Conservation of Energy Calculator

Conservation of Energy: Definition, Formula, and Practical Use

The Conservation of Energy principle says that energy cannot be created or destroyed—only transformed or transferred. For an isolated system with no energy exchange across its boundaries, the total energy remains constant over time. In real engineering problems, we track how energy shifts between kinetic, gravitational potential, elastic (spring), and rotational forms, plus any non-conservative work such as friction, air resistance, or motor input.

\( \displaystyle \boxed{E_{\text{initial}} + W_{\text{nc}} = E_{\text{final}}} \quad\text{with}\quad E = K + U_g + U_s (+\,K_{\text{rot}} + \text{other forms}) \)

If the system is isolated, the non-conservative work term vanishes \((W_{\text{nc}}=0)\) and \( \displaystyle E_{\text{total}} = \text{constant} \). When losses or inputs exist (e.g., friction, drag, motors, brakes), they appear as negative or positive values in \(W_{\text{nc}}\).

Core Equations and Rearrangements

\( \displaystyle K = \tfrac{1}{2} m v^2, \qquad U_g = m g h, \qquad U_s = \tfrac{1}{2} k x^2, \qquad K_{\text{rot}} = \tfrac{1}{2} I \omega^2 \)

\( \displaystyle (K_i + U_{g,i} + U_{s,i} + K_{\text{rot},i} + \cdots) + W_{\text{nc}} \;=\; (K_f + U_{g,f} + U_{s,f} + K_{\text{rot},f} + \cdots) \)

Two useful rearrangements that your calculator’s outputs may mirror:

\( \displaystyle v_f = \sqrt{\,v_i^2 + \frac{2}{m}\Big[(U_i – U_f) + (K_{\text{rot},i}-K_{\text{rot},f}) + W_{\text{nc}}\Big]\,} \)

\( \displaystyle h_{\max} = \frac{K_i + U_{g,i} + U_{s,i} + K_{\text{rot},i} + W_{\text{nc}} – (K_f + U_{s,f} + K_{\text{rot},f})}{m g} \)

Variables and Units

\(m\) — Mass (kg, SI).
\(v\) — Linear speed (m/s); \(\omega\) — angular speed (rad/s).
\(g\) — Gravitational acceleration (\(\approx 9.80665\,\text{m/s}^2\) near Earth’s surface).
\(h\) — Height relative to a chosen reference (m). Only differences in \(h\) matter.
\(k\) — Spring constant (N/m); \(x\) — spring compression/extension (m).
\(I\) — Mass moment of inertia (kg·m\(^2\)).
\(W_{\text{nc}}\) — Work by non-conservative forces (J). Positive if energy added to the system; negative if removed.

Common Energy Forms (at a glance)

Energy Form	Symbol	Formula	Typical Context
Kinetic (translation)	\(K\)	\( \tfrac{1}{2} m v^2 \)	Moving bodies, vehicles, projectiles
Gravitational potential	\(U_g\)	\( m g h \)	Ramps, lifts, roller coasters
Elastic (spring)	\(U_s\)	\( \tfrac{1}{2} k x^2 \)	Springs, compliant mechanisms
Rotational kinetic	\(K_{\text{rot}}\)	\( \tfrac{1}{2} I \omega^2 \)	Wheels, pulleys, flywheels
Thermal/internal	\(Q,\,U_{\text{int}}\)	Process-dependent	Frictional heating, electrical losses

Step-by-Step Method (pairs perfectly with the calculator above)

Define the system and states. Decide what’s “inside.” Pick initial and final states and a zero height and zero spring length.
List energy terms at each state. Include \(K, U_g, U_s, K_{\text{rot}}\) as needed; ignore terms that are clearly zero.
Account for non-conservative work. Add motor/brake work; subtract losses (friction, drag, eddy current braking) as negative \(W_{\text{nc}}\).
Write the energy balance. \( E_i + W_{\text{nc}} = E_f \).
Solve algebraically for the unknown (often \(v_f\), \(h\), or \(x\)).
Check units and limits. Results should be physically plausible; test special cases like \(W_{\text{nc}}=0\) or \(k=0\).

Worked Examples

Example 1 — Sled Down a Hill (No Friction)

A \(m=40\,\text{kg}\) sled starts from rest at \(h_i=12\,\text{m}\) and arrives at \(h_f=2\,\text{m}\) with negligible losses. Find the final speed.

\( \displaystyle K_i+U_{g,i} = K_f + U_{g,f} \Rightarrow 0 + m g h_i = \tfrac{1}{2} m v_f^2 + m g h_f \)

\( \displaystyle v_f = \sqrt{2 g (h_i-h_f)} = \sqrt{2(9.80665)(10)} \approx 14.00\,\text{m/s}. \)

Example 2 — Block Compressing a Spring with Kinetic Friction

A \(m=3.0\,\text{kg}\) block slides on a rough horizontal surface (\(\mu_k=0.25\)) into a spring (\(k=400\,\text{N/m}\)) with initial speed \(v_i=2.6\,\text{m/s}\). What compression \(x\) stops the block?

Friction does negative work over the compression distance: \( W_{\text{fr}} = -\mu_k m g x \).

\( \displaystyle \tfrac{1}{2} m v_i^2 + W_{\text{fr}} = U_{s,f} \ \Rightarrow\ \tfrac{1}{2} m v_i^2 – \mu_k m g x = \tfrac{1}{2} k x^2 \)

Substitute: \( \tfrac{1}{2}(3)(2.6)^2 – (0.25)(3)(9.80665) x = \tfrac{1}{2}(400) x^2 \). Numerically, \(10.14 – 7.355x = 200x^2 \Rightarrow 200x^2 + 7.355x – 10.14 = 0\). The positive root gives \( x \approx 0.20\,\text{m} \).

Example 3 — Roller Coaster with Energy Loss

A \(m=500\,\text{kg}\) car drops from \(h_i=30\,\text{m}\) to \(h_f=5\,\text{m}\). Losses remove \(W_{\text{loss}}=25\,\text{kJ}\). Find the speed at \(h_f\).

\( \displaystyle m g h_i + W_{\text{nc}} = \tfrac{1}{2} m v_f^2 + m g h_f,\quad W_{\text{nc}}=-W_{\text{loss}} \)

\( m g (h_i-h_f) = 500(9.80665)(25) = 122{,}583\,\text{J} \). After losses: \( \tfrac{1}{2} m v_f^2 = 122{,}583 – 25{,}000 = 97{,}583\,\text{J} \Rightarrow v_f \approx \sqrt{(2\cdot 97{,}583)/500} \approx 19.76\,\text{m/s}. \)

Example 4 — Rolling Without Slipping

A solid cylinder rolls from height \(h\) without slipping. What is \(v_f\) at the bottom?

\( \displaystyle m g h = K + K_{\text{rot}} = \tfrac{1}{2} m v_f^2 + \tfrac{1}{2} I \omega^2,\quad I=\tfrac{1}{2} m R^2,\ \omega=\tfrac{v_f}{R} \Rightarrow m g h = \tfrac{3}{4} m v_f^2 \)

\( \displaystyle v_f = \sqrt{\tfrac{4}{3} g h} \) (slower than a frictionless sliding block because some energy is in rotation).

Why Use Conservation of Energy?

Fast speed/height estimates without summing forces over time.
Design sanity checks for ramps, coasters, ziplines, drop towers, and braking systems.
Spring-mass systems where compliance stores and releases energy.
Rotational systems (pulleys, flywheels) combining translational and rotational energies in one balance.
Loss budgeting — quantify how much mechanical energy becomes heat, noise, or deformation.

Assumptions and Limitations

System boundaries matter: Energy is always conserved overall, but if you exclude a motor, battery, or environment from your system, you must include their effects as \(W_{\text{nc}}\).
Non-conservative forces don’t “break” conservation: Friction and drag convert mechanical energy to internal/thermal energy; include them as negative work or with a power-loss model.
Path independence applies only to conservative forces: \(U_g\) and \(U_s\) depend only on endpoints; frictional loss depends on the path length and conditions.
Beyond classical regime: Near relativistic speeds or at microscopic scales, classical formulas (e.g., \(K=\tfrac12 m v^2\)) no longer suffice.
Variable mass/time-varying parameters: Rockets (mass ejection), changing \(k(t)\), or time-dependent inputs may require work/power integrals or momentum methods.

Conservation of Energy — FAQ

Is energy always conserved?

Yes, for the universe as a whole. In your chosen system, total energy is conserved if no energy crosses the boundary or if all transfers are accounted for (via \(W_{\text{nc}}\) or heat).

How is this different from conservation of momentum?

Energy conservation tracks scalar energy forms. Momentum conservation involves vectors and requires zero net external impulse. Perfectly inelastic collisions conserve momentum but not mechanical energy; some energy becomes heat and sound.

When can I ignore friction and drag?

When surfaces are smooth, speeds are low, or travel is short. Otherwise include a loss term (negative \(W_{\text{nc}}\)) or use a drag model (e.g., \(F_D \propto v\) or \(v^2\)) to estimate dissipation.

What’s the work–energy theorem?

The net work on a particle equals the change in its kinetic energy: \( \displaystyle W_{\text{net}} = \Delta K \). Combining this with potential energies for conservative forces yields the general balance \(E_i + W_{\text{nc}} = E_f\).

Which zero height should I use?

Any consistent reference level works; only differences in \(m g h\) matter.

How do I include rotation or rolling?

Add \(K_{\text{rot}}=\tfrac12 I\omega^2\) at each state. For rolling without slipping, relate \(v\) and \(\omega\) via \(v=R\omega\) and include both translation and rotation in the energy sum.

Practice Problems (with brief answers)

Problem A — Projectile Speed at Lower Height (No Drag)

A \(0.20\,\text{kg}\) ball is launched horizontally at \(v_i=8.0\,\text{m/s}\) from \(h_i=20\,\text{m}\). What is its speed at \(h_f=5.0\,\text{m}\)?

\( \displaystyle \tfrac12 m v_f^2 = \tfrac12 m v_i^2 + m g (h_i – h_f) \Rightarrow v_f = \sqrt{8.0^2 + 2(9.80665)(15)} \approx 19.5\,\text{m/s}. \)

Problem B — Spring Launcher Height (No Losses)

A \(0.10\,\text{kg}\) puck is launched by a spring (\(k=300\,\text{N/m}\)) compressed \(x=0.12\,\text{m}\). Find \(h_{\max}\) above the launch point.

\( \displaystyle \tfrac12 k x^2 = m g h_{\max} \Rightarrow h_{\max} = \frac{k x^2}{2 m g} \approx 2.20\,\text{m}. \)

Problem C — Ramp with Quadratic Drag (Approximate Work)

A \(m=60\,\text{kg}\) rider descends \(\Delta h=8\,\text{m}\) on a bike. If estimated drag/rolling losses total \(W_{\text{loss}}=1.8\,\text{kJ}\), find \(v_f\) at the bottom starting from rest.

\( \displaystyle m g \Delta h – W_{\text{loss}} = \tfrac12 m v_f^2 \Rightarrow v_f = \sqrt{\frac{2(m g \Delta h – W_{\text{loss}})}{m}} = \sqrt{2 g \Delta h – \frac{2 W_{\text{loss}}}{m}} \approx \sqrt{2(9.80665)(8) – \frac{3600}{60}} \approx 10.9\,\text{m/s}. \)

Key Takeaways

Master balance: \(E_i + W_{\text{nc}} = E_f\). Set \(W_{\text{nc}}=0\) for isolated systems.
Track the right forms: Include \(K, U_g, U_s, K_{\text{rot}}\) as appropriate for your scenario.
System definition drives bookkeeping: Decide what’s inside and model what crosses the boundary as work or heat.
Great for quick results: Energy methods often beat force-by-force approaches when the path is complicated but the states are simple.
Losses don’t “break” conservation: They convert mechanical energy into heat/sound; include them as negative work or via power models.