Mathematics (9709) | Energy, work and power

Hello, Future Engineer! Welcome to Energy, Work, and Power!

In this chapter, we step away from just focusing on forces and motion, and look at the "why" and "how much" behind them. Energy, Work, and Power are essential concepts that connect kinematics (how things move) and dynamics (why they move). Mastering this section of Mechanics (Paper 4) will simplify many complex problems, especially those involving friction or motion on inclined planes.

Don't worry if these terms sound physics-heavy; we will break them down into simple, usable mathematical relationships!

Key Takeaway from the Start: Units Matter!

We use the SI unit for energy and work: the Joule (J). The SI unit for power is the Watt (W).

1. Work Done by a Constant Force (W)

Work is simply the energy transferred to or from a system by a force. If you push something, you do work on it. If friction slows something down, friction does negative work on it.

1.1 Definition and Basic Calculation

Work is defined as the product of the force component parallel to the displacement and the distance moved in that direction.

• If the force \(F\) acts in the same direction as the displacement \(d\):
  Work Done \(W = Fd\)

Analogy: If you push a box 5 meters with a constant force of 10 Newtons, the work done is \(W = 10 \times 5 = 50 \text{ J}\).

1.2 Work Done When Force is Not Parallel to Displacement

This is the most crucial part of the definition. If a constant force \(F\) acts at an angle \(\theta\) to the direction of motion (displacement \(d\)), only the component of the force *in the direction of motion* does work.

• The component of \(F\) parallel to \(d\) is \(F \cos \theta\).
• The component of \(F\) perpendicular to \(d\) does zero work.

The general formula for work done by a constant force \(F\) over displacement \(d\) is:

\[W = Fd \cos \theta\]

Important Note for 9709 Mechanics:

The syllabus confirms that while you must use the formula \(W = Fd \cos \theta\), Use of the scalar product (dot product) is NOT required. Just remember to use the trigonometric component of the force parallel to the motion!

Common Mistake to Avoid:

Students often forget to take the component of the force. If you pull a sled with a rope 30° above the horizontal, and the sled moves horizontally, you must use \(F \cos(30^\circ)\) in your work calculation.

2. Kinetic Energy (KE)

Kinetic Energy (KE) is the energy a particle possesses due to its motion. If it's moving, it has KE.

2.1 The Formula

Kinetic energy depends on the particle's mass \(m\) and its velocity \(v\):

\[KE = \frac{1}{2}mv^2\]

Since velocity \(v\) is squared, even if the direction of velocity changes (e.g., moving up or down a hill), the kinetic energy is always positive (a scalar quantity).

3. Gravitational Potential Energy (GPE)

Gravitational Potential Energy (GPE) is the stored energy a particle possesses due to its position within a gravitational field (i.e., its height).

3.1 The Formula

GPE depends on mass, gravity, and height:

\[GPE = mgh\]

• \(m\) is the mass (kg).
• \(g\) is the acceleration due to gravity (\(10 \text{ m s}^{-2}\) is commonly used in 9709).
• \(h\) is the vertical height above a defined reference level (m).

3.2 The Importance of the Reference Level

The absolute value of GPE doesn't matter; only the change in GPE matters. Therefore, you can choose *any* convenient point as your zero height (\(h=0\)) reference level (e.g., the ground, the lowest point of motion). Be consistent throughout the problem!

Example: If a ball goes from 5m above the ground to 10m above the ground, \(\Delta GPE\) is the same whether you set the ground as \(h=0\) or the ceiling as \(h=0\).

4. The Principle of Conservation of Energy

Energy cannot be created or destroyed, only transferred from one form to another (or dissipated, usually as heat/sound).

4.1 The Energy Equation (The Master Tool)

When solving Mechanics problems involving height, speed, and forces, the most effective approach is often using the energy principle, which links work and energy changes:

\[\text{Initial Energy} + \text{Work Done by External Forces} = \text{Final Energy}\]

We usually write this using the specific forms of energy and work:

\[(KE_1 + GPE_1) + W_{\text{driving/tractive}} - W_{\text{resistance}} = (KE_2 + GPE_2)\]

Where:

• \(KE_1\) and \(GPE_1\) are the kinetic and potential energies at the start.
• \(KE_2\) and \(GPE_2\) are the kinetic and potential energies at the end.
• \(W_{\text{driving}}\) is the work done by forces helping the motion (e.g., engine force).
• \(W_{\text{resistance}}\) is the work done against resisting forces (e.g., friction, air resistance).

4.2 The Pure Conservation Case (Smooth Systems)

If there are no non-gravitational forces doing work (e.g., if the surface is 'smooth' and there is no air resistance or driving force), then the total mechanical energy is conserved:

\[KE_1 + GPE_1 = KE_2 + GPE_2\]

This is very useful for sliding down smooth slides or objects falling freely.

Step-by-Step for Energy Problems

1. Identify State 1 and State 2: Determine the start and end points of the motion you are analyzing.
2. Choose a Reference Level: Set \(h=0\) (usually the lowest point in the problem).
3. Calculate Initial Energy: Find \(KE_1\) (\(\frac{1}{2}mv_1^2\)) and \(GPE_1\) (\(mgh_1\)).
4. Calculate Final Energy: Find \(KE_2\) (\(\frac{1}{2}mv_2^2\)) and \(GPE_2\) (\(mgh_2\)).
5. Calculate External Work: Find the work done by any forces *other than gravity* (e.g., friction, driving force) using \(W = Fd \cos \theta\).
6. Apply the Energy Equation: Substitute all values into the formula and solve for the unknown variable.

5. Power (P)

Power measures how fast work is being done or how quickly energy is being transferred.

5.1 Power Definition and Average Power

Power is the rate at which work is done over time:

\[P = \frac{\text{Work Done}}{\text{Time Taken}} \quad \text{or} \quad P = \frac{W}{t}\]

If the force or velocity is constant over time, this gives the Average Power.

5.2 Instantaneous Power and the P=Fv Formula

In many Mechanics problems (especially those involving cars or engines), you need the instantaneous power (the power being generated at a specific moment). This is related to the instantaneous force and instantaneous velocity.

For a force \(F\) acting in the direction of motion with instantaneous velocity \(v\):

\[P = Fv\]

The force \(F\) here is the tractive force (the engine force or driving force), NOT the resultant force.

Connecting P, F, and V (Crucial for A-Level)

In constant speed motion (zero acceleration), the driving force \(F\) exactly balances the total resistance \(R\) (e.g., friction and air resistance). In this case:

\[P = Fv \implies P = R_{\text{total}} v\]

If the power \(P\) is constant, the force \(F\) must change as velocity \(v\) changes (since \(F = P/v\)). This is common for cars at high speeds.

Solving Complex Problems (Car on a Hill)

When solving problems involving power and motion (especially when the vehicle is accelerating or decelerating), you often need to use both Newton's Second Law (\(F=ma\)) and the Power formula (\(P=Fv\)).

Step-by-Step for Problems Involving Acceleration:

1. Find the Driving Force: If you know the instantaneous power \(P\) and velocity \(v\), calculate the instantaneous driving force \(F_{\text{drive}}\) using: \[F_{\text{drive}} = \frac{P}{v}\]
2. Apply Newton's Second Law (F=ma): Calculate the Resultant Force \(F_{\text{resultant}}\) acting on the mass \(m\). This force causes acceleration \(a\).
   \(F_{\text{resultant}} = F_{\text{drive}} - R - W_{\text{component}}\)
3. Find Acceleration: Use \(a = \frac{F_{\text{resultant}}}{m}\) to find the instantaneous acceleration.

Where \(R\) is the total resistance, and \(W_{\text{component}}\) is the component of weight opposing the motion (if moving uphill).