Work done, potential energy and kinetic energy

Welcome to Work, Energy, and Power

Welcome to one of the most fundamental and exciting topics in Physics! Whether you are watching a rollercoaster dive down a steep drop, kicking a football into the top corner, or simply walking up a flight of stairs, the concepts of work done, kinetic energy, and potential energy are constantly at play.

Don't worry if physics formulas sometimes feel overwhelming. In this guide, we will break down every single idea into simple, manageable steps with clear examples and memory tricks. By the end of these notes, you will feel confident tackling exam questions on this topic!

1. Work Done by a Force

What is "Work" in Physics?

In everyday language, "work" means doing a job or studying for an exam. But in physics, work done has a very precise scientific meaning: work is done whenever a force moves an object through a distance in the direction of the force.

If you push against a solid brick wall with all your strength until you are sweating, you might feel exhausted, but in physics, you have done zero work on the wall because the wall did not move!

The Work Done Formula (Force Parallel to Motion)

When a constant force \(F\) acts directly along the line of motion of an object, causing it to move a displacement \(s\), the work done \(W\) is calculated using:

\(W = F s\)

Where:

• \(W\) = Work done, measured in Joules (\(\text{J}\))
• \(F\) = Applied force, measured in Newtons (\(\text{N}\))
• \(s\) = Displacement in the direction of the force, measured in metres (\(\text{m}\))

Definition of the Joule: One Joule (\(1\text{ J}\)) is defined as the work done when a force of \(1\text{ N}\) moves an object through a displacement of \(1\text{ m}\) in the direction of the force. Therefore, \(1\text{ J} = 1\text{ N m}\). In SI base units, \(1\text{ J} = 1\text{ kg m}^2\text{ s}^{-2}\).

What Happens When the Force is at an Angle?

Imagine pulling a suitcase along a flat airport floor using a handle tilted at an angle \(\theta\) to the ground. Only the horizontal component of your pull actually moves the suitcase forward! The vertical component simply reduces the contact force with the floor.

When the force \(F\) acts at an angle \(\theta\) to the direction of displacement \(s\), we resolve the force into the direction of motion:

\(W = F s \cos\theta\)

Special Angles to Remember:

1. Force in the exact direction of motion (\(\theta = 0^\circ\)):
Since \(\cos(0^\circ) = 1\), the formula simplifies to \(W = F s\). (Maximum positive work done).

2. Force perpendicular to motion (\(\theta = 90^\circ\)):
Since \(\cos(90^\circ) = 0\), the work done is \(W = 0\text{ J}\).
Example: The Moon orbiting the Earth experiences a gravitational force directed towards Earth's centre, at \(90^\circ\) to its circular path. Therefore, gravity does no work on the Moon in a circular orbit!

3. Force opposing motion (\(\theta = 180^\circ\)):
Since \(\cos(180^\circ) = -1\), the work done is negative: \(W = -F s\).
Example: Friction and air resistance act in the opposite direction to velocity, doing negative work and removing kinetic energy from the system (converting it into thermal energy).

Work Done from Force-Displacement Graphs

In the real world, forces are not always constant (think of stretching a spring or the thrust of a rocket). To find the work done from a graph of Force (\(F\)) against Displacement (\(s\)):

Work Done = Area under the Force–Displacement graph

• Constant Force: The area is a simple rectangle: \(\text{Area} = \text{base} \times \text{height} = s \times F\).
• Linearly Increasing Force (e.g., stretching a spring): The area is a triangle: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} F s\).

Key Takeaway for Section 1: Work done is energy transferred by a force (\(W = F s \cos\theta\)). Always ensure you use the component of force parallel to the displacement, and remember that the area under a force-displacement graph represents work done.

2. Kinetic Energy (\(E_k\))

What is Kinetic Energy?

Kinetic energy (\(E_k\)) is the energy possessed by an object due to its motion. Any object with mass that has speed carries kinetic energy.

The standard equation for kinetic energy is:

\(E_k = \frac{1}{2} m v^2\)

Where:

• \(E_k\) = Kinetic energy in Joules (\(\text{J}\))
• \(m\) = Mass of the object in kilograms (\(\text{kg}\))
• \(v\) = Speed of the object in metres per second (\(\text{m s}^{-1}\))

Step-by-Step Derivation of \(E_k = \frac{1}{2}mv^2\)

CCEA AS Physics frequently asks you to derive physical relationships. Here is how we derive the kinetic energy equation step-by-step using Newton's second law and the equations of uniform acceleration:

Step 1: Start with the definition of work done on an object by a resultant force \(F\) accelerating it over a displacement \(s\):
\(W = F s\)

Step 2: Apply Newton's Second Law (\(F = m a\)):
\(W = (m a) s\)

Step 3: Use the kinematic equation \(v^2 = u^2 + 2as\). Rearrange this equation to express acceleration multiplied by displacement (\(a s\)):
\(v^2 - u^2 = 2as \implies a s = \frac{v^2 - u^2}{2}\)

Step 4: Substitute \((a s)\) into our work equation:
\(W = m(a s) = m\left(\frac{v^2 - u^2}{2}\right) = \frac{1}{2} m v^2 - \frac{1}{2} m u^2\)

Step 5: If the object starts from rest (\(u = 0\text{ m s}^{-1}\)), the initial kinetic energy is zero. The work done on the object directly equals the final kinetic energy gained:
\(E_k = \frac{1}{2} m v^2\)

The Power of the Velocity-Squared (\(v^2\)) Term

Notice that kinetic energy is proportional to the square of the speed (\(E_k \propto v^2\)):

• If you double the speed (\(\times 2\)): Kinetic energy increases by \(2^2 = 4\) times.
• If you triple the speed (\(\times 3\)): Kinetic energy increases by \(3^2 = 9\) times!

Real-world connection: This is why speeding in a car is so dangerous. A car travelling at \(60\text{ mph}\) has four times as much kinetic energy to dissipate during braking compared to a car travelling at \(30\text{ mph}\), requiring a much longer braking distance.

Key Takeaway for Section 2: Kinetic energy is the energy of movement (\(E_k = \frac{1}{2}mv^2\)). Because velocity is squared, small increases in speed cause large increases in energy.

3. Gravitational Potential Energy (\(\Delta E_p\))

What is Gravitational Potential Energy?

Gravitational potential energy (\(E_p\)) is the energy stored in an object due to its vertical position within a gravitational field. When you lift an object upwards, you do work against the downward pull of gravity. That work done is stored as gravitational potential energy.

The Formula for \(\Delta E_p\)

In a uniform gravitational field (such as near Earth's surface where \(g\) is constant at \(9.81\text{ m s}^{-2}\) or \(9.81\text{ N kg}^{-1}\)), the change in gravitational potential energy is given by:

\(\Delta E_p = m g \Delta h\)

Where:

• \(\Delta E_p\) = Change in gravitational potential energy in Joules (\(\text{J}\))
• \(m\) = Mass of the object in kilograms (\(\text{kg}\))
• \(g\) = Acceleration due to gravity (\(9.81\text{ m s}^{-2}\) or \(9.81\text{ N kg}^{-1}\))
• \(\Delta h\) = Vertical height change in metres (\(\text{m}\))

Step-by-Step Derivation of \(\Delta E_p = m g \Delta h\)

Step 1: The upward force \(F\) needed to lift an object at constant speed must balance its weight \(W_{\text{weight}}\):
\(F = m g\)

Step 2: Work done to lift this object vertically through a height \(\Delta h\) is:
\(\text{Work done} = F \times \Delta h\)

Step 3: Substitute \(F = m g\) into the work equation:
\(\text{Work done} = (m g) \Delta h = m g \Delta h\)

Step 4: Since the work done is transferred into stored potential energy:
\(\Delta E_p = m g \Delta h\)

Important Tip: Only the vertical height matters! If a trolley travels up a smooth slope of length \(10\text{ m}\) to reach a vertical height of \(3\text{ m}\), you use \(\Delta h = 3\text{ m}\) when calculating the gain in gravitational potential energy.

Key Takeaway for Section 3: Gravitational potential energy depends strictly on mass, the strength of the gravitational field, and the vertical change in height (\(\Delta E_p = mg\Delta h\)).

4. The Principle of Conservation of Energy

The Law of Conservation of Energy

The principle of conservation of energy states that: Energy cannot be created or destroyed; it can only be transformed from one form into another.

In a closed, isolated system, the total amount of energy remains completely constant:

\(E_{\text{total, initial}} = E_{\text{total, final}}\)

The Interchange Between \(E_p\) and \(E_k\) (Ideal Systems)

Consider an object of mass \(m\) dropped from rest at a height \(h\) above the ground with no air resistance:

• At the top: The object has maximum gravitational potential energy and zero kinetic energy (\(E_p = mgh\), \(E_k = 0\)).
• While falling: As height decreases, \(E_p\) is continuously converted into \(E_k\).
• Just before impact: All potential energy has been converted into kinetic energy (\(E_p = 0\), \(E_k = \frac{1}{2}mv^2\)).

By equating the loss in \(E_p\) to the gain in \(E_k\):

\(m g h = \frac{1}{2} m v^2\)

Notice that the mass \(m\) cancels out on both sides!

\(g h = \frac{1}{2} v^2 \implies v = \sqrt{2 g h}\)

Did you know? Because mass cancels out, in the absence of air resistance, a bowling ball and a feather dropped from the same height hit the ground with the exact same speed!

Real-World Systems (Accounting for Resistive Forces)

In real situations, friction and air resistance do work against the moving object. This work removes mechanical energy and converts it into thermal energy (heat) and sound.

The energy balance equation becomes:

\(\text{Initial Total Energy} = \text{Final Total Energy} + \text{Work Done against Resistive Forces}\)

For an object falling or sliding down a slope:

\(m g h = \frac{1}{2} m v^2 + W_{\text{friction}}\)

Where the work done against friction is given by:

\(W_{\text{friction}} = f \times d\)

(\(f\) is the average frictional force in \(\text{N}\), and \(d\) is the distance travelled along the surface in \(\text{m}\)).

Key Takeaway for Section 4: Energy is always conserved. In ideal mechanical systems, \(\Delta E_p = \Delta E_k\). In real systems with friction, some mechanical energy is converted into heat: \(\Delta E_p = \Delta E_k + W_{\text{friction}}\).

5. Quick Review & Common Exam Pitfalls

Quick Formula Reference

• Work Done (at angle \(\theta\)): \(W = F s \cos\theta\)
• Kinetic Energy: \(E_k = \frac{1}{2} m v^2\)
• Gravitational Potential Energy: \(\Delta E_p = m g \Delta h\)
• Work done against friction: \(W = f \times d\)
• Free-fall speed from height \(h\) (no drag): \(v = \sqrt{2gh}\)

Common Mistakes to Avoid in Exams

1. Forgetting to square velocity: When calculating \(\frac{1}{2}mv^2\), always calculate \(v^2\) first. Don't multiply \(m\) by \(v\) and then square the whole thing!

2. Using the wrong distance for \(\Delta E_p\): Always use the vertical height (\(h\)) for gravitational potential energy, never the diagonal length of the ramp unless calculating work against friction along the surface.

3. Unit conversions: Make sure mass is in kilograms (\(\text{kg}\)), displacement is in metres (\(\text{m}\)), and speed is in metres per second (\(\text{m s}^{-1}\)) before putting numbers into formulas (e.g., convert grams to kilograms by dividing by \(1000\), and \(\text{km h}^{-1}\) to \(\text{m s}^{-1}\) by dividing by \(3.6\)).

4. Confusing Force with Work: Remember that force is measured in Newtons (\(\text{N}\)) and energy/work is measured in Joules (\(\text{J}\)). Work is force multiplied by distance!