--- title: "Conservation of energy; block on incline" subject: "Applied Mechanics (Dynamics)" chapter: 2 key: "semester_3/applied_mechanics_dynamics:principle-conservation-energy-block-incline" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/principle-conservation-energy-block-incline site: BCT Engineering --- # Conservation of energy; block on incline Applied Mechanics (Dynamics) · Chapter 2: Kinetics of particles: Energy and Momentum Methods - Asked in: 2079 Baishakh, 2+6 marks - Video: [Physics 8 Work, Energy, and Power (7 of 37) Inclined Plane (Friction)](https://www.youtube.com/watch?v=qwIUqtITHEE) — Michel van Biezen - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/principle-conservation-energy-block-incline - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-2 ## Question What is the principle of conservation of energy of a system? Illustrate it with suitable example. A small block starts from rest at point A and slides down the inclined plane as shown. What distance along the horizontal plane will it travel before coming to rest? The coefficient of static and kinetic friction between the block and either plane are 0.35 and 0.3 respectively. Assume that the initial velocity with which it starts to move along BC is of the same magnitude as that gained sliding from A to B. [Figure: block at A at the top of an incline 3 m high and 4 m horizontal ending at B; horizontal plane BC continues from B.] ## Short answer The total mechanical energy of a system remains constant if only conservative forces act. For the block, using work–energy on the incline gives speed 5.94 m/s at B; on the horizontal plane, kinetic energy is dissipated by friction, giving a stopping distance of 6.0 m. ## Answer #### Principle of conservation of energy The total mechanical energy (kinetic + potential) of a system remains constant when only conservative forces (gravity, springs) do work: $$ T_1+V_1=T_2+V_2 $$ **Example:** a body of mass $m$ dropped from height $h$. At the top $T_1=0$ and $V_1=mgh$; just before it hits the ground $V_2=0$ and $T_2=\tfrac12mv^{2}$. So $mgh=\tfrac12mv^{2}$ and $v=\sqrt{2gh}$: the potential energy changes completely into kinetic energy. When friction acts, mechanical energy is not conserved, and the work done by friction must be subtracted. #### Block on the incline **Data:** incline 3 m high and 4 m horizontal, so length $AB=5$ m and $\cos\theta=4/5=0.8$. $\mu_k=0.3$ (the block slides because $\tan\theta=0.75>\mu_s=0.35$). ```text A |\ | \ 5 m 3 m| \ | \ theta +----B==========C friction ``` **A to B (work-energy):** $$ mg(3)-\mu_k(mg\cos\theta)(5)=\tfrac12mv_B^{2} $$ $$ v_B^{2}=2g\left[3-0.3(0.8)(5)\right]=2(9.81)(1.8)=35.32\ \text{m}^2/\text{s}^2 $$ $$ v_B=5.943\ \text{m/s} $$ **B to C (comes to rest):** $$ \tfrac12mv_B^{2}=\mu_k mg\,d\ \Rightarrow\ d=\frac{v_B^{2}}{2\mu_kg}=\frac{35.32}{2(0.3)(9.81)}=6.0\ \text{m} $$ **Answer: the block travels $6.0$ m along the horizontal plane before stopping.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When is total mechanical energy of a system conserved? A. When only conservative forces such as gravity and springs do work B. Always, including with friction C. Only at rest D. Only for horizontal motion Answer: A. With friction, its negative work is subtracted from the energy. (Key point: Principle statement) 2. A body is dropped from height h. What is its speed on reaching the ground? A. sqrt(g h) B. sqrt(2 g h) C. 2 g h D. g h / 2 Answer: B. mgh = (1/2) m v^2 gives v = sqrt(2gh). (Key point: Example) 3. For the 3 m high, 4 m horizontal incline, which friction value and normal force are used on AB? A. mu_k = 0.3 and N = mg cos(theta) with AB = 5 m B. mu_s = 0.35 and N = mg C. mu_k = 0.3 and N = mg sin(theta) D. mu_s = 0.35 and N = mg cos(theta) Answer: A. The block slides, so kinetic friction acts; N = mg(0.8) on the 5 m incline. (Key point: Incline length) 4. How is the horizontal stopping distance found? A. (1/2) m v_B^2 = mu_k m g d, so d = v_B^2 / (2 mu_k g) = 6.0 m B. d = v_B^2 / g C. d = v_B t D. d = mu_k v_B Answer: A. On the horizontal plane friction mu_k m g removes all the kinetic energy at B. 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