Conservation of mechanical energy

Conservation of mechanical energy sits inside the CBSE Class 11 unit Work, Energy and Power, immediately after work by constant and variable forces, kinetic energy, the work-energy theorem, and potential energy (including the spring potential energy $U=\tfrac12kx^2$ and gravitational potential energy $U=mgh$ near Earth). The prescribed syllabus explicitly lists conservative versus non-conservative forces, the conservation law itself, and motion in a vertical circle as core content, so every part of this article — the derivation, the vertical-circle application, and both worked examples — is squarely inside the 2025-26 board scope. One thing to keep honest: the syllabus asks you to treat the coefficient of restitution only qualitatively (it belongs properly to collisions, the next topic), so this article does not build a formal restitution law even though a taut string briefly resembles an elastic bounce.

Every mechanical system carries two kinds of energy that can be measured at any instant: kinetic energy $K$, the energy of motion, and potential energy $U$, the energy stored by position (height in a gravitational field, or compression/extension in a spring). Whenever the only forces doing work are conservative, these two are locked together like water sloshing between two connected tanks — whatever drains from one fills the other, and the combined level never changes.

where: $E$ is the total mechanical energy of the system (joules), $K = \tfrac12 mv^2$ is kinetic energy, and $U$ is potential energy — either gravitational, $U = mgh$ (height $h$ measured from a chosen reference level), or elastic, $U = \tfrac12 kx^2$ (spring constant $k$, displacement $x$ from natural length). The physical reading is simple: $E$ is a bookkeeping total, and the law of conservation of mechanical energy is the claim that this total is the same number at every instant of the motion, as long as gravity and springs are the only things doing work.

where: subscripts $i$ and $f$ label any two instants — initial and final — of the same motion. This is the working form of the law used in every problem: pick a convenient reference level for $U=0$, write down $K$ and $U$ at the state you know, write down $K$ and $U$ at the state you want, and set the two totals equal.

Conservative versus non-conservative forces. A force is conservative when the work it does on an object moving between two points does not depend on the path taken — only on the endpoints — which is exactly what lets a potential energy $U$ be defined for it in the first place. Gravity and the ideal spring force pass this test: lift a ball straight up or along a zig-zag path to the same height, and gravity does the same (negative) work either way. Friction fails the test completely — a longer path always means more work done against friction, so no single potential energy function can capture it, and any energy it removes is gone from the mechanical books for good, converted to heat and a little sound.

A second geometry the syllabus singles out is a ball or bob moving in a vertical circle of radius $r$, on a string or inside a circular track. Two forces act: gravity, which does work as height changes, and tension or the track's normal reaction, which always points along the radius, perpendicular to the instantaneous velocity.

where: the work done by a force perpendicular to displacement is $F\,d\cos 90^\circ = 0$ at every instant, so tension (or normal reaction) never appears in the energy equation at all — only gravity's potential energy $U=mgh$ needs to be tracked between any two points on the loop, with $h$ measured against a convenient reference such as the lowest point of the circle.

Board level - a ball dropped from height

A ball of mass $0.4\ \text{kg}$ is dropped from a height of $5\ \text{m}$ above the ground. Using conservation of mechanical energy, find its speed just before hitting the ground (ignore air resistance). Take $g = 9.8\ \text{m/s}^2$.

Step 1. Choose the ground as the reference level, $U=0$ there, and write down energies at the release point. At the top the ball starts from rest, so $K_i = 0$, and its height is $5\ \text{m}$, so $U_i = mgh = 0.4 \times 9.8 \times 5 = 19.6\ \text{J}$.

Step 2. Write down energies at the ground. Here $U_f = 0$ by choice of reference, and $K_f = \tfrac12 m v^2$, with $v$ the unknown speed just before impact.

Step 3. Apply $K_i + U_i = K_f + U_f$: $0 + 19.6 = \tfrac12(0.4)v^2 + 0$, which gives $0.2\,v^2 = 19.6$.

Step 4. Solve for $v$: $v^2 = 98$, so $v \approx \mathbf{9.9\ m/s}$. This matches the pure-kinematics result $v=\sqrt{2gh}=\sqrt{2\times9.8\times5}\approx 9.9\ \text{m/s}$, confirming that energy conservation and Newton's-laws kinematics agree, as they must for a conservative force.

Competitive level - vertical circular swing with a spring launch

A block of mass $0.5\ \text{kg}$ is pushed against a spring of force constant $k = 200\ \text{N/m}$, compressing it by $x = 0.2\ \text{m}$, on a smooth horizontal table. The block is released and slides (frictionlessly) until it reaches the bottom of a vertical circular loop of radius $r = 0.8\ \text{m}$, at the same level as the table, and goes around the inside of the loop. Find the block's speed at the top of the loop, and check whether it completes the loop at all (minimum speed needed at the top is $v_{min}=\sqrt{gr}$).

Step 1. Find the energy released by the spring, using $U_{spring} = \tfrac12 k x^2 = \tfrac12(200)(0.2)^2 = 4\ \text{J}$. Since the table is smooth (frictionless) and the spring's force is conservative, this entire $4\ \text{J}$ converts into the block's kinetic energy at the bottom of the loop: $\tfrac12 m v_b^2 = 4\ \text{J}$, giving $v_b^2 = \dfrac{2\times4}{0.5} = 16\ \text{m}^2/\text{s}^2$, so $v_b = 4\ \text{m/s}$.

Step 2. Set up conservation of mechanical energy between the bottom (b) and the top (t) of the loop, with the bottom as the reference level ($U_b=0$). The height gained is $h = 2r = 1.6\ \text{m}$, and the track's normal reaction does zero work throughout (perpendicular to velocity), so only gravity needs tracking: $\tfrac12 m v_b^2 + 0 = \tfrac12 m v_t^2 + mg(2r)$.

Step 3. Cancel mass $m$ from every term and substitute numbers: $\tfrac12(16) = \tfrac12 v_t^2 + (9.8)(1.6)$, i.e. $8 = \tfrac12 v_t^2 + 15.68$.

Step 4. This gives $\tfrac12 v_t^2 = 8 - 15.68 = -7.68$, a negative kinetic energy — physically impossible. The correct reading is that the block never reaches the top at all: it runs out of kinetic energy partway up the loop and leaves the track (or, if constrained by a groove, reverses) well below the top, because the $4\ \text{J}$ the spring supplied is less than the $mg(2r) = 0.5\times9.8\times1.6 = 7.84\ \text{J}$ needed just to climb to top-height, let alone retain the minimum speed there.

Step 5. Compare against the minimum-completion condition to see how far short the design falls. At the top, completing the loop needs at least $v_{min}=\sqrt{gr}=\sqrt{9.8\times0.8}\approx 2.8\ \text{m/s}$, so the minimum energy required at the bottom is $\tfrac12 m v_{min}^2 + mg(2r) = \tfrac12(0.5)(7.84) + 7.84 \approx 1.96+7.84 = \mathbf{9.8\ J}$ — more than double the $4\ \text{J}$ the spring actually delivers, confirming the block cannot complete the loop with this compression. Timing note: on a competitive-exam clock, do not grind through Step 4's algebra to a "negative $v_t^2$" surprise — the instant you compute the bottom kinetic energy (Step 1), compare it directly against the standard benchmark $mg(2r)$ (needed just to reach top height) or, better, $\tfrac52 mgr$ (the true minimum energy including $v_{min}=\sqrt{gr}$ at the top); a single inequality check like this takes seconds and tells you immediately whether "find $v_t$" even has a physical answer, before you invest in solving for it.

• Applying $K+U=\text{constant}$ when the question mentions a rough surface, air resistance, or any stated friction — mechanical energy is only conserved when every force doing work is conservative; with friction present you must use $\Delta K + \Delta U = W_{friction}$ instead.
• Forgetting that normal reaction and string tension do zero work in vertical-circle and track problems because they act perpendicular to velocity — students who "track" these forces in the energy equation get the wrong answer even though the physics of the geometry hasn't required it.
• Choosing an inconsistent reference level for $U=mgh$ between the initial and final states, or forgetting to keep the sign of $h$ consistent when an object goes both up and down in the same problem — always fix one reference level for the whole solution.
• Assuming a spring or vertical-loop system automatically completes its intended motion (a block "makes it" around a loop, or a spring "launches" a block over a hill) without checking a minimum-speed or minimum-energy condition first, as in the second worked example above.
• Mixing up which state has more potential energy without first sketching heights — a quick sketch showing which point is higher prevents sign errors in $U_i$ versus $U_f$.

• State the law of conservation of mechanical energy and derive it from the work-energy theorem and the definition of potential energy.
• Distinguish conservative from non-conservative forces with concrete examples, and explain why only conservative forces admit a potential energy.
• Apply $K_i+U_i=K_f+U_f$ to free-fall and spring-launch problems, choosing a consistent reference level for potential energy.
• Apply energy conservation to vertical circular motion, explaining why tension or normal reaction never appears in the energy equation.
• Check a minimum-speed or minimum-energy condition (such as $v_{min}=\sqrt{gr}$ at the top of a loop) before assuming a motion completes.
• Recognise when mechanical energy is not conserved (friction, air resistance present) and switch to the extended work-energy balance.

Ask the tutor: Still unsure why a negative value under a square root in an energy equation is a real physical answer rather than a mistake — ask the AI tutor to walk through the spring-and-loop worked example step by step and show exactly where the energy falls short.