Conservation of Mechanical Energy

AP Physics 1· AP Physics 1 CED — Energy· 14 min read

1. What Is Conservation of Mechanical Energy?★★☆☆☆⏱ 3 min

Conservation of Mechanical Energy (CoME) is a core principle in AP Physics 1 Unit 4, which accounts for 20–25% of the total AP exam score. It appears in both MCQ and FRQ sections, almost always combined with other core concepts like forces or momentum to test multi-step reasoning.

📘 Definition

Total Mechanical Energy

E_{mech}

The sum of all kinetic energy (energy of motion) and all potential energy (stored energy due to position or configuration of a system).

Example:

$E_{m ec h} = K + U_{g} + U_{s}$ for systems with gravity and ideal springs

CoME states that if only conservative forces do work on a system, the total mechanical energy remains constant over time, or $Δ E_{m ec h} = 0$ . Unlike the universal law of conservation of total energy (which holds for any isolated system, even when non-conservative forces act), CoME only applies under specific conditions, which makes it a frequent testing point on the AP exam. It also provides a much faster method for relating speed and position than kinematics, especially for curved or non-uniform paths where acceleration changes continuously.

2. Conditions for Valid Application of CoME★★☆☆☆⏱ 4 min

Before you can use CoME to solve any problem, you must first confirm that two key conditions are met. First, you must correctly define your system to include all objects that exert conservative forces on each other. For example, to count gravitational potential energy as part of the system’s internal energy, you must include both the interacting object (e.g., a falling ball) and Earth in your system. Second, the net work done by all non-conservative forces on the system must be zero ( $W_{n c} = 0$ ).

Non-conservative forces are forces where work done depends on the path taken, including friction, air resistance, tension from an external rope, and applied force from a hand. If non-conservative forces do work, the general relation $W_{n c} = Δ E_{m ec h}$ holds, and CoME ( $E_{ini t ia l} = E_{f ina l}$ ) is just the special case when $W_{n c} = 0$ .

Common scenarios where CoME applies include pendulum motion with negligible air resistance, a mass sliding down a frictionless ramp, projectile motion with negligible air resistance, and oscillation of a mass on an ideal frictionless spring.

📐 Worked Example

A student wants to use CoME to find the speed of a block sliding down a rough ramp (friction is present). The student defines the system as block + ramp + Earth. Can the student use CoME for this problem? Justify your answer.

1
List all forces doing work on the system: gravity (conservative, internal), normal force (does no work, perpendicular to displacement), and kinetic friction (non-conservative, internal).
2
Kinetic friction does negative non-zero work on the block, converting mechanical energy to thermal energy that is not counted in $E_{m ec h}$ , so
3
$W_{nc} \neq 0$
4
Even though all objects are included in the system, the non-conservative friction force removes mechanical energy from the system.
5
Conclusion: CoME cannot be used for this problem.

Exam tip:

When justifying whether CoME applies on an FRQ, always explicitly state whether $W_{n c} = 0$ and reference your system definition to earn full points; AP readers require explicit justification, not just a yes/no answer.

3. CoME with Gravitational Potential Energy Only★★★☆☆⏱ 3 min

When the only potential energy in the system is gravitational potential energy, CoME simplifies to a straightforward relation between initial and final energy. For problems near Earth’s surface, we use the linear approximation:

U_g = mgy

where $y$ is measured from an arbitrary reference point. The choice of reference does not affect the final result because we only care about changes in potential energy, not absolute values. The full CoME formula for this case is:

\frac{1}{2}mv_i^2 + mgy_i = \frac{1}{2}mv_f^2 + mgy_f

Notice that mass $m$ appears in every term, so it cancels out completely. This means the final speed of the object does not depend on its mass, a result that is frequently tested in MCQ distractors. CoME is particularly useful for this type of problem because it skips calculating acceleration or time, which is required for kinematic solutions, especially for curved paths like a roller coaster track.

📐 Worked Example

A roller coaster cart starts from rest at the top of a 42 m high frictionless hill, then rolls down to the bottom of the hill at ground level. What is the speed of the cart at the bottom, assuming negligible air resistance?

1
Confirm conditions: Track is frictionless, no air resistance, so $W_{n c} = 0$ , CoME applies. Define system as cart + Earth, and set reference level $y = 0$ at the bottom of the hill.
2
Calculate initial energy: The cart starts from rest, so $K_{i} = 0$ . Initial height $y_{i} = 42 m$ , so $U_{g, i} = m g (42)$ . Total initial energy
3
$E_i = 42mg$
4
Calculate final energy: Final height $y_{f} = 0$ , so $U_{g, f} = 0$ . Final kinetic energy $K_{f} = \frac{1}{2} m v_{f}^{2}$ . Total final energy
5
$E_f = \frac{1}{2}mv_f^2$
6
Set $E_{i} = E_{f}$ and solve:
7
$42mg = \frac{1}{2}mv_f^2 \\ v_f = \sqrt{2(9.8)(42)} \approx 29 \ m/s$

Exam tip:

Always cancel mass term-by-term when it appears in all terms of CoME; not only does this simplify calculation, it also reinforces that mass does not affect speed in gravity-only, frictionless problems, a common point of confusion.

4. CoME with Combined Gravitational and Elastic Potential Energy★★★★☆⏱ 4 min

Many AP exam problems involve both gravitational potential energy and elastic potential energy from an ideal spring. For an ideal spring, elastic potential energy is:

U_s = \frac{1}{2}kx^2

where $k$ is the spring constant and $x$ is displacement from the spring’s unstretched (equilibrium) position. Unlike gravitational potential energy, the reference for $x$ is not arbitrary: $x$ must always be measured relative to the unstretched length.

The CoME formula extends naturally to include both potential energy terms:

K_i + U_{g,i} + U_{s,i} = K_f + U_{g,f} + U_{s,f}

A common problem type for this case is a block falling onto a vertical spring, where we need to find maximum compression. At both the initial release and maximum compression, the block is at rest, so $K_{i} = K_{f} = 0$ , which simplifies the calculation.

📐 Worked Example

A 0.5 kg block is held at rest 1.2 m above the top of a vertical relaxed spring with spring constant $k = 40 N / m$ . The block is released from rest, falls onto the spring, and compresses it. What is the maximum compression $x$ of the spring, assuming no friction or air resistance?

1
Confirm conditions: No friction/air resistance, so $W_{n c} = 0$ , CoME applies. Define system as block + Earth + spring, and set $y = 0$ at the top of the relaxed spring.
2
Initial energy: $K_{i} = 0$ , $U_{g, i} = m g (1.2)$ , $U_{s, i} = 0$ , so $E_{i} = 1.2 m g$ .
3
Final energy at maximum compression: $K_{f} = 0$ , $U_{g, f} = m g (- x)$ (the block is $x$ below the reference point), $U_{s, f} = \frac{1}{2} k x^{2}$ , so total final energy
4
$E_f = -mgx + \frac{1}{2}kx^2$
5
Set $E_{i} = E_{f}$ and substitute values to get the quadratic equation:
6
$1.2mg = -mgx + \frac{1}{2}kx^2 \\ 20x^2 - 4.9x - 5.88 = 0$
7
Solve and select the positive root (the only physical solution for compression):
8
$x = \frac{4.9 + \sqrt{4.9^2 + 4(20)(5.88)}}{40} \approx 0.68 \ m$

Exam tip:

Don’t forget to include the gravitational potential energy change during spring compression! Many students only count the fall above the spring, not the additional $x$ fall after contact, leading to an answer that is too small.

5. AP Style Practice Problems★★★★☆⏱ 5 min

📐 Worked Example

A pendulum bob is pulled to the side, held at rest at a height $h$ above its lowest position, then released. It swings through the lowest position, where its speed is measured to be $v$ . If the same pendulum is released from rest at a height $2 h$ above the lowest position, what is its new speed at the lowest position? Neglect air resistance.
Options: A) $v$ | B) $2 v$ | C) $2 v$ | D) $4 v$

1
Neglecting air resistance, only gravity does work, so $W_{n c} = 0$ and CoME applies. Set potential energy to zero at the lowest position: initial energy is $m g h$ , final energy is $\frac{1}{2} m v^{2}$ .
2
Equating gives $v = 2 g h$ . For a starting height of $2 h$ , the new speed is:
3
$v' = \sqrt{2g(2h)} = \sqrt{2} \sqrt{2gh} = \sqrt{2}v$
4
The correct answer is B.

📐 Worked Example

A 2.0 kg block slides along a horizontal track with an initial speed of 5.0 m/s. At the end of the track is a horizontal ideal spring with spring constant $k = 20 N / m$ , fixed to a wall at its other end. (a) If the track is frictionless, what is the maximum compression of the spring when the block runs into it? (b) The block compresses the spring, then is pushed back onto the horizontal track. What is the speed of the block when it leaves the spring, for the frictionless case? (c) Suppose the track is not frictionless, and the measured maximum compression is 1.4 m instead of the value you found in (a). How much work is done by friction during compression?

1
(a) For the frictionless case, $W_{n c} = 0$ so CoME applies. Initial kinetic energy converts entirely to elastic potential energy at maximum compression ( $K_{f} = 0$ ):
2
$\frac{1}{2}mv_i^2 = \frac{1}{2}kx_{max}^2 \\ x_{max} = v_i\sqrt{\frac{m}{k}} = 5.0\sqrt{\frac{2.0}{20}} \approx 1.6 m$
3
(b) CoME still applies, no energy is lost. All elastic potential energy converts back to kinetic energy when the block leaves the spring, so $v_{f} = v_{i} = 5.0 m / s$ .
4
(c) Use the general relation $W_{n c} = Δ E_{m ec h} = E_{f} - E_{i}$ : $E_{i} = \frac{1}{2} m v_{i}^{2} = 25 J$ , $E_{f} = \frac{1}{2} k x^{2} = 0.5 (20) (1.4)^{2} = 19.6 J$ , so $W_{f r i c t i o n} = 19.6 - 25 = - 5.4 J$ . The negative sign indicates friction removes 5.4 J of mechanical energy from the system.

6. Common Pitfalls

Wrong move:

Forgetting that elastic potential energy $U_{s}$ is quadratic in $x$ , so compressing a spring twice as far stores four times as much energy, not twice.

Why:

Students confuse the linear spring force $F = k x$ with the quadratic potential energy $U_{s} = \frac{1}{2} k x^{2}$ .

Correct move:

Always write $U_{s} = \frac{1}{2} k x^{2}$ explicitly before plugging in values, and double-check that you squared $x$ .

Wrong move:

Measuring $x$ for elastic potential energy from an arbitrary reference point, instead of the spring’s unstretched equilibrium.

Why:

Students get used to arbitrary gravitational references and incorrectly assume the same rule applies.

Correct move:

Always define $x$ as displacement from the unstretched spring length, no exceptions.

Wrong move:

Using CoME ( $E_{i} = E_{f}$ ) when friction is present, without accounting for work done by non-conservative forces.

Why:

Students remember 'energy is conserved' and apply CoME even when the problem explicitly states the surface is rough.

Correct move:

Always check for non-conservative work first; if $W_{n c} \neq = 0$ , use $W_{n c} = Δ E_{m ec h}$ instead of CoME.

Wrong move:

Double-counting conservative forces by adding both potential energy and work done by gravity or the spring.

Why:

Students mix the work-energy theorem for external forces with CoME for internal conservative forces.

Correct move:

If you include potential energy for a conservative force in $E_{m ec h}$ , do not count its work again in $W_{n c}$ .

Wrong move:

Automatically canceling mass when it does not appear in all terms (e.g., when elastic potential energy is present).

Why:

Students get used to mass canceling in gravity-only problems and cancel it by habit.

Correct move:

Cancel mass term-by-term, only crossing it out if it appears in every term.

Wrong move:

Selecting the negative root when solving a quadratic for distance, leading to a negative compression or height.

Why:

Students forget that quadratic roots are mathematical, and only one matches the physical scenario.

Correct move:

Always select the positive root for any length or distance problem.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Total Mechanical Energy	$E_{m ec h} = K + U_{g} + U_{s}$	Sum of kinetic and all potential energy in the system
General Work-Energy Relation	$W_{n c} = E_{m ec h, f} - E_{m ec h, i}$	$W_{n c}$ = net work done by non-conservative forces
Conservation of Mechanical Energy	$E_{m ec h, i} = E_{m ec h, f}$	Only applies when $W_{n c} = 0$
Kinetic Energy	$K = \frac{1}{2} m v^{2}$	Valid for all AP Physics 1 problems
Gravitational Potential Energy (Near Earth)	$U_{g} = m g y$	$y$ measured from arbitrary reference; only $Δ U_{g}$ matters
Elastic Potential Energy (Ideal Spring)	$U_{s} = \frac{1}{2} k x^{2}$	$x$ measured from unstretched spring equilibrium
CoME (Gravity Only)	$\frac{1}{2} m v_{i}^{2} + m g y_{i} = \frac{1}{2} m v_{f}^{2} + m g y_{f}$	Mass cancels from all terms
CoME (Gravity + Elastic)	$\frac{1}{2} m v_{i}^{2} + m g y_{i} + \frac{1}{2} k x_{i}^{2} = \frac{1}{2} m v_{f}^{2} + m g y_{f} + \frac{1}{2} k x_{f}^{2}$	Mass does not cancel automatically; check term-by-term

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Pendulum speed change problem
2022 · FRQ
Block on vertical spring problem

What's Next

Conservation of Mechanical Energy is the foundation for all energy-based problem solving in AP Physics 1, and it is a prerequisite for the next topics in Unit 4, and later for combining energy with momentum to solve collision problems. Without mastering the conditions for CoME and how to apply it, multi-concept FRQ problems that combine energy and forces or energy and momentum will be very difficult to solve correctly, as you will be forced to use slower kinematic methods that introduce more calculation errors. This topic also feeds into the bigger picture of conservation laws across all of physics, which are the core of AP Physics 1 and all subsequent physics study.