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Work-Energy
Theorem

Master how net work done on an object equals its change in kinetic energy, and apply this powerful principle to solve real-world physics problems.

High School Physics

What is Work in Physics?

In physics, work has a precise meaning: it is the energy transferred to or from an object by a force acting over a displacement. Work is done only when a force has a component along the direction of motion.

SI Unit
Joule (J) = \(1\,\text{N} \cdot \text{m}\)
Scalar Quantity
Work has magnitude but no direction

Force, Displacement, and Angle

\(m\) \(d\) \(F\) \(\theta\) \(F\cos\theta\)

What is Kinetic Energy?

Kinetic energy is the energy an object possesses due to its motion. Any object with mass and velocity has kinetic energy. It is always a non-negative scalar quantity.

Doubles Speed?
\( KE \) quadruples because \( v \) is squared
At Rest?
\( v = 0 \implies KE = 0 \)

The Work-Energy Theorem

The Work-Energy Theorem states that the net work done on an object equals its change in kinetic energy. This is a direct consequence of Newton's Second Law applied over a displacement.

\( W_{\text{net}} > 0 \)
Object speeds up: kinetic energy increases
\( W_{\text{net}} < 0 \)
Object slows down: kinetic energy decreases
\( W_{\text{net}} = 0 \)
Speed is constant: no change in \( KE \)

Why Does It Work?

1
Start with Newton's Second Law: \( F_{\text{net}} = ma \)
2
Multiply both sides by displacement \( d \): \( F_{\text{net}} \cdot d = m \cdot a \cdot d \)
3
Apply kinematics: \( v_f^2 = v_i^2 + 2ad \), so \( ad = \frac{v_f^2 - v_i^2}{2} \)
4
Substitute: \( W_{\text{net}} = m \cdot \frac{v_f^2 - v_i^2}{2} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 \)

Net Work from Multiple Forces

\(m\) \(F_{\text{app}}\) \(f_k\) \(N\) \(mg\) \(d\)

Work vs Energy: Key Distinction

Work Kinetic Energy
What is it? Energy transfer process Energy state of an object
Sign Can be positive, negative, or zero Always \( \geq 0 \)
Depends on Force, displacement, angle Mass, speed
Formula \( W = Fd\cos\theta \) \( KE = \frac{1}{2}mv^2 \)
Unit Joule (J) Joule (J)

How to Apply the Theorem

1
Identify the object and the interval of motion (start and end points).
2
Draw a free-body diagram showing all forces acting on the object.
3
Calculate each force's work: \( W = Fd\cos\theta \) for each force.
4
Sum all works to get \( W_{\text{net}} \).
5
Set equal to \( \Delta KE \): \( W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 \). Solve for the unknown.

Example 1: Pushing a Crate

A 10 kg crate starts from rest. A horizontal force of \(F = 50\,\text{N}\) pushes it across a frictionless floor for \(d = 6\,\text{m}\). Find the final speed.

\(10\,\text{kg}\) \(50\,\text{N}\) \(6\,\text{m}\) \(v_i = 0\) \(v_f = ?\)

Example 2: Friction Slowing a Sled

A 15 kg sled is moving at \(v_i = 8\,\text{m/s}\). Kinetic friction (\(f_k = 30\,\text{N}\)) opposes motion over \(d = 10\,\text{m}\). Find the final speed.

Example 3: Force at an Angle

A 5 kg box starts at rest. A force of \(F = 40\,\text{N}\) is applied at \(\theta = 30^\circ\) above horizontal over \(d = 4\,\text{m}\) on a frictionless surface. Find \( v_f \).

Work as Area Under the Curve

\(x\,\text{(m)}\) \(F\,\text{(N)}\) 2 4 6 8 20 40 Constant: \(W = F \cdot d\) Variable: \(W = \text{Area}\) 0

Positive, Negative, and Zero Work

Positive Work
\( 0^\circ \le \theta < 90^\circ \)
Force has a component along motion. Object speeds up.
Negative Work
\( 90^\circ < \theta \le 180^\circ \)
Force has a component opposing motion. Object slows down.
Zero Work
\( \theta = 90^\circ \)
Force is perpendicular to motion. No energy transfer.

Common Mistakes

✕ Forgetting \( \cos\theta \)
Using \( W = Fd \) when force is at an angle. Always include the angle factor: \( W = Fd\cos\theta \).
✕ Using total force
Applying the theorem with the work from just one force. The theorem requires the net work, which is the sum of the work done by all individual forces.
✕ Sign errors
Forgetting that friction does negative work. When \( \theta = 180^\circ \), \( \cos 180^\circ = -1 \).

Summary of Key Formulas

Concept Formula When to Use
Work (constant force) \( W = Fd\cos\theta \) Force at angle to displacement
Kinetic Energy \( KE = \frac{1}{2}mv^2 \) Energy of a moving object
Work-Energy Theorem \( W_{\text{net}} = \Delta KE \) Relating force/distance to speed change
Net Work (sum) \( W_{\text{net}} = \sum W_i \) Multiple forces on one object
Work (variable force) \( W = \int F \, dx \) Force changes with position

Question 1 of 5

A 4 kg object is pushed with a horizontal force of 20 N over a distance of 5 m on a frictionless surface, starting from rest. What is its final kinetic energy?

Question 2 of 5

A car has a kinetic energy of 8000 J. The brakes apply a friction force of 400 N until the car stops. How far does it travel while braking?

Question 3 of 5

A force of 60 N is applied at \( 60^\circ \) above the horizontal to pull a sled 10 m across flat ground. How much work does this force do?

Question 4 of 5

A 2 kg ball is thrown upward with \( v_i = 10 \) m/s. Using the Work-Energy Theorem, how much work does gravity do as it rises to its highest point?

Question 5 of 5

A 3 kg block initially at rest on a rough surface is pushed by a 24 N horizontal force over 10 m. The friction force is 9 N. What is the final speed?