Work, energy and power An understanding of the forms of energy and energy transfers from Cambridge IGCSE/O Level Physics or equivalent is assumed.

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Equilibrium of Forces, Work, Energy and Power

Equilibrium of Forces

In mechanics a body is said to be in equilibrium when the vector sum of all the forces acting on it is zero. This condition is expressed as

\$\sum \vec{F}=0\$

Two types of equilibrium are recognised:

  • Static equilibrium: the body is at rest (velocity = 0).

  • Dynamic equilibrium: the body moves with constant velocity (acceleration = 0).

Conditions for Equilibrium

  1. Resultant force in the horizontal direction must be zero: \$\sum F_x = 0\$.

  2. Resultant force in the vertical direction must be zero: \$\sum F_y = 0\$.

  3. If the body can rotate, the resultant torque about any axis must also be zero: \$\sum \tau = 0\$.

These three equations are sufficient to solve most A‑Level equilibrium problems.

Suggested diagram: Free‑body diagram of a block on an inclined plane showing weight, normal reaction, friction and a pulling force.

Work, Energy and Power

Work

Work is the transfer of energy when a force causes a displacement. The scalar work done by a constant force \$\vec{F}\$ acting through a displacement \$\vec{s}\$ is

\$W = \vec{F}\cdot\vec{s}=Fs\cos\theta\$

where \$\theta\$ is the angle between the force and the displacement direction. The SI unit of work is the joule (J), where \$1\;\text{J}=1\;\text{N·m}\$.

Kinetic Energy

The kinetic energy of a body of mass \$m\$ moving with speed \$v\$ is

\$E_k = \frac{1}{2}mv^{2}\$

If the net work done on a body is \$W_{\text{net}}\$, the work‑energy theorem states

\$W{\text{net}} = \Delta Ek = \frac{1}{2}m(vf^{2}-vi^{2})\$

Potential Energy

Potential energy is stored energy associated with the position of a body in a force field.

  • Gravitational potential energy near Earth’s surface: \$E_g = mgh\$.

  • Elastic potential energy in a spring (Hooke’s law): \$E_s = \frac{1}{2}kx^{2}\$, where \$k\$ is the spring constant and \$x\$ the extension/compression.

Conservation of Mechanical Energy

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

\$E{\text{total}} = Ek + E_{\text{potential}} = \text{constant}\$

Power

Power is the rate at which work is done (or energy is transferred). It is defined as

\$P = \frac{dW}{dt} = \frac{\Delta E}{\Delta t}\$

For a constant force moving at constant speed \$v\$, power can also be written as

\$P = Fv\cos\theta\$

The SI unit of power is the watt (W), where \$1\;\text{W}=1\;\text{J·s}^{-1}\$.

Linking Equilibrium to Energy Concepts

Even when a body is in equilibrium, forces can do work if the point of application moves. Typical examples include:

  • Raising a weight at constant speed – the upward force equals the weight, net force zero, but work \$W = mgh\$ is done against gravity.

  • Pulling a block across a frictionless surface at constant speed – the pulling force balances friction, net force zero, yet power \$P = Fv\$ is required.

Example Problem

Problem: A 5.0 kg block is pulled up a smooth 30° incline by a horizontal force of 40 N. Determine (a) the tension in the rope, (b) the work done by the pulling force after the block moves 2.0 m along the incline, and (c) the power required if the motion takes 4.0 s.

Solution Sketch:

  1. Resolve the horizontal pulling force into components parallel and perpendicular to the incline.

    \$F_{\parallel}=40\cos30^{\circ}=34.6\;\text{N}\$

    \$F_{\perp}=40\sin30^{\circ}=20.0\;\text{N}\$

  2. Write the equilibrium condition parallel to the plane (no acceleration):

    \$\sum F{\parallel}=0 \;\Rightarrow\; T\cos30^{\circ}+F{\parallel}=mg\sin30^{\circ}\$

    Solve for the tension \$T\$.

  3. Work done by the pulling force:

    \$W = \vec{F}\cdot\vec{s}=Fs\cos\phi\$

    where \$\phi\$ is the angle between the horizontal force and the displacement (the incline direction). \$\phi = 30^{\circ}\$, so

    \$W = 40\;(2.0)\cos30^{\circ}=69.3\;\text{J}\$

  4. Power:

    \$P = \frac{W}{t}= \frac{69.3}{4.0}=17.3\;\text{W}\$

Summary Table of Key Equations

ConceptFormulaUnits
Equilibrium (resultant force)\$\displaystyle \sum \vec{F}=0\$Newtons (N)
Equilibrium (resultant torque)\$\displaystyle \sum \tau =0\$Newton‑metre (N·m)
Work\$\displaystyle W = \vec{F}\cdot\vec{s}=Fs\cos\theta\$Joule (J)
Kinetic Energy\$\displaystyle E_k = \frac{1}{2}mv^{2}\$Joule (J)
Gravitational Potential Energy\$\displaystyle E_g = mgh\$Joule (J)
Elastic Potential Energy\$\displaystyle E_s = \frac{1}{2}kx^{2}\$Joule (J)
Power\$\displaystyle P = \frac{dW}{dt}= \frac{\Delta E}{\Delta t}=Fv\cos\theta\$Watt (W)

Key Points to Remember

  • Equilibrium requires both translational and rotational conditions to be satisfied.

  • Work is only done by the component of force parallel to the displacement.

  • Energy can be transferred between kinetic, gravitational, and elastic forms, but the total mechanical energy is conserved only when non‑conservative forces (e.g., friction) do no work.

  • Power quantifies how quickly energy is transferred; high power does not necessarily mean large total work.