 20 m F = 50 N F_∥ = F cos(θ) F_⊥ θ = 30° Work = F_∥ * d W = Fd cos(θ) W = (50 N)(20 m)cos(30°) W = 866 J Only the component of the force parallel to the displacement (F_∥) does work on the object. The perpendicular component (F_⊥) does zero work.
This diagram shows a box on a flat surface being pulled by a force at a 30-degree angle. The force vector is broken down into its horizontal (parallel) and vertical (perpendicular) components. The diagram illustrates that only the parallel component of the force contributes to the work done on the box as it moves horizontally, reinforcing the formula W = Fd cos(θ).

Concept map

flowchart TD
    A[Start: A force acts on an object] --> B{Does the object move? (d > 0)};
    B -- No --> C[No work is done. W = 0];
    B -- Yes --> D{What is the angle θ between F and d?};
    D --> E{Is θ = 90°?};
    E -- Yes --> C;
    E -- No --> F[Calculate W = Fd cos(θ)];
    F --> G{Is W > 0?};
    G -- Yes --> H[Positive Work: Energy is added to the system];
    G -- No --> I{Is W < 0?};
    I -- Yes --> J[Negative Work: Energy is removed from the system];
    I -- No --> K[Zero Work: No change in energy from this force];

Core explanation

Hello everyone! I'm Saavi, and I'm so glad you're here. Today, we're tackling a word you use every day—work—and giving it a precise, powerful new meaning.

What is Work, Really?

In everyday language, "work" can mean your job, your homework, or any strenuous effort. But in physics, the definition is much stricter:

Work is done on an object when a force causes that object to move some distance.

If there's no force, there's no work. If there's a force but no movement (like pushing on that stubborn wall), there's no work. Both force and displacement must happen.

Think of work as an energy transaction. When you do positive work on a system, you are depositing energy into it. When you do negative work, you are withdrawing energy. This makes work a scalar quantity, not a vector. It has a magnitude and a sign (positive, negative, or zero), but no direction.

The Magic Formula: W = Fd cos(θ)

So how do we calculate work? The formula is your new best friend:

W = F_∥d = Fd cos(θ)

Let's break this down:

  • W is the work done, measured in Joules (J).
  • F is the magnitude of the constant force you're applying, in Newtons (N).
  • d is the magnitude of the object's displacement, in meters (m).
  • θ (theta) is the angle between the force vector and the displacement vector.

This cos(θ) part is everything. It tells us that only the component of the force that is parallel to the direction of motion does work.

Imagine you're pulling a rolling suitcase through the airport. The handle is at an angle, but the suitcase moves horizontally. The part of your pull that's directed forward is doing positive work and adding kinetic energy. The part of your pull that's lifting the suitcase slightly upward is perpendicular to the motion, so it does zero work.

  • If force and displacement are in the same direction, θ = 0°. Since cos(0°) = 1, work is positive and at its maximum: W = Fd.
  • If force and displacement are in opposite directions, θ = 180°. Since cos(180°) = -1, work is negative: W = -Fd. This happens with friction, which always opposes motion and removes energy from the system.
  • If force and displacement are perpendicular, θ = 90°. Since cos(90°) = 0, no work is done. Think of the normal force on a box sliding horizontally. It pushes up, the box moves sideways. No work is done by the normal force.

The Work-Energy Theorem: The Grand Unifier

This is one of the most important concepts in the entire course. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.

ΔK = W_net

Where W_net is the sum of the work done by all the forces acting on the object (ΣWᵢ).

Let's say you push a cart on a flat surface. You apply a force, and friction applies a force in the opposite direction.

  • W_net = W_you + W_friction
  • If W_net is positive, the cart's kinetic energy increases (it speeds up).
  • If W_net is negative, its kinetic energy decreases (it slows down).
  • If W_net is zero, its kinetic energy is constant (it moves at a steady speed).

This theorem is powerful because it connects the ΣF = ma world of forces directly to the energy of the system, without you having to calculate acceleration or time.

Conservative vs. Nonconservative Forces

This is where many students slip up, but you can master this. Forces can be sorted into two teams: conservative and nonconservative.

Team Conservative:

  • Examples: Gravity, the force from an ideal spring.
  • Key Idea: The work they do is path-independent. It only depends on the start and end points.
  • Analogy: Imagine hiking from the base of a mountain to the summit. The work done by gravity on you is the same whether you take the steep, direct trail or the long, winding switchbacks. It only depends on your change in elevation.
  • If you start and end at the same point (a round trip), the net work done by a conservative force is zero.
  • These are the only forces that have a potential energy associated with them (like gravitational potential energy).

Team Nonconservative:

  • Examples: Friction, air resistance, the force from your hand pushing something.
  • Key Idea: The work they do is path-dependent. The longer the path, the more work they do.
  • Analogy: Think about pushing a box across a sandy beach. The work you do against friction depends entirely on the path you take. A longer, winding path will lose much more energy to friction (as heat) than a short, direct path.
  • The energy "lost" to friction is calculated as W_friction = -F_f * d, where F_f is the force of friction. This energy doesn't vanish; it's dissipated, usually as heat.

Work from a Graph

The AP exam loves graphical analysis. If you see a graph of Force vs. Displacement (F_∥ vs. d), you need to know one thing:

Work is the area under the curve of a Force-Displacement graph.

  • If the force is constant, the area will be a rectangle (Area = base × height = d × F).
  • If the force changes linearly, the area will be a triangle (Area = ½ × base × height) or a trapezoid.
  • If the force is below the axis (a negative force), the area is negative, and it represents negative work.

You just calculate the geometric area of the sections to find the total work done.

That's the core of it! Work isn't just effort; it's a force causing displacement and changing a system's energy. By understanding the formula, the work-energy theorem, and the two types of forces, you've built a powerful new toolkit for solving physics problems.

Worked examples

Let's put these ideas into practice. The key is to be methodical: identify forces, check angles, and apply the right formula.

Example 1

Pulling a Sled

Priya pulls her little brother on a sled across a snowy, flat field. She pulls with a force of 50 N on a rope angled at 30° above the horizontal. How much work does she do if she pulls the sled a distance of 20 meters?

1. Visualize and Identify:

  • Force F = 50 N
  • Displacement d = 20 m
  • Angle θ = 30° (This is the angle between her force and the sled's motion)

2. Choose the Right Equation: We have force, distance, and the angle between them. The perfect tool is the work formula: W = Fd cos(θ)

3. Plug and Solve: W = (50 N) * (20 m) * cos(30°) Remember from your calculator, cos(30°) ≈ 0.866. W = 1000 * 0.866 W = 866 J

Example 2

The Work-Energy Theorem in Action

A 10 kg crate is initially at rest. Carlos pushes it horizontally with a 100 N force for 5 meters across a floor with a kinetic friction force of 30 N. What is the final speed of the crate?

1. Identify all forces doing work:

  • Applied Force (Carlos)
    F_app = 100 N. It's in the direction of motion, so θ = 0°.
  • Friction Force
    F_f = 30 N. It's opposite the direction of motion, so θ = 180°.
  • Gravity and Normal Force
    Both are perpendicular to the motion (θ = 90°), so they do zero work. We can ignore them for the work calculation.

2. Calculate the work done by each force:

  • W_app = F_app * d * cos(0°) = (100 N)(5 m)(1) = +500 J
  • W_friction = F_f * d * cos(180°) = (30 N)(5 m)(-1) = -300 J

3. Calculate the Net Work: W_net = W_app + W_friction = 500 J + (-300 J) = 200 J

4. Apply the Work-Energy Theorem: W_net = ΔK = K_final - K_initial The crate starts from rest, so K_initial = 0. 200 J = K_final - 0 K_final = 200 J

5. Solve for the final speed: We know K = ½mv². 200 J = ½ * (10 kg) * v_final² 200 = 5 * v_final² 40 = v_final² v_final = √40 ≈ 6.32 m/s

Try it yourself

Ready to test your understanding? Give these a shot.

Problem 1: A shopper pushes a 15 kg grocery cart with a constant force of 40 N, directed 25° below the horizontal. The cart moves 10 m down a long aisle. The force of friction opposing the motion is 12 N. a) How much work does the shopper do? b) How much work does friction do? c) What is the net work done on the cart?

Hint: Draw a free-body diagram first! Remember that the work done by friction is always negative because it opposes the direction of motion.

Problem 2: The graph below shows the net force applied to a model rocket as it moves along a track. How much work is done on the rocket from x = 0 m to x = 10 m? What is the rocket's change in kinetic energy over this interval?

(Imagine a graph where the force is 20 N from x=0 to x=4, then drops linearly from 20 N to 0 N from x=4 to x=10.)

Hint: Work is the area under the F-vs-x graph. Break the area into a rectangle and a triangle. How does the total work relate to the change in kinetic energy?

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