Welcome to Work, Energy, and Power!
Welcome to one of the most practical and satisfying chapters in AS Mechanics 1! In previous chapters, you may have used Newton's second law (\(F = ma\)) and the constant acceleration equations (the standard suvat formulae) to track how objects speed up, slow down, and move around. While those tools are great, they can sometimes make problems with changing directions or slopes quite lengthy.
That is where Work and Energy come to the rescue! Energy methods often give you a neat shortcut to solve complex mechanics problems without having to calculate acceleration at every single step. Don't worry if this seems new or challenging at first — we will break down every single formula and idea into easy, bite-sized steps.
Did you know? Energy cannot be created or destroyed; it only changes from one form to another. Whether you are riding a rollercoaster, pedalling a bicycle uphill, or driving a car, the exact same energy rules apply!
1. Work Done by a Constant Force
What is Work Done?
In everyday language, "doing work" might mean doing your homework or lifting weights at the gym. In mechanics, work done has a very strict definition: it measures the amount of energy transferred when a force moves an object through a distance.
If a constant force of magnitude \(F\) acts in the exact same direction as the displacement \(s\), the work done \(W\) is defined as:
\(W = F s\)
Where:
• \(W\) is the work done, measured in Joules (\(\text{J}\)) or Newton-metres (\(\text{N}\,\text{m}\)).
• \(F\) is the constant force applied, measured in Newtons (\(\text{N}\)).
• \(s\) is the distance moved in the direction of the force, measured in metres (\(\text{m}\)).
Force at an Angle to Motion
What happens if the force pulls at an angle \(\theta\) to the direction of motion? Only the component of the force acting parallel to the displacement does work. The perpendicular component does zero work!
\(W = (F \cos\theta) s = F s \cos\theta\)
Analogy: Imagine pulling a sledge across flat snow using a rope tilted at an angle \(\theta\) to the ground. The upward pull balances part of the weight, but only the horizontal part (\(F \cos\theta\)) actually pulls the sledge forward along the snow.
Work Done Against Specific Forces
1. Work Done Against Gravity: When lifting an object of mass \(m\) vertically upwards through a height \(h\) at constant speed, the force required equals its weight (\(mg\)). Therefore:
\(\text{Work done against gravity} = m g h\)
2. Work Done Against Friction/Resistance: If a constant resistive force \(R\) opposes the motion over a distance \(s\), the work done against this resistance is:
\(\text{Work done against resistance} = R s\)
Common Mistakes to Avoid
• Forgetting the angle: Always check whether the force is parallel to the displacement. If it is angled, remember to multiply by \(\cos\theta\).
• Perpendicular forces: Forces acting at \(90^\circ\) to the direction of motion (such as the normal contact force \(R\) on a flat surface) do no work because \(\cos(90^\circ) = 0\).
Key Takeaway for Section 1
Work Done = Force in the direction of motion \(\times\) Distance moved (\(W = F s \cos\theta\)). Work is a scalar quantity measured in Joules (\(\text{J}\)).
2. Kinetic Energy and Gravitational Potential Energy
Kinetic Energy (\(E_k\))
Kinetic energy is the energy possessed by an object due to its motion. Any moving particle with mass has kinetic energy.
\(E_k = \frac{1}{2} m v^2\)
Where:
• \(m\) is the mass of the particle in kilograms (\(\text{kg}\)).
• \(v\) is the speed of the particle in metres per second (\(\text{ms}^{-1}\)).
• \(E_k\) is the kinetic energy in Joules (\(\text{J}\)).
Memory Tip: Notice that speed is squared (\(v^2\)). If you double your speed, your kinetic energy doesn't just double — it quadruples (\(2^2 = 4\))!
Gravitational Potential Energy (\(E_p\))
Gravitational Potential Energy is the energy stored in an object due to its vertical position (height) in a gravitational field.
\(E_p = m g h\)
Where:
• \(m\) is the mass in \(\text{kg}\).
• \(g\) is the acceleration due to gravity (usually \(9.8\,\text{ms}^{-2}\) in CCEA examinations).
• \(h\) is the vertical height in metres (\(\text{m}\)) above an agreed reference level (often called the datum line).
Choosing a Datum: You can choose any convenient level as your baseline where \(h = 0\) (e.g., the lowest point in the problem, or the ground). Any height above this line has positive \(E_p\), and any height below it has negative \(E_p\).
Key Takeaway for Section 2
Kinetic energy depends on speed: \(E_k = \frac{1}{2}mv^2\). Potential energy depends on vertical height: \(E_p = mgh\). Both are scalar quantities measured in Joules (\(\text{J}\)).
3. The Work-Energy Principle & Conservation of Energy
The Conservation of Mechanical Energy
When an object moves under the action of gravity alone, and there are no external driving forces or resistive forces (like friction or air resistance), the total mechanical energy remains constant throughout the motion:
\(\text{Total Initial Mechanical Energy} = \text{Total Final Mechanical Energy}\)
\(E_{k,\text{initial}} + E_{p,\text{initial}} = E_{k,\text{final}} + E_{p,\text{final}}\)
\(\frac{1}{2} m u^2 + m g h_1 = \frac{1}{2} m v^2 + m g h_2\)
Rollercoaster Analogy: At the top of the track, you have maximum potential energy and minimum kinetic energy. As you plunge downward, potential energy converts directly into kinetic energy — you speed up! At the bottom, kinetic energy is at its peak.
The General Work-Energy Principle
In the real world, engines provide driving forces and friction causes energy losses. The general Work-Energy Principle accounts for everything in a single, simple accounting balance:
\(\text{Initial Energy} + \text{Work Done by Driving Forces} = \text{Final Energy} + \text{Work Done Against Resistances}\)
\((E_{k,1} + E_{p,1}) + W_{\text{driving}} = (E_{k,2} + E_{p,2}) + W_{\text{resistance}}\)
Alternatively, the net work done on a particle by all forces equals its change in kinetic energy:
\(\text{Work done by resultant force} = \Delta E_k = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\)
Step-by-Step Method for Solving Problems
Step 1: Draw a clear diagram showing all forces, distances, and heights.
Step 2: Choose a reference level (datum) for zero gravitational potential energy.
Step 3: Identify the start point (1) and end point (2). Calculate \(E_{k,1}\), \(E_{p,1}\), \(E_{k,2}\), and \(E_{p,2}\).
Step 4: Calculate any work done by driving forces (\(F \times s\)) or against resistance (\(R \times s\)).
Step 5: Substitute your expressions into the master equation:
\(\text{Initial } (E_k + E_p) + W_{\text{driving}} = \text{Final } (E_k + E_p) + W_{\text{resistance}}\)
Step 6: Solve for the unknown quantity.
Worked Example: Motion on an Inclined Plane
A box of mass \(4\,\text{kg}\) is released from rest at the top of a rough plane inclined at \(30^\circ\) to the horizontal. The box slides \(5\,\text{m}\) down the slope. A constant frictional force of \(6\,\text{N}\) opposes the motion. Find the speed of the box at the bottom of the slope. (Take \(g = 9.8\,\text{ms}^{-2}\)).
Solution:
• Vertical height lost: \(h = 5 \sin(30^\circ) = 5 \times 0.5 = 2.5\,\text{m}\).
• Initial state (top of slope): Box is at rest \(\implies u = 0 \implies E_{k,1} = 0\). Taking the bottom as datum: \(E_{p,1} = mgh = 4 \times 9.8 \times 2.5 = 98\,\text{J}\).
• Final state (bottom of slope): Box has speed \(v \implies E_{k,2} = \frac{1}{2}(4)v^2 = 2v^2\). Height is zero \(\implies E_{p,2} = 0\).
• Work done against friction: \(W_{\text{fric}} = \text{Friction} \times s = 6 \times 5 = 30\,\text{J}\).
• No driving force: \(W_{\text{driving}} = 0\).
Setting up the energy equation:
\(\text{Initial } (E_k + E_p) = \text{Final } (E_k + E_p) + W_{\text{fric}}\)
\(0 + 98 = 2v^2 + 0 + 30\)
\(2v^2 = 68 \implies v^2 = 34 \implies v = \sqrt{34} \approx 5.83\,\text{ms}^{-1}\)
Key Takeaway for Section 3
Energy is always conserved! The master balance is: Initial Mechanical Energy + Work In (Driving) = Final Mechanical Energy + Work Out (Losses/Friction).
4. Power
What is Power?
Power is the rate at which work is done, or the rate at which energy is transferred with respect to time.
\(P = \frac{W}{t}\)
Where:
• \(P\) is Power in Watts (\(\text{W}\)), where \(1\,\text{W} = 1\,\text{J}\,\text{s}^{-1}\).
• \(W\) is work done in Joules (\(\text{J}\)).
• \(t\) is time taken in seconds (\(\text{s}\)).
Note: In many exam questions, power is given in kilowatts (\(\text{kW}\)). Always remember to convert: \(1\,\text{kW} = 1000\,\text{W}\)!
Power and Velocity (\(P = F v\))
When an engine exerts a forward tractive (driving) force \(F\) on a vehicle moving with instantaneous speed \(v\), the power delivered is:
\(P = F v\)
Rearranging this formula gives the driving force generated by the engine:
\(F = \frac{P}{v}\)
Vehicles Moving at Maximum (Terminal) Speed
When a vehicle reaches its maximum speed on a road or incline, it is no longer accelerating. Therefore, the acceleration is zero (\(a = 0\)), which means the forces acting along the line of motion are in equilibrium (balanced).
• On a level horizontal road: Forward Driving Force = Total Resistance
\(\frac{P}{v_{\text{max}}} = R \implies P = R v_{\text{max}}\)
• Travelling uphill on an incline of angle \(\theta\): Forward Driving Force must overcome both resistance and the downhill component of weight (\(mg\sin\theta\)):
\(F = R + mg\sin\theta \implies \frac{P}{v_{\text{max}}} = R + mg\sin\theta\)
• Travelling downhill on an incline of angle \(\theta\): Gravity helps the vehicle downhill:
\(F + mg\sin\theta = R \implies \frac{P}{v_{\text{max}}} = R - mg\sin\theta\)
Worked Example: Car on an Incline
A car of mass \(1200\,\text{kg}\) travels up a hill inclined at \(\arcsin\left(\frac{1}{14}\right)\) to the horizontal against a constant resistance to motion of \(400\,\text{N}\). The engine works at a constant rate of \(24\,\text{kW}\). Find the maximum speed of the car up the hill. (Take \(g = 9.8\,\text{ms}^{-2}\)).
Solution:
• Convert power to standard units: \(P = 24\,\text{kW} = 24000\,\text{W}\).
• Given: \(\sin\theta = \frac{1}{14}\), \(m = 1200\,\text{kg}\), \(R = 400\,\text{N}\).
• Downhill component of weight: \(mg\sin\theta = 1200 \times 9.8 \times \frac{1}{14} = 840\,\text{N}\).
• At maximum speed, acceleration \(a = 0\), so the driving force \(F\) balances all opposing forces:
\(F = R + mg\sin\theta = 400 + 840 = 1240\,\text{N}\).
• Using \(P = F v_{\text{max}}\):
\(24000 = 1240 \times v_{\text{max}}\)
\(v_{\text{max}} = \frac{24000}{1240} \approx 19.35\,\text{ms}^{-1}\) (or \(19.4\,\text{ms}^{-1}\) to 3 s.f.).
Key Takeaway for Section 4
Power is the rate of energy transfer: \(P = \frac{W}{t}\) or \(P = Fv\). At maximum speed, \(a = 0\), so the driving force \(F = \frac{P}{v}\) exactly balances the opposing forces.
Quick Reference Summary & Exam Checklist
Before sitting your CCEA Mechanics 1 examination, make sure you can confidently use each of these relationships:
• Work Done: \(W = F s \cos\theta\) (in Joules, \(\text{J}\))
• Kinetic Energy: \(E_k = \frac{1}{2}mv^2\)
• Gravitational Potential Energy: \(E_p = mgh\)
• Work-Energy Equation: \(\text{Initial } (E_k + E_p) + W_{\text{driving}} = \text{Final } (E_k + E_p) + W_{\text{resistances}}\)
• Power: \(P = \frac{W}{t}\) and \(P = Fv\) (in Watts, \(\text{W}\))
• Acceleration when non-zero: Use \(F_{\text{net}} = ma \implies \frac{P}{v} - R - mg\sin\theta = ma\)
Keep your units in standard SI (\(\text{kg}\), \(\text{m}\), \(\text{s}\), \(\text{N}\), \(\text{J}\), \(\text{W}\)), always draw a clear diagram, and you will master this topic with ease!