Introduction to Work, Energy, and Power
Welcome to one of the most fundamental chapters in AS 1: Forces, Energy and Electricity. Whether you are lifting a backpack, sliding down a ramp, or watching a roller coaster loop, the physics of work, potential energy, and kinetic energy describe how motion happens and how energy transfers from one form to another.
Don't worry if equations or derivations have felt daunting in the past. We will break every concept down into clear, manageable steps so you can tackle any question on your CCEA AS 1 paper with confidence.
1. Work Done (\(W\))
What is Work Done?
In physics, work has a very precise definition. Work done (\(W\)) by a constant force is defined as the product of the magnitude of the force \(F\) and the displacement \(s\) (or distance moved) in the direction of the force.
When the applied force acts parallel to the direction of motion, we use the formula:
\(W = F s\)
• \(W\) = work done in Joules (\(\text{J}\))
• \(F\) = constant force in Newtons (\(\text{N}\))
• \(s\) = displacement in metres (\(\text{m}\))
Force at an Angle (\(\theta\))
Often, a force pulls or pushes at an angle to the direction in which the object actually moves (for example, pulling a sledge by a rope angled upwards). Only the component of the force that acts along the line of displacement does work.
\(W = F s \cos\theta\)
Here, \(\theta\) is the angle between the line of action of the force and the direction of the displacement.
Analogy: Imagine pulling a suitcase on wheels along a flat airport floor. Your pulling force points up and forward along the handle. The upward part of your pull simply reduces the contact force with the floor; only the forward horizontal part (\(F\cos\theta\)) actually moves the suitcase forward across distance \(s\).
Units and the Scalar Nature of Work
• The SI unit of work is the Joule (\(\text{J}\)), where \(1\text{ J} = 1\text{ N m} = 1\text{ kg}\cdot\text{m}^2\text{s}^{-2}\).
• Work done is a scalar quantity. It has magnitude, but no direction, even though it is calculated from two vectors (force and displacement).
Work Done by a Variable Force: Graphical Interpretation
Forces are not always constant. When a force varies over a distance (such as stretching a spring or a rocket burning fuel):
• On a Force vs. Displacement (\(F\)–\(s\)) graph, the area under the graph represents the total work done.
Quick Review & Key Takeaway: Always check the direction of motion. If force and displacement are parallel, \(W = F s\). If they are at an angle \(\theta\), use \(W = F s \cos\theta\). On any \(F\)–\(s\) graph, find the area underneath to find work done.
2. Gravitational Potential Energy (\(\Delta E_p\))
Definition and Formula
Gravitational Potential Energy (\(E_p\)) is the energy stored in an object due to its position within a gravitational field.
When an object is raised or lowered near the Earth's surface, the change in gravitational potential energy is given by:
\(\Delta E_p = m g \Delta h\)
• \(m\) = mass of the object in kilograms (\(\text{kg}\))
• \(g\) = acceleration due to gravity / gravitational field strength (\(9.81\text{ m s}^{-2}\) or \(9.81\text{ N kg}^{-1}\))
• \(\Delta h\) = vertical height gained or lost in metres (\(\text{m}\))
Derivation of \(\Delta E_p = m g \Delta h\)
CCEA exams may ask you to show how this formula is derived from first principles:
Step 1: To lift an object of mass \(m\) vertically upwards at a constant speed, you must apply an upward force \(F\) equal and opposite to its weight: \(F = mg\).
Step 2: The work done \(W\) to lift it through a vertical height \(\Delta h\) is: \(W = F \cdot \Delta h\).
Step 3: Substituting \(F = mg\) into the work equation gives: \(W = mg\Delta h\).
Step 4: Since the energy transferred by this work is stored as gravitational potential energy: \(\Delta E_p = m g \Delta h\).
Common Exam Pitfall: Always use the vertical height (\(\Delta h\)), never the distance travelled along an inclined slope (\(s\)), when calculating \(\Delta E_p\). If an object moves a distance \(s\) along a slope inclined at angle \(\alpha\) to the horizontal, \(\Delta h = s\sin\alpha\).
Quick Review & Key Takeaway: \(\Delta E_p\) depends strictly on vertical displacement (\(\Delta h\)). If an object returns to its original vertical height, its net change in \(E_p\) is zero, regardless of the path taken.
3. Kinetic Energy (\(E_k\))
Definition and Formula
Kinetic Energy (\(E_k\)) is the energy an object possesses by virtue of its motion.
\(E_k = \frac{1}{2} m v^2\)
• \(m\) = mass in kilograms (\(\text{kg}\))
• \(v\) = speed or velocity in metres per second (\(\text{m s}^{-1}\))
Derivation from Equations of Constant Acceleration
You need to be able to derive this formula using Newton's Second Law and the kinematic equations:
Step 1: Consider a constant resultant force \(F\) acting on a mass \(m\), accelerating it uniformly from initial velocity \(u\) to final velocity \(v\) over a displacement \(s\). From Newton's Second Law: \(F = ma\).
Step 2: Use the kinematic equation relating velocity, acceleration, and displacement: \(v^2 = u^2 + 2as\).
Step 3: Rearrange for \(as\):
\(as = \frac{v^2 - u^2}{2}\)
Step 4: Write the equation for work done \(W = F s\) and substitute \(F = ma\):
\(W = (ma)s = m(as)\)
Step 5: Substitute the expression for \(as\):
\(W = m\left(\frac{v^2 - u^2}{2}\right) = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\)
Step 6: If the object starts from rest (\(u = 0\)), the total work done on the object equals its kinetic energy:
\(W = E_k = \frac{1}{2}mv^2\)
Quick Review & Key Takeaway: Kinetic energy depends on the square of the speed (\(v^2\)). If you double an object's speed, its kinetic energy increases by a factor of four (\(2^2 = 4\)).
4. Conservation of Mechanical Energy
The Principle
The Principle of Conservation of Mechanical Energy states that in a closed, isolated system where only conservative forces act (ignoring external resistive forces such as air resistance and friction), the total mechanical energy remains constant:
\(E_{\text{total}} = E_k + E_p = \text{constant}\)
Free Fall and Energy Exchange
When an object of mass \(m\) falls freely under gravity from rest through a vertical height \(\Delta h\), all of its lost gravitational potential energy is converted into kinetic energy:
\(\Delta E_p = \Delta E_k\)
\(m g \Delta h = \frac{1}{2} m v^2\)
Notice that the mass \(m\) cancels out from both sides:
\(g \Delta h = \frac{1}{2} v^2 \implies v^2 = 2 g \Delta h \implies v = \sqrt{2 g \Delta h}\)
Did you know? Because mass cancels out, in the absence of air resistance, a bowling ball and a feather dropped from the same height reach the exact same speed just before hitting the ground!
Systems with Friction or Resistive Forces
In real-world situations, resistive forces (friction and drag) act on moving objects. When this happens, mechanical energy is converted into thermal energy:
\(\text{Initial Mechanical Energy} = \text{Final Mechanical Energy} + \text{Work Done against friction/drag}\)
The work done against resistive forces is given by:
\(W_{\text{resistive}} = F_{\text{drag}} \cdot s\)
For example, for a trolley sliding down a slope of length \(s\) and vertical height \(\Delta h\):
\(m g \Delta h = \frac{1}{2} m v^2 + (F_{\text{friction}} \cdot s)\)
Quick Review & Key Takeaway: If there is no friction, equate \(\Delta E_p = \Delta E_k\). If friction is present, remember that some mechanical energy is lost as work done against resistive forces (\(W = F s\)).
5. Power and Efficiency
Power (\(P\))
Power is defined as the rate of doing work or the rate of transferring energy.
\(P = \frac{W}{t} = \frac{\Delta E}{t}\)
• \(P\) = power in Watts (\(\text{W}\))
• \(W\) or \(\Delta E\) = work done or energy transferred in Joules (\(\text{J}\))
• \(t\) = time taken in seconds (\(\text{s}\))
• Unit: \(1\text{ Watt} = 1\text{ J s}^{-1}\)
Power at Constant Velocity (\(P = F v\))
When a vehicle or object moves at a constant velocity \(v\) against a total resistive force \(F\), the engine must supply a forward force equal to \(F\). Since \(W = F s\):
\(P = \frac{F s}{t} = F \left(\frac{s}{t}\right) = F v\)
• \(P\) = power output (\(\text{W}\))
• \(F\) = driving force / resistive force (\(\text{N}\))
• \(v\) = constant speed (\(\text{m s}^{-1}\))
Efficiency
Devices do not convert 100% of input energy into useful work due to energy wasted as heat and sound. Efficiency is the ratio of useful output to total input:
\(\text{Efficiency} = \left(\frac{\text{Useful energy output}}{\text{Total energy input}}\right) \times 100\%\)
Alternatively, using power values:
\(\text{Efficiency} = \left(\frac{\text{Useful power output}}{\text{Total power input}}\right) \times 100\%\)
Efficiency can be written as a percentage or as a decimal between 0 and 1.
Quick Review & Key Takeaway: Power is the speed of energy transfer (\(\text{J s}^{-1}\)). For a vehicle at constant velocity, use \(P = F v\). Efficiency compares useful output to total input.
6. Summary of Common Exam Mistakes to Avoid
1. Incorrect Angle in \(W = F s \cos\theta\):
Always ensure \(\theta\) is the angle between the line of action of the force and the displacement vector. Do not automatically use whatever angle is drawn on the exam diagram without checking where it lies.
2. Confusing Slope Distance with Vertical Height:
In \(\Delta E_p = m g \Delta h\), \(\Delta h\) must strictly be the vertical height. Never substitute the length along a ramp directly into this formula.
3. Neglecting Friction in Energy Conservation:
If the question mentions friction or air drag, do not use \(m g \Delta h = \frac{1}{2} m v^2\). You must add the work done against friction: \(m g \Delta h = \frac{1}{2} m v^2 + F s\).
4. Algebraic Rearrangement of \(E_k\):
When solving for \(v\) in \(E_k = \frac{1}{2} m v^2\), remember that \(v = \sqrt{\frac{2 E_k}{m}}\). A common error is forgetting to take the square root or mistakenly squaring the mass.
5. Vector vs. Scalar Confusion:
Remember that work, kinetic energy, and potential energy are all scalars. When adding energies, you add them directly as ordinary numbers—never resolve them into vector components.