--- title: "Work-energy principle, power and efficiency" subject: "Engineering Mechanics" chapter: 3 key: "semester_3/engineering_mechanics:work-energy-principle-power-efficiency" source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/work-energy-principle-power-efficiency site: BCT Engineering --- # Work-energy principle, power and efficiency Engineering Mechanics · Chapter 3: Kinetics of Particles: Work Energy Principles - Asked in: year not given, 3+3 marks - Video: [Dynamics: Lesson 27- Work and Energy Balance with Power and Efficiency](https://www.youtube.com/watch?v=ggS4eq1GrJY) — Jeff Hanson - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/work-energy-principle-power-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-3 ## Question (a) State and prove the principle of work and energy for a particle. (b) Define power and mechanical efficiency and give the relation between power, force and velocity. ## Short answer Work done by all forces equals change in kinetic energy: U_{1→2} = ½ m v₂² – ½ m v₁². Proof: integrate F_t ds = m v dv. Power is rate of work: P = F·v. Efficiency η = output power / input power. ## Answer #### (a) Principle of work and energy **Statement:** the work done by all forces acting on a particle as it moves from position 1 to position 2 equals the change in its kinetic energy. $$ U_{1\to2} = T_2 - T_1 = \tfrac12 m v_2^2 - \tfrac12 m v_1^2 $$ **Proof:** Let the resultant force $\mathbf{F}$ act on a particle of mass $m$ moving along a path with tangential acceleration $a_t$. Newton's second law along the tangent gives $F_t = m a_t = m\,\dfrac{dv}{dt}$. Using $a_t = v\,\dfrac{dv}{ds}$: $$ F_t\,ds = m v\,dv $$ Integrating from position 1 ($s_1$, $v_1$) to position 2 ($s_2$, $v_2$): $$ \int_{s_1}^{s_2} F_t\,ds = m\int_{v_1}^{v_2} v\,dv = \tfrac12 m v_2^2 - \tfrac12 m v_1^2 $$ The left side is the work $U_{1\to2}$ of the resultant force, so $T_1 + U_{1\to2} = T_2$. The normal component of the force does no work, because it is perpendicular to the displacement. #### (b) Power and efficiency **Power** is the rate of doing work: $$ P = \frac{dU}{dt} = \mathbf{F}\cdot\mathbf{v} = Fv\cos\alpha $$ where $\alpha$ is the angle between the force and the velocity. Unit: watt (W = J/s); 1 hp = 746 W. For rotation, $P = M\omega$ (torque times angular velocity). **Mechanical efficiency** is the ratio of useful output power (or work) to input power (or work): $$ \eta = \frac{\text{output power}}{\text{input power}} = \frac{P_{out}}{P_{in}} < 1 $$ The difference between input and output is lost mainly to friction and appears as heat. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The principle of work and energy for a particle states that the work of all forces between two positions equals: A. The change in momentum B. The change in kinetic energy C. The change in potential energy only D. The final kinetic energy Answer: B. U(1 to 2) = T2 - T1. (Key point: Statement) 2. In the proof, which substitution lets F_t ds be written as m v dv? A. a_t = v dv/ds B. a_t = ds/dt C. a_t = v^2/r D. a_t = dv/ds Answer: A. Using a_t = v dv/ds, F_t = m v dv/ds so F_t ds = m v dv, which integrates to the kinetic energy change. (Key point: Proof step) 3. Why do normal forces not appear in the work-energy equation? A. They are always zero B. They are perpendicular to the displacement and do no work C. They are conservative D. They cancel the weight Answer: B. Work is force times displacement along the force, which is zero for a perpendicular force. (Key point: Normal forces) 4. Which pair of statements is correct? A. P = F.v and efficiency = input/output B. P = F.v = F v cos(alpha) and efficiency = output/input, which is less than 1 C. P = F/v and efficiency = output x input D. P = F v sin(alpha) and efficiency is above 1 Answer: B. Power is the rate of work F.v and efficiency is useful output over input, always below 1 because of friction losses. 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