Conservation of Energy: Worked Examples
Conservation of energy examples in physics share one core structure: write down everything the system starts with, write down everything it ends with, and set them equal. That sounds simple, but the step most students skip is the energy bookkeeping that sits between those two lines. They jump straight to the formula without identifying what the initial and final states actually are, and that is where marks disappear.
The four problems below are graded by difficulty. Each one introduces one new layer of complexity, so by the time you reach the friction-loss incline, you have already seen every piece of it in a simpler context. This follows the approach outlined in OpenStax University Physics (Volume 1, Chapter 8), which treats energy bookkeeping as the central skill, not the formula recall.
What Is Conservation of Energy?
Conservation of energy states that the total mechanical energy of an isolated system remains constant. Mechanical energy is the sum of kinetic energy (KE) and potential energy (PE). An object falling, swinging, rolling, or being launched by a spring converts energy between these forms, but the total does not change unless an external force like friction removes energy as heat.
The law itself is broader: all forms of energy, including thermal, chemical, and electromagnetic, sum to a constant across a closed system. In introductory physics problems, you focus on the mechanical subset and track any losses explicitly.
Energy Bookkeeping: The Core Method
Every conservation of energy problem follows the same five-step structure. The structure is worth learning before the first example, because it applies unchanged across all four problems here.
| Step | Action | What to write |
|---|---|---|
| 1 | Define the system | Name every object; pick a reference height (h = 0) at the lowest point |
| 2 | Write initial energy | KE_i = (1/2)mv_i^2, PE_i = mgh_i, add elastic PE if spring present |
| 3 | Write final energy | KE_f = (1/2)mv_f^2, PE_f = mgh_f, add elastic PE if spring present |
| 4 | Include losses | W_friction = friction force times distance; subtract from right side |
| 5 | Solve and check | Isolate unknown, cancel mass if possible, verify units and sign |
This five-step method works on every conservation of energy problem. Step 4 is zero for friction-free systems.
KE and PE: The Two Forms You Need First
Kinetic energy is the energy of motion: KE = (1/2)mv2. The units are joules (J). Mass in kilograms, speed in metres per second. A 2 kg object moving at 4 m/s carries KE = (1/2)(2)(16) = 16 J.
Gravitational potential energy measures height above a chosen reference: PE = mgh. Set h = 0 at the lowest point in the problem, and all PE values come out positive. That same 2 kg object sitting 3 m above the ground holds PE = (2)(9.8)(3) = 58.8 J. The reference height choice is arbitrary; what matters is that you use the same reference throughout the whole problem.
Always set h = 0 at the lowest point in the problem. This keeps all PE values non-negative and prevents sign errors. If two positions are at the same height, their PE contributions cancel and you can ignore height in that part of the equation entirely.
Example 1: Free-Fall KE-PE Exchange
A 3 kg ball is released from rest at a height of 5 m above the ground. Find its speed just before it hits the ground. This is the most direct conservation of energy example: pure KE-PE exchange with no friction.
Setting Up the Energy Equation
Set the reference height at the ground (h = 0). At the release point, the ball is at rest (v_i = 0) and 5 m high. At the moment before impact, the ball is at ground level (h_f = 0) and moving at unknown speed v_f.
Initial state: KE_i = 0 (ball is at rest), PE_i = mgh_i = (3)(9.8)(5) = 147 J. Final state: KE_f = (1/2)mv_f2, PE_f = 0 (h = 0 at reference level).
Step-by-Step Solution
Write the conservation equation: KE_i + PE_i = KE_f + PE_f. Substitute: 0 + 147 = (1/2)(3)v_f2 + 0. Simplify: 147 = 1.5 v_f2. Divide: v_f2 = 98. Take the square root: v_f = 9.9 m/s.
Notice that mass cancels if you work symbolically first. Starting from 0 + mgh = (1/2)mv2 + 0, divide both sides by m: gh = (1/2)v2. Rearrange: v = sqrt(2gh) = sqrt(2 x 9.8 x 5) = sqrt(98) = 9.9 m/s. The 3 kg mass was irrelevant. A 1 kg ball or a 100 kg ball dropped from 5 m reaches the same speed. That is the cancellation that shows up on almost every friction-free problem.
Example 2: Pendulum Swing
A pendulum has a string length of 1.5 m. The bob is released from rest at an angle of 30 degrees from vertical. Find its speed at the lowest point. This is a conservation of energy example that requires a small geometry step before the energy equation.
Height Geometry of the Pendulum
The tricky part of pendulum problems is finding the height the bob drops. Set h = 0 at the lowest point (bottom of the swing). At angle theta = 30 degrees, the bob sits at a height h = L(1 - cos theta) above the lowest point.
Calculate: h = 1.5 x (1 - cos 30) = 1.5 x (1 - 0.866) = 1.5 x 0.134 = 0.201 m. That 0.201 m is the only height change that matters. The pendulum's mass and string length are both known; the unknown is the speed at the bottom.
Finding Speed at the Bottom
Initial state: bob is at rest at height 0.201 m. KE_i = 0, PE_i = mgh = m(9.8)(0.201) = 1.97m J (keeping mass symbolic). Final state: bob is at the bottom (h = 0), speed is v_f. KE_f = (1/2)mv_f2, PE_f = 0.
Set equal: 0 + mgh = (1/2)mv_f2 + 0. Cancel mass: gh = (1/2)v_f2. Solve: v_f = sqrt(2gh) = sqrt(2 x 9.8 x 0.201) = sqrt(3.94) = 1.98 m/s.
The formula v = sqrt(2gL(1 - cos theta)) is worth memorising for pendulum conservation of energy practice problems. Here it gives sqrt(2 x 9.8 x 1.5 x 0.134) = sqrt(3.94) = 1.98 m/s, matching the step-by-step result.
The height formula h = L(1 - cos theta) measures theta from the vertical, not from the horizontal. A pendulum swung out to 30 degrees from vertical gives h = L(1 - cos 30) = 0.134L. If you measured 30 degrees from horizontal (i.e., 60 degrees from vertical) you would get h = L(1 - cos 60) = 0.5L, a height more than three times larger and completely wrong. Always confirm: theta is the angle between the string and the straight-down vertical position.
Example 3: Block Down an Incline with Friction
A 4 kg block slides from rest down a 6 m ramp inclined at 25 degrees. The coefficient of kinetic friction between block and ramp is 0.15. Find the speed of the block at the bottom. This is the first conservation of energy practice problem that requires a friction loss term, and it is the format most common in university physics exams.
Accounting for Thermal Energy Loss
The vertical drop is h = 6 x sin 25 = 6 x 0.423 = 2.54 m. The initial PE is mgh = (4)(9.8)(2.54) = 99.6 J. Without friction, the block would reach the bottom with all 99.6 J converted to KE.
Friction steals energy. The friction force on an incline equals mu_k times the normal force. The normal force on a slope is N = mg cos theta = (4)(9.8)(cos 25) = (4)(9.8)(0.906) = 35.5 N. The friction force is then f = 0.15 x 35.5 = 5.33 N. Over the 6 m path length (not the vertical drop), friction removes W_friction = 5.33 x 6 = 32.0 J.
Solving with the Work-Energy Theorem
Write the modified conservation equation: KE_i + PE_i = KE_f + PE_f + W_friction. Substitute the known values: 0 + 99.6 = (1/2)(4)v_f2 + 0 + 32.0. Solve: 67.6 = 2v_f2. Then v_f2 = 33.8, and v_f = 5.82 m/s.
Without friction the answer would have been v_f = sqrt(2 x 9.8 x 2.54) = sqrt(49.8) = 7.06 m/s. Friction reduced the final speed from 7.06 m/s to 5.82 m/s, a 17.6% reduction. That is the kind of cross-check that catches sign errors: friction always lowers the final speed.
| Quantity | Symbol | Value | How calculated |
|---|---|---|---|
| Ramp length | d | 6 m | given |
| Incline angle | theta | 25 deg | given |
| Vertical height | h | 2.54 m | d sin theta = 6 x 0.423 |
| Initial PE | PE_i | 99.6 J | mgh = (4)(9.8)(2.54) |
| Normal force | N | 35.5 N | mg cos theta = (4)(9.8)(0.906) |
| Friction force | f | 5.33 N | mu_k N = 0.15 x 35.5 |
| Energy lost to friction | W_f | 32.0 J | f x d = 5.33 x 6 |
| Final KE | KE_f | 67.6 J | PE_i - W_f = 99.6 - 32.0 |
| Final speed | v_f | 5.82 m/s | sqrt(2 KE_f / m) = sqrt(33.8) |
Full energy ledger for the incline problem. Every number traces to a single calculation step.
Friction force acts along the surface, so multiply by the path length d (6 m here), not the vertical drop h (2.54 m). Gravitational PE uses the vertical height. Mixing these up is the single most common error in incline conservation of energy problems and typically overstates or understates energy loss by a factor of sin(theta) or cos(theta).
The energy calculator on the university resources page can verify each step here. Enter the mass, height, and friction parameters and confirm your manual calculation matches. Try it for the no-friction version first (enter mu_k = 0) to see the baseline 7.06 m/s, then add the friction coefficient to see the 5.82 m/s result.
Physics Energy Calculator
Enter mass, height, and friction coefficient to check conservation of energy problems step by step.
Example 4: Spring Launch and Projectile
A 0.2 kg ball rests against a spring with spring constant k = 800 N/m. The spring is compressed 0.05 m from its natural length. The spring releases the ball horizontally from the edge of a 1.2 m high table. Find the ball's speed as it leaves the spring and the horizontal distance it travels before hitting the floor. This problem chains two energy conservation steps together.
Elastic PE to KE at Launch
Elastic potential energy stored in the compressed spring: PE_elastic = (1/2)kx2 = (1/2)(800)(0.05)2 = (1/2)(800)(0.0025) = 1.0 J. When the spring releases the ball to its natural length, all that elastic PE converts to KE (assuming the spring is horizontal, so gravitational PE does not change during the launch phase).
Set elastic PE = KE at launch: 1.0 = (1/2)(0.2)v_launch2. Then v_launch2 = 10, and v_launch = 3.16 m/s. That is the horizontal speed leaving the spring.
Full Solution with Projectile Check
Once the ball leaves the table, it is in projectile motion with initial horizontal speed 3.16 m/s, initial vertical speed zero, and height 1.2 m to fall. The time to reach the floor comes from the vertical: h = (1/2)gt2, so t = sqrt(2h/g) = sqrt(2 x 1.2 / 9.8) = sqrt(0.245) = 0.495 s.
Horizontal distance: x = v_horizontal x t = 3.16 x 0.495 = 1.56 m. As a cross-check, apply conservation of energy from the compressed spring to the moment before floor impact. Initial energy: elastic PE = 1.0 J plus gravitational PE from height 1.2 m, which is (0.2)(9.8)(1.2) = 2.35 J. Total initial energy = 3.35 J. Just before hitting the floor, all energy is KE: (1/2)(0.2)v_impact2 = 3.35 J, so v_impact = sqrt(33.5) = 5.79 m/s. That cross-check is optional in the exam but worth 30 seconds to catch algebraic errors.
When a problem has two distinct energy conversions, solve them sequentially. Step 1: spring to KE (elastic PE converts to kinetic). Step 2: KE at table edge to final KE plus any gravitational PE lost during flight. Each step uses the same bookkeeping structure. The output of step 1 (launch speed) becomes the input of step 2.
Common Errors That Cost Marks
Four mistakes appear across conservation of energy practice problems at university level. Knowing them by name means you can check each one in under 30 seconds before submitting.
| Error | What happens | How to avoid it |
|---|---|---|
| Wrong reference height | PE values come out negative or inconsistent | Set h = 0 at the lowest point in the problem before writing any equations |
| Distance vs height in friction | Friction energy loss is 20-30% wrong on inclines | Friction acts along the path (use ramp length d); gravity acts vertically (use height h) |
| Forgetting mass cancellation check | Extra algebra, error-prone | After setting up the equation, look for m on both sides and cancel before substituting numbers |
| Using sin vs cos for normal force | Normal force and friction force are wrong | On an incline at angle theta from horizontal: N = mg cos theta (not sin) |
| Ignoring elastic PE in spring problems | Spring energy simply disappears from the calculation | Include (1/2)kx^2 in the initial energy state whenever a spring is compressed or stretched |
Each of these errors produces a physically plausible but wrong answer. A quick pre-submission check catches all five.
The most costly error in exams is the distance vs height confusion in friction problems. A block on a 30-degree slope with ramp length 4 m falls only h = 4 sin 30 = 2 m vertically but travels 4 m along the surface. Using 2 m instead of 4 m for the friction distance understates the energy lost by 50% and produces a final speed that is substantially too high. Examiners write questions specifically to catch this, and it accounts for a disproportionate share of partial-credit deductions on mechanics papers.
The OpenStax University Physics chapter on conservation of energy provides additional worked examples including rotational kinetic energy and multi-body systems. MIT OpenCourseWare 8.01 (Classical Mechanics) has problem sets with full solutions that extend these four examples to more complex scenarios.
For more mechanics problems solved step by step, the free body diagram guide covers the force analysis that feeds into incline and friction calculations. The u-substitution worked examples post uses the same graded-difficulty structure for calculus if you want to build the same kind of step-by-step pattern across subjects. For chemistry, the resonance structures worked examples post applies the same progressive-difficulty approach to organic chemistry.
If you want to practice these problems with an AI tutor that checks your energy bookkeeping line by line and flags the exact step where errors appear:
The subject calculators hub also has tools for mechanics, thermodynamics, and other physics topics that can verify your manual calculations. For broader revision across modules, the university resources hub organises tools and guides by subject area.
Key Takeaways
- Conservation of energy states that KE + PE = constant in an isolated system. When friction acts, the equation becomes KE_i + PE_i = KE_f + PE_f + W_friction, where W_friction is the energy converted to heat.
- In friction-free problems, mass cancels because both KE and gravitational PE contain m as a factor. The speed depends only on the height change and g: v = sqrt(2gh).
- For pendulum problems, the height drop is h = L(1 - cos theta), where theta is measured from the vertical. This is the only geometry step needed before applying the standard energy equation.
- For incline problems with friction, the friction force equals mu_k times mg cos theta, and the energy lost equals that force times the ramp length (not the vertical height). Using vertical height instead of ramp length overstates or understates friction loss by roughly 1/sin(theta).
- The work-energy theorem (W_net = delta KE) is the same principle as conservation of energy restated: non-conservative forces like friction do negative work that reduces mechanical energy by exactly the thermal energy they produce.
- Spring problems add elastic PE = (1/2)kx^2. At maximum compression, all KE converts to elastic PE. At full release, elastic PE converts back to KE. Chain the steps when a spring launches an object into further motion.
- The five-item pre-submission checklist (reference height, friction path vs height, mass cancellation, normal force formula, spring PE) catches the most common errors in under 30 seconds.
