Q.A block of mass $m = 1\ \text{kg}$, moving on a horizontal surface with speed $v_i = 2\ \text{m s}^{-1}$ enters a rough patch ranging from $x = 0.10\ \text{m}$ to $x = 2.01\ \text{m}$. The retarding force $F_r$ on the block in this range is inversely proportional to $x$ over this range,

[!FORMULA] $F_r = -\frac{k}{x} \quad \text{for } 0.1 < x < 2.01\ \text{m}$

[!FORMULA] $= 0 \quad \text{for } x < 0.1\ \text{m and } x > 2.01\ \text{m}$

where $k = 0.5\ \text{J}$. What is the final kinetic energy and speed $v_f$ of the block as it crosses this patch?

The Work-Energy Theorem: From Intuition to Precision

Imagine pushing a heavy box across the floor. The harder you push and the farther it slides, the faster it moves when you let go. That connection — the push (force) over a distance (displacement) changing the box's speed — is exactly what the Work-Energy Theorem captures.

The Intuition First

Think of work as the "currency" that buys motion. When you do work on an object, you transfer energy to it. That energy shows up as kinetic energy — the energy of motion. The more work you do, the more the object's kinetic energy changes.

If you push a stationary ball, it starts moving. If you push a moving ball in the same direction, it speeds up. If you push against its motion, it slows down. In every case, the work done equals the change in the ball's kinetic energy.

The Precise Statement

Where:

• $W_{\text{net}}$ is the net work done on the object (the total work from all forces combined)
• $K_f$ is the final kinetic energy
• $K_i$ is the initial kinetic energy

And kinetic energy is defined as:

So the theorem can also be written as:

Why "Net" Work Matters

This is the most common point of confusion. The theorem uses net work — the work done by the net force (the vector sum of all forces). If you push a box and friction opposes it, the net work is the work you do minus the work friction does. Only that net amount changes the kinetic energy.

A Simple Derivation (for constant force)

Consider a constant net force $F_{\text{net}}$ acting on an object of mass $m$ over a displacement $s$. From Newton's second law:

From kinematics (constant acceleration):

Multiply both sides by $\frac{1}{2}m$:

But $mas = F_{\text{net}} s = W_{\text{net}}$, so:

Rearranging:

What It Tells You (and What It Doesn't) …