Mechanics 1: Power

Welcome to the study guide on Power for CCEA AS 2 Further Mathematics (Mechanics 1). Whether you are looking to secure full marks or trying to wrap your head around the mechanics of engines and pumps, this guide will break down each concept step by step.

In everyday language, we often use words like "force", "energy", and "power" interchangeably. In mechanics, however, they mean very specific things. Power tells us how fast work is being done or how rapidly energy is being transferred. Let's dive in!

1. Fundamental Concepts & Definitions

What is Power?

Power (\(P\)) is defined as the rate of doing work with respect to time, or the rate of energy transfer.

Mathematically, instantaneous power is the derivative of work with respect to time:

\(P = \frac{\mathrm{d}W}{\mathrm{d}t}\)

If work is done at a steady rate over a time interval \(\Delta t\), we use Average Power:

\(P_{\text{avg}} = \frac{\text{Total Work Done}}{\text{Total Time Taken}} = \frac{\Delta W}{\Delta t}\)

Units of Power

The standard SI unit for power is the Watt (\(\text{W}\)).

\(1\text{ Watt} = 1\text{ Joule per second} = 1\text{ J s}^{-1} = 1\text{ N m s}^{-1} = 1\text{ kg m}^2\text{ s}^{-3}\)

In examination questions, power is very frequently given in larger multiples:

• Kilowatt (\(\text{kW}\)): \(1\text{ kW} = 10^3\text{ W} = 1000\text{ W}\)
• Megawatt (\(\text{MW}\)): \(1\text{ MW} = 10^6\text{ W} = 1\,000\,000\text{ W}\)

Exam Note: For all CCEA Mechanics units, take the acceleration due to gravity as \(g = 9.8\text{ m s}^{-2}\) unless a question explicitly states otherwise.

The Power of a Moving Force: \(P = Fv\)

When a constant or instantaneous force \(\mathbf{F}\) moves an object at a velocity \(\mathbf{v}\), the rate at which that force does work is:

\(P = \mathbf{F} \cdot \mathbf{v} = Fv\cos(\theta)\)

When the driving force acts directly along the line of motion (\(\theta = 0^{\circ}\) and \(\cos(0^{\circ}) = 1\)), this simplifies to:

\(P = Fv\)

Where:
• \(P\) is the instantaneous power in Watts (\(\text{W}\))
• \(F\) is the driving or tractive force in Newtons (\(\text{N}\))
• \(v\) is the instantaneous speed in metres per second (\(\text{m s}^{-1}\))

Key Takeaway: Power is the product of driving force and speed (\(P = Fv\)). Rearranging this gives the driving force as \(F = \frac{P}{v}\).

2. Vehicles in Motion on Horizontal Roads

Tractive Force and Newton's Second Law

When a vehicle moves, its engine generates a forward mechanical force known as the tractive force (\(F_{\text{tract}}\) or simply \(F\)). At an instantaneous speed \(v\), an engine delivering power \(P\) produces a tractive force:

\(F = \frac{P}{v}\)

As the vehicle moves forward, it experiences resistive forces \(R\) (such as air resistance, road friction, and internal mechanical drag). Applying Newton's Second Law (\(F_{\text{net}} = ma\)) in the direction of motion gives:

\(F_{\text{tract}} - R = ma\)

\(\frac{P}{v} - R = ma\)

Where \(m\) is the mass of the vehicle in \(\text{kg}\), and \(a\) is the instantaneous acceleration in \(\text{m s}^{-2}\).

Maximum (Terminal) Speed on Level Ground

As a vehicle speeds up under constant power, its tractive force \(F = \frac{P}{v}\) decreases. Eventually, the forward tractive force becomes exactly equal to the total resistance \(R\). At this point, the acceleration becomes zero (\(a = 0\)), and the vehicle reaches its maximum speed (\(v_{\text{max}}\)):

\(\frac{P_{\text{max}}}{v_{\text{max}}} - R = 0 \implies \frac{P_{\text{max}}}{v_{\text{max}}} = R \implies P_{\text{max}} = R \cdot v_{\text{max}}\)

CRITICAL WARNING: Why SUVAT Does Not Apply

Don't fall into this common trap: if a vehicle accelerates under constant power, the driving force \(F = \frac{P}{v}\) decreases as speed \(v\) increases. Consequently, the acceleration \(a = \frac{\frac{P}{v} - R}{m}\) is not constant. Therefore, you cannot use the constant acceleration equations (SUVAT) when an engine is operating at constant power!

Worked Example 1: Horizontal Motion

A car of mass \(1200\text{ kg}\) travels along a straight horizontal road against a constant resistance of \(600\text{ N}\). The engine works at a constant rate of \(48\text{ kW}\).
(a) Find the acceleration of the car when its speed is \(15\text{ m s}^{-1}\).
(b) Determine the maximum speed of the car.

Solution:
Part (a):
1. Convert power to Watts: \(P = 48\text{ kW} = 48 \times 10^3\text{ W} = 48\,000\text{ W}\).
2. Find the tractive force at \(v = 15\text{ m s}^{-1}\):
\(F = \frac{P}{v} = \frac{48\,000}{15} = 3200\text{ N}\).
3. Apply Newton's Second Law (\(F - R = ma\)):
\(3200 - 600 = 1200a\)
\(2600 = 1200a\)
\(a = \frac{2600}{1200} = 2.17\text{ m s}^{-2}\) (to 3 s.f.).

Part (b):
At maximum speed, acceleration \(a = 0\), so tractive force equals resistance (\(F = R\)):
\(\frac{P}{v_{\text{max}}} = R \implies \frac{48\,000}{v_{\text{max}}} = 600\)
\(v_{\text{max}} = \frac{48\,000}{600} = 80\text{ m s}^{-1}\).

Key Takeaway: Always convert \(\text{kW}\) to \(\text{W}\) first. Use \(\frac{P}{v} - R = ma\) for accelerating motion, and set \(a = 0\) for maximum/steady speed.

3. Vehicles on Inclined Slopes

When a vehicle moves on a hill inclined at an angle \(\alpha\) to the horizontal, gravity enters the equation. The weight of the vehicle, \(mg\), acts vertically downwards. Resolving this weight gives:

• Parallel to the slope (acting down the slope): \(mg\sin(\alpha)\)
• Perpendicular to the slope: \(mg\cos(\alpha)\)

Case 1: Moving Uphill

When travelling uphill, gravity opposes the motion alongside the non-gravitational resistance \(R\):

\(F_{\text{tract}} - R - mg\sin(\alpha) = ma\)

\(\frac{P}{v} - R - mg\sin(\alpha) = ma\)

For steady or maximum speed uphill (\(a = 0\)):

\(\frac{P}{v} = R + mg\sin(\alpha)\)

Case 2: Moving Downhill Under Power

When travelling downhill, the component of gravity acts in the direction of motion, helping the engine:

\(F_{\text{tract}} - R + mg\sin(\alpha) = ma\)

\(\frac{P}{v} - R + mg\sin(\alpha) = ma\)

For steady or maximum speed downhill (\(a = 0\)):

\(\frac{P}{v} = R - mg\sin(\alpha)\)

Case 3: Free-Wheeling Downhill (Engine Off)

If the engine is switched off or the vehicle is "free-wheeling", the engine delivers zero power (\(P = 0 \implies F_{\text{tract}} = 0\)):

\(mg\sin(\alpha) - R = ma\)

If the vehicle rolls down at a steady speed (\(a = 0\)), then \(R = mg\sin(\alpha)\).

Worked Example 2: Motion on an Incline

A lorry of mass \(5000\text{ kg}\) travels up a slope inclined at an angle \(\alpha\) to the horizontal, where \(\sin(\alpha) = \frac{1}{14}\). The non-gravitational resistance to motion is constant at \(1500\text{ N}\). The lorry's engine operates at a power of \(70\text{ kW}\).
Find the acceleration of the lorry when its speed is \(10\text{ m s}^{-1}\).

Solution:
1. Convert power to Watts: \(P = 70\text{ kW} = 70\,000\text{ W}\).
2. Find the tractive force at \(v = 10\text{ m s}^{-1}\):
\(F = \frac{P}{v} = \frac{70\,000}{10} = 7000\text{ N}\).
3. Calculate the component of weight parallel to the slope (using \(g = 9.8\text{ m s}^{-2}\)):
\(mg\sin(\alpha) = 5000 \times 9.8 \times \frac{1}{14} = 3500\text{ N}\).
4. Set up Newton's Second Law up the slope:
\(F - R - mg\sin(\alpha) = ma\)
\(7000 - 1500 - 3500 = 5000a\)
\(2000 = 5000a\)
\(a = \frac{2000}{5000} = 0.4\text{ m s}^{-2}\).

Key Takeaway: On a slope, always include the resolved weight component \(mg\sin(\alpha)\). It opposes motion when going uphill and assists motion when going downhill.

4. Variable Resistances

In more realistic scenarios, resistance is not constant; it increases as the vehicle moves faster (e.g., due to aerodynamic drag). The question will often specify that resistance is proportional to speed or the square of speed:

• Linear resistance: \(R = kv\)
• Quadratic resistance: \(R = kv^2\)

When the vehicle reaches a steady/maximum speed \(v\) on level ground, tractive force equals resistance:

\(\frac{P}{v} = R\)

• If \(R = kv\), then \(\frac{P}{v} = kv \implies P = kv^2\).
• If \(R = kv^2\), then \(\frac{P}{v} = kv^2 \implies P = kv^3\).

Did you know? Because air resistance grows quadratically (\(R \propto v^2\)), the engine power required to double a car's top speed increases eightfold (\(2^3 = 8\))!

5. Rate of Work Done & Power in Pumps

In CCEA AS 2 Mechanics, power problems also include systems that move fluids, such as water pumps. A pump does work in two main ways:
1. Lifting fluid to a higher level (increasing its Gravitational Potential Energy).
2. Discharging fluid at a certain speed (imparting Kinetic Energy).

Mass Flow Rate

The rate at which mass passes through the pump is the mass flow rate (\(\frac{\mathrm{d}m}{\mathrm{d}t}\)):

\(\frac{\mathrm{d}m}{\mathrm{d}t} = \rho \times (\text{Volume Flow Rate})\)

Where \(\rho\) is the density of the fluid (for water, \(\rho = 1000\text{ kg m}^{-3}\)).

Components of Pump Power

1. Rate of work done against gravity (raising mass through a vertical height \(h\)):

\(P_{\text{PE}} = \frac{\mathrm{d}(\text{PE})}{\mathrm{d}t} = \left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)gh\)

2. Rate of kinetic energy imparted (ejecting mass with speed \(v\)):

\(P_{\text{KE}} = \frac{\mathrm{d}(\text{KE})}{\mathrm{d}t} = \frac{1}{2}\left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)v^2\)

Total Power Output of a Pump

Combining both terms gives the total rate of work done by the pump:

\(P_{\text{total}} = \left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)gh + \frac{1}{2}\left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)v^2 = \left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)\left(gh + \frac{1}{2}v^2\right)\)

Worked Example 3: Power of a Pump

A pump raises water from a well of depth \(15\text{ m}\) at a rate of \(20\text{ kg s}^{-1}\) and discharges it through a nozzle with a speed of \(8\text{ m s}^{-1}\).
Calculate the power output of the pump.

Solution:
Here, \(\frac{\mathrm{d}m}{\mathrm{d}t} = 20\text{ kg s}^{-1}\), \(h = 15\text{ m}\), \(v = 8\text{ m s}^{-1}\), and \(g = 9.8\text{ m s}^{-2}\).
1. Rate of work done against gravity:
\(P_{\text{PE}} = \left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)gh = 20 \times 9.8 \times 15 = 2940\text{ W}\).
2. Rate of kinetic energy imparted:
\(P_{\text{KE}} = \frac{1}{2}\left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)v^2 = \frac{1}{2} \times 20 \times 8^2 = 10 \times 64 = 640\text{ W}\).
3. Total power output:
\(P_{\text{total}} = 2940 + 640 = 3580\text{ W} = 3.58\text{ kW}\).

Key Takeaway: For pumps, check whether the water is both raised (\(mgh\)) and projected (\(\frac{1}{2}mv^2\)), or just one of the two.

6. Common Pitfalls & Examiner Checklist

Ensure you avoid these frequent mistakes highlighted in examiner reports:

1. Confusing Driving Force with Resultant Force:
When using \(P = Fv\), the force \(F\) is the forward tractive force produced by the engine, NOT the net accelerating force \(ma\). Always calculate \(F = \frac{P}{v}\) first, and then substitute that \(F\) into \(F - R = ma\).

2. Forgetting Unit Conversions (\(\text{kW}\) to \(\text{W}\)):
Never substitute \(48\text{ kW}\) as \(48\) into your formulas! Always convert to Watts by multiplying by \(1000\) (\(48\text{ kW} = 48\,000\text{ W}\)).

3. Slope Angles and Trigonometric Components:
Remember that the weight component acting along the slope is always \(mg\sin(\alpha)\), while the component perpendicular to the slope is \(mg\cos(\alpha)\).

4. Downhill Signs:
When moving downhill, gravity is working with the engine. The equation of motion is \(\frac{P}{v} - R + mg\sin(\alpha) = ma\). Do not accidentally subtract the weight component when descending!

5. Misapplying SUVAT:
Constant power means variable driving force, which means variable acceleration. Never use SUVAT equations when power is constant.

Quick Reference Summary Box

• Instantaneous Power: \(P = Fv\)
• Tractive Force: \(F = \frac{P}{v}\)
• Level Road: \(\frac{P}{v} - R = ma\)
• Uphill: \(\frac{P}{v} - R - mg\sin(\alpha) = ma\)
• Downhill: \(\frac{P}{v} - R + mg\sin(\alpha) = ma\)
• Max Speed Condition: \(a = 0\)
• Pump Power: \(P = \left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)\left(gh + \frac{1}{2}v^2\right)\)