Energy, work and power: kinetic and potential energy, conservation, work done, power

Mechanics (M1) – Energy, Work and Power

1. Introduction

This section follows the Cambridge International AS & A Level Mathematics (9709) syllabus (Topic 4.5). You will learn:

• Definitions and formulas for kinetic and potential energy.
• Work as a scalar (dot) product and the work‑energy theorem.
• Power and its relation to work and kinetic energy.
• Conservation of mechanical energy, the exact sign convention ΔU = –W_conservative, and when to use g = 9.81 m s⁻² or the approximation g = 10 m s⁻².
• How to treat variable forces using line integrals.

2. Kinetic Energy

The kinetic energy of a particle of mass m moving with speed v is

$$E_k=\frac12\,mv^{2}$$

• Scalar quantity (always ≥ 0).
• Depends only on the magnitude of the velocity, not its direction.
• For a system of particles, total kinetic energy is the sum of the individual contributions.

3. Potential Energy

3.1 Gravitational Potential Energy

Choosing a reference level at height h = 0, the gravitational potential energy is

$$U_g=mgh$$

Use g = 9.81 m s⁻² for accurate work and g = 10 m s⁻² for quick estimates (syllabus allowance).

3.2 Elastic (Spring) Potential Energy

For a spring obeying Hooke’s law (spring constant k) displaced by x from its natural length,

$$U_s=\frac12\,kx^{2}$$

3.3 General Potential Energy (Conservative Forces)

If a force \(\mathbf F\) is conservative, it can be written as the negative gradient of a scalar potential U:

$$\mathbf F = -abla U$$

The change in potential energy between points A and B is defined by the syllabus as

$$\Delta U = U_B-U_A = -\int_{A}^{B}\mathbf F\!\cdot\!d\mathbf r\qquad\text{(so } \Delta U = -W_{\text{conservative}}\text{)}$$

This explicit wording is required for all energy‑conservation problems.

4. Work

4.1 Definition – scalar (dot) product

For a constant force \(\mathbf F\) acting through a displacement \(\mathbf s\), the work done is

$$W = \mathbf F\!\cdot\!\mathbf s = Fs\cos\theta$$

Only the component of the force parallel to the displacement contributes to the work.

4.2 Work‑Energy Theorem (net work)

The net work done by **all** forces on a particle equals the change in its kinetic energy:

$$W_{\text{net}} = \Delta E_k = E_{k,2}-E_{k,1}$$

The theorem follows directly from Newton’s second law and is the bridge between dynamics and energy methods. Note that net work, not the work of a single force, appears here.

4.3 Variable‑force work

When the force varies with position, work must be evaluated by a line integral:

$$W = \int_{C}\mathbf F\!\cdot\!d\mathbf r$$

Example (syllabus style):

$$\begin{aligned} F(x) &= 5x\ \text{N}\\[2mm] W &= \int_{0}^{3}5x\,dx = \left[\frac{5}{2}x^{2}\right]_{0}^{3} = 22.5\ \text{J} \end{aligned}$$

Never replace a variable force by an “average” value unless the force is actually constant.

5. Power

Power is the rate at which work is done (or energy is transferred):

$$P = \frac{dW}{dt}= \mathbf F\!\cdot\!\mathbf v = Fv\cos\theta$$

Using the kinetic‑energy form:

$$P = \frac{dE_k}{dt}$$

Units: 1 W = 1 J s⁻¹. The expression P = Fv is valid only when the force is parallel to the motion.

6. Conservation of Mechanical Energy

If **only conservative forces** do work, the total mechanical energy is constant:

$$E_{\text{mech}} = E_k + U = \text{constant}$$

Between two positions 1 and 2:

$$\frac12 mv_1^{2}+U_1 = \frac12 mv_2^{2}+U_2$$

Because the syllabus states ΔU = –W_conservative, the sign of the potential‑energy change must be handled exactly as written.

7. Worked Examples

Example 1 – Frictionless incline (energy conservation)

Problem: A 2 kg block slides down a frictionless incline of height 5 m. Find its speed at the bottom.

1. Initial mechanical energy: U = mgh.
2. Final mechanical energy: E_k = \frac12 mv^{2}.
3. Conservation:
   $$mgh = \frac12 mv^{2}$$
4. Insert numbers (choose g = 9.81 m s⁻² or 10 m s⁻²).
   $$2\times9.81\times5 = \tfrac12\times2\times v^{2}$$
5. Solve:
   $$v^{2}=9.81\times5\;\Rightarrow\;v\approx7.0\ \text{m s}^{-1}$$

Example 2 – Dot‑product work calculation

Problem: A 20 N force acts on a crate that moves 4 m. The force makes an angle of 30° with the direction of motion. Find the work done.

Using the scalar product:

$$W = Fs\cos30^{\circ}=20\times4\times0.866\approx69.3\ \text{J}$$

Example 3 – Variable force (integral)

Calculate the work done by F(x)=5x N from x=0 m to x=3 m.

$$W=\int_{0}^{3}5x\,dx = 22.5\ \text{J}$$

Example 4 – Quick estimate with g≈10 m s⁻²

A 0.5 kg ball is thrown vertically upward with speed 10 m s⁻¹. Estimate the maximum height.

$$\frac12 mv^{2}=mgh\;\Rightarrow\;h=\frac{v^{2}}{2g} =\frac{10^{2}}{2\times10}=5\ \text{m}$$

Example 5 – Power of an accelerating car

A 1500 kg car accelerates from rest to 20 m s⁻¹ in 10 s. Assuming the engine provides the net work, the average power is

$$\Delta E_k = \tfrac12 m v^{2}= \tfrac12(1500)(20)^{2}=3.0\times10^{5}\ \text{J}$$ $$P_{\text{avg}} = \frac{\Delta E_k}{\Delta t}= \frac{3.0\times10^{5}}{10}=3.0\times10^{4}\ \text{W}=30\ \text{kW}$$

8. Summary Table

Quantity	Symbol	Formula	Units
Kinetic Energy	\(E_k\)	\(\displaystyle\frac12 mv^{2}\)	J
Gravitational Potential Energy	\(U_g\)	\(mgh\)	J
Elastic Potential Energy	\(U_s\)	\(\displaystyle\frac12 kx^{2}\)	J
General Potential Energy (conservative)	\(U\)	\(-\displaystyle\int_{A}^{B}\mathbf F\!\cdot\!d\mathbf r\)	J
Work (constant force)	\(W\)	\(Fs\cos\theta\)	J
Work (variable force)	\(W\)	\(\displaystyle\int_{C}\mathbf F\!\cdot\!d\mathbf r\)	J
Power	\(P\)	\(\displaystyle\frac{dW}{dt}= \mathbf F\!\cdot\!\mathbf v\)	W
Mechanical Energy (conserved)	\(E_{\text{mech}}\)	\(E_k+U\)	J

9. Common Pitfalls (syllabus focus)

• Work‑energy theorem vs. energy conservation: The theorem uses net work (W_net = ΔE_k). Conservation of mechanical energy applies only when non‑conservative work (e.g., friction) is absent.
• Sign of potential‑energy change: The syllabus wording is ΔU = –W_conservative. Forgetting the minus sign gives the wrong energy balance.
• Dot product requirement: Always include the cosine of the angle between force and displacement (or velocity). P = Fv is a special case of P = Fv cosθ.
• Variable forces: Do not replace a variable force by an average value. Write the force as a function of the relevant coordinate and integrate.
• Choosing g: Use 9.81 m s⁻² for accurate results; the syllabus permits 10 m s⁻² for quick estimates – state clearly which value you are using.
• Units: Energy in joules (J), power in watts (W). Remember that 1 W = 1 J s⁻¹.

10. Suggested Further Practice

1. Calculate the work done by a variable force F(x)=5x N as a particle moves from x=0 m to x=3 m (answer = 22.5 J).
2. A 0.5 kg ball is thrown vertically upward with an initial speed of 10 m s⁻¹. Determine the maximum height using both g=9.81 m s⁻² and the approximation g=10 m s⁻².
3. A 1500 kg car accelerates from rest to 20 m s⁻¹ in 10 s. Compute the average power output of the engine, ignoring losses.
4. Two springs with constants k₁=200 N m⁻¹ and k₂=300 N m⁻¹ are attached in series and compressed by 0.1 m. Find the total elastic potential energy stored.
5. Work out the work done by a 30 N force acting at 45° to a displacement of 5 m. Verify the result using the dot‑product definition.
6. A block slides down a rough incline (coefficient of kinetic friction μₖ = 0.15) of height 3 m. Use the work‑energy theorem (including the work of friction) to find the speed at the bottom.

11. Suggested Diagram