Kinetic Energy and the Work-Energy Theorem

Overview

This page develops two central ideas in mechanics:

• Kinetic energy: energy associated with motion
• Work-energy theorem: net work done on an object equals change in kinetic energy

These ideas provide a powerful alternative to repeated use of Dynamics equations, especially when force acts through a displacement.

This page extends the overview in Work, Energy and Power.

Why It Matters

The work-energy theorem is often the fastest route from forces over distance to changes in speed, especially when time is not involved.

Definition

Kinetic energy is the energy associated with motion:

$$E_k=\frac{1}{2}m{v}^2$$

It is scalar and cannot be negative. The speed $v$ is squared, so two objects moving in opposite directions with the same speed have the same kinetic energy.

Key Representations

$$W=\vec{F}\cdot \vec{s}$$ $$E_k=\frac{1}{2}m{v}^2$$ $$W_{\text{net}}=\Delta E_k$$ $$W_{\text{net}}=E_{k,f}-E_{k,i}$$

Scalar and Vector Distinction

Be precise.

Vector quantities

• Force: $\vec{F}$
• Displacement: $\vec{s}$
• Velocity: $\vec{v}$
• Acceleration: $\vec{a}$

Scalar quantities

• Work: $W$
• Kinetic energy: $E_k$
• Speed: $v$
• Mass: $m$

Kinetic energy depends on speed (magnitude of velocity), not direction.

Kinetic Energy

Definition

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

$$E_k=\frac{1}{2}m{v}^2$$

where:

• $m$ in kg
• $v$ in m s${}^{-1}$

Unit:

$$1\text{ J}=1{\text{ kg m}}^2{\text{s}}^{-2}$$

Key Features

Always Non-Negative

Since $v^2\ge 0$:

$$E_k\ge 0$$

Depends Strongly on Speed

If speed doubles:

$$E_k\propto v^2$$

So kinetic energy becomes four times larger.

Scalar Quantity

Kinetic energy has magnitude only.

Derivation of $E_k=\frac{1}{2}m{v}^2$

Consider an object of mass $m$ accelerating from rest under a constant resultant force.

From dynamics:

$$F=ma$$

From kinematics:

$$v^2=u^2+2as$$

Starting from rest:

$$u=0$$

So:

$$v^2=2as$$

Hence:

$$a=\frac{v^2}{2s}$$

Then:

$$F=ma=m\frac{v^2}{2s}$$

Work done by the force:

$$W=Fs$$ $$W=\left(m\frac{v^2}{2s}\right)s$$ $$W=\frac{1}{2}m{v}^2$$

Thus, the work required to accelerate an object from rest to speed $v$ is:

$$E_k=\frac{1}{2}m{v}^2$$

Work-Energy Theorem

Statement

The net work done on an object equals the change in kinetic energy.

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

That is:

$$W_{\text{net}}=E_{k,f}-E_{k,i}$$

Meaning

If net work is positive:

$$W_{\text{net}}>0$$

then kinetic energy increases and the object speeds up.

If net work is negative:

$$W_{\text{net}}<0$$

then kinetic energy decreases and the object slows down.

If net work is zero:

$$W_{\text{net}}=0$$

then kinetic energy remains constant.

Net Work from Multiple Forces

If several forces act:

$$W_{\text{net}}=\sum W$$

This is equivalent to work done by the resultant force.

Examples:

• engine force positive
• friction negative
• drag negative

Relationship with Dynamics

Dynamics Method

Use:

$$\sum \vec{F}=m\vec{a}$$

when acceleration is required.

Work-Energy Method

Use:

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

when speed, stopping distance or energy change is required.

Often faster and cleaner.

See Dynamics

Common Situations

Constant Driving Force

Engine does positive work, speed increases.

Braking

Braking force does negative work, speed decreases.

Rough Surface

Friction removes mechanical energy.

Falling Object

Gravity does positive work as object moves downward.

Worked Examples

Example 1: Speed Increase from Net Work

A $2.0\text{ kg}$ object starts from rest. Net work done is $50\text{ J}$.

$$\Delta E_k=50$$

Since initial kinetic energy is zero:

$$\frac{1}{2}m{v}^2=50$$ $$\frac{1}{2}(2.0)v^2=50$$ $$v^2=50$$ $$v=7.07{\text{ m s}}^{-1}$$

Example 2: Braking Distance

A $1000\text{ kg}$ car travels at $20{\text{ m s}}^{-1}$. Braking force is $5000\text{ N}$.

Initial kinetic energy:

$$E_k=\frac{1}{2}(1000)(20^2)=2.0\times 10^5\text{ J}$$

Work done by brakes:

$$W=-Fs$$

Set equal to loss in kinetic energy:

$$-5000s=-2.0\times 10^5$$ $$s=40\text{ m}$$

Example 3: Resultant Force Over Distance

A block experiences a constant resultant force of $12\text{ N}$ over $5.0\text{ m}$.

Net work:

$$W_{\text{net}}=Fs=60\text{ J}$$

Hence:

$$\Delta E_k=60\text{ J}$$

Sign Interpretation

Positive Work

Adds kinetic energy.

Examples:

• pulling
• thrust
• gravity on falling body

Negative Work

Removes kinetic energy.

Examples:

• friction
• drag
• braking

Zero Work

No change in kinetic energy from that force.

Examples:

• normal reaction perpendicular to motion
• centripetal force in uniform circular motion

Common Exam Pitfalls

1. Forgetting Net Work

Use total work from all forces, not one force only.

2. Wrong Sign for Friction

Friction usually does negative work.

3. Using Velocity Sign in $E_k$

Use speed magnitude:

$$E_k=\frac{1}{2}m{v}^2$$

always non-negative.

4. Mixing Work-Energy with Mechanical Energy Conservation

If friction exists, pure mechanical energy conservation may fail.

But:

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

still remains valid.

5. Forgetting Units

Energy in J, force in N, speed in m s${}^{-1}$.

Summary

• Kinetic energy is energy of motion:

$$E_k=\frac{1}{2}m{v}^2$$

• Net work done changes kinetic energy:

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

• Positive net work increases speed.
• Negative net work decreases speed.
• This method is powerful for stopping-distance and speed-change problems.

Links

• Work, Energy and Power
• Potential Energy and Conservative Forces
• Energy Forms and Conservation
• Power and Efficiency
• Dynamics
• Kinematics
• Vectors