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United StatesPhysics C: MechanicsSyllabus dot point

How does conservation of mechanical energy let us solve problems, and how do we account for energy dissipated by non-conservative forces?

Topic 3.4 Conservation of Energy: apply conservation of mechanical energy for conservative systems, and extend the energy balance to include the work done by non-conservative forces.

A focused answer to AP Physics C: Mechanics Topic 3.4, covering conservation of mechanical energy in conservative systems, the work-energy bookkeeping when non-conservative forces such as friction dissipate energy, choosing a system and reference, and applying the energy balance to incline, spring and pendulum problems, with worked examples.

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  1. What this topic is asking
  2. Conservation of mechanical energy
  3. Including non-conservative forces
  4. Choosing a system and a reference
  5. Try this

What this topic is asking

The College Board (Topic 3.4) wants you to apply conservation of mechanical energy in systems where only conservative forces act, and to extend the energy balance to account for non-conservative forces such as friction that dissipate mechanical energy. Energy conservation is one of the most powerful problem-solving tools in the course, letting you connect speeds and positions at different points without tracking the detailed forces in between.

Conservation of mechanical energy

When gravity and springs are the only forces doing work, energy simply shifts between kinetic and potential forms while the total stays fixed. A ball dropped from height hh trades all its gravitational PE (mghmgh) for kinetic energy (12mv2\tfrac{1}{2}mv^2), giving v=2ghv = \sqrt{2gh}, the same result as a frictionless ramp or a swinging pendulum through the same height. The strategy is to pick two instants, write the total energy at each, and set them equal. The mass often cancels, and you never need the acceleration or the time.

Including non-conservative forces

Real situations usually involve non-conservative forces, chiefly friction, which dissipate mechanical energy as heat. The energy balance then has an extra term:

Ki+Ui+Wnc=Kf+Uf,K_i + U_i + W_{nc} = K_f + U_f,

where WncW_{nc} is the work done by non-conservative forces. For friction acting over a distance dd, this work is negative, Wfriction=βˆ’fkdW_{friction} = -f_k d, and represents mechanical energy lost from the system (it becomes thermal energy). An applied force can also do positive non-conservative work, adding energy. The accounting is the same: start with the initial mechanical energy, add the non-conservative work (signed), and you get the final mechanical energy.

Choosing a system and a reference

Two setup choices make energy problems clean. First, choose the system so that as many forces as possible become internal conservative forces; including the spring and the Earth in the system lets you use UU instead of tracking their work. Second, choose a reference level for potential energy (often the lowest point of the motion), since only changes in UU matter. With those fixed, you write the total mechanical energy at the start and end, include any friction or applied work, and solve. This approach shines in multi-stage problems, a ramp leading to a rough floor leading to a spring, where forces change but energy bookkeeping stays simple.

Try this

Q1. A 0.200.20 kg ball is thrown straight up at 1212 m/s. Ignoring air resistance, calculate the maximum height (g=9.8g = 9.8 m/s squared). [2 points]

  • Cue. 12mv2=mghβ‡’h=v2/(2g)=144/19.6=7.3\tfrac{1}{2}mv^2 = mgh \Rightarrow h = v^2/(2g) = 144/19.6 = 7.3 m.

Q2. A 4.04.0 kg crate slides 5.05.0 m across a floor with ΞΌk=0.20\mu_k = 0.20 before stopping. Calculate the mechanical energy dissipated (g=9.8g = 9.8 m/s squared). [2 points]

  • Cue. ∣Wfriction∣=ΞΌkmg d=(0.20)(4.0)(9.8)(5.0)=39|W_{friction}| = \mu_k mg\,d = (0.20)(4.0)(9.8)(5.0) = 39 J.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)6 marksSection II (FRQ). A 2.02.0 kg block is released from rest at the top of a frictionless ramp 3.03.0 m high, then slides onto a rough horizontal floor (ΞΌk=0.25\mu_k = 0.25) and compresses a spring (k=800k = 800 N/m). Take g=9.8g = 9.8 m/s squared. (a) Determine the block's speed at the bottom of the ramp. (b) If the block travels 4.04.0 m across the floor before reaching the spring, determine its speed at the spring. (c) Determine the maximum compression of the spring.
Show worked answer β†’

A 6-point energy-conservation FRQ with a dissipative segment.

(a) Speed at the bottom (2 points): the ramp is frictionless, so mgh=12mv2β‡’v=2gh=2(9.8)(3.0)=7.7mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2(9.8)(3.0)} = 7.7 m/s.
(b) Speed at the spring (2 points): friction removes fkd=ΞΌkmg d=(0.25)(2.0)(9.8)(4.0)=19.6f_k d = \mu_k mg\,d = (0.25)(2.0)(9.8)(4.0) = 19.6 J. Energy at the spring: 12mvspring2=12mvbottom2βˆ’19.6=12(2.0)(7.7)2βˆ’19.6=58.8βˆ’19.6=39.2\tfrac{1}{2}mv_{spring}^2 = \tfrac{1}{2}mv_{bottom}^2 - 19.6 = \tfrac{1}{2}(2.0)(7.7)^2 - 19.6 = 58.8 - 19.6 = 39.2 J, so vspring=39.2=6.3v_{spring} = \sqrt{39.2} = 6.3 m/s.
(c) Maximum compression (2 points): at maximum compression the block stops; its kinetic energy plus the friction over the compression xx goes into the spring: approximating friction over xx as small or, more simply, equating the kinetic energy at the spring to 12kx2\tfrac{1}{2}kx^2 gives 39.2=12(800)x2β‡’x=0.098=0.3139.2 = \tfrac{1}{2}(800)x^2 \Rightarrow x = \sqrt{0.098} = 0.31 m.

Markers reward tracking energy through each stage and subtracting the friction work as dissipated energy.

AP 2021 (style)1 marksSection I (multiple choice). A pendulum bob is released from rest at a height hh above its lowest point. Ignoring air resistance, its speed at the lowest point is... (A) ghgh (B) 2gh\sqrt{2gh} (C) 2gh2gh (D) gh\sqrt{gh}. Justify your reasoning.
Show worked answer β†’

A 1-point energy-conservation MCQ. The answer is (B).

With no air resistance, mechanical energy is conserved: the gravitational potential energy at the top converts entirely to kinetic energy at the bottom, mgh=12mv2mgh = \tfrac{1}{2}mv^2, so v=2ghv = \sqrt{2gh}. The mass cancels. The trap (A) forgets the factor of 12\tfrac{1}{2} and the square root. This is the same result as a frictionless ramp or free fall through the same height.

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