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Physics · Work, Energy, Momentum and Circular Motion

Conservation of Mechanical Energy

Conservation of mechanical energy derived from the work-energy theorem, applied via a stated reference level to curved tracks, pendulums and spring launches, with friction as an energy-accounting extension.

Conservation of Mechanical Energy

The last three chapters built the pieces: work transfers energy, kinetic energy is the energy of motion, potential energy is the energy of configuration. This chapter assembles them into the single most useful principle in mechanics — that in the absence of dissipation, the total mechanical energy of a system does not change, however complicated the motion in between.

The power of the statement lies in what it lets you ignore. A ball rolling down a curved ramp experiences a normal force that changes direction continuously and a gravitational component that varies with the slope. Solving that with Newton's laws would be formidable. Conservation of energy gives the speed at the bottom in one line, because it compares only two moments and never asks what happened between them.

The principle. If only conservative forces do work on a system, its total mechanical energy is constant:

Ek+Ep=constant⟺ΔEk+ΔEp=0E_k + E_p = \text{constant} \qquad\Longleftrightarrow\qquad \Delta E_k + \Delta E_p = 0

1. Derivation from the Work–Energy Theorem

Nothing new is being assumed. Start from the work–energy theorem, which holds for any force:

Wnet=ΔEkW_{\text{net}} = \Delta E_k

Split the net work into contributions from conservative and non-conservative forces:

Wcons+Wnon-cons=ΔEkW_{\text{cons}} + W_{\text{non-cons}} = \Delta E_k

By the definition of potential energy, Wcons=−ΔEpW_{\text{cons}} = -\Delta E_p. Substituting,

−ΔEp+Wnon-cons=ΔEk-\Delta E_p + W_{\text{non-cons}} = \Delta E_k  Wnon-cons=ΔEk+ΔEp \boxed{\ W_{\text{non-cons}} = \Delta E_k + \Delta E_p\ }

This is the general statement, and it is worth committing to memory in this form, because it covers both cases at once. When no non-conservative force does work — no friction, no drag, no external push — the left-hand side vanishes and

ΔEk+ΔEp=0⟺12mv2+mgh=12mu2+mgh0\Delta E_k + \Delta E_p = 0 \qquad\Longleftrightarrow\qquad \tfrac{1}{2}mv^{2} + mgh = \tfrac{1}{2}mu^{2} + mgh_0

Mechanical energy is conserved. When friction is present, the same equation still applies, with Wnon-consW_{\text{non-cons}} negative and equal in magnitude to the thermal energy generated.

2. Using the Principle

Every conservation-of-energy problem follows the same three steps.

  1. Choose a reference level for gravitational potential energy and state it. Any choice works; the lowest point of the motion is usually the most convenient, since it makes Ep=0E_p = 0 there.
  2. Write the total mechanical energy at the initial and final points. Include every relevant store: kinetic, gravitational, elastic.
  3. Set them equal (if no dissipation) or relate them through Wnon-consW_{\text{non-cons}} (if there is).

Worked example — a curved ramp. A ball of mass mm is released from rest at a height hh on a smooth curved track. Find its speed at the bottom.

Taking the bottom as the reference level:

12m(0)2+mgh⏟top=12mv2+0⏟bottom\underbrace{\tfrac{1}{2}m(0)^{2} + mgh}_{\text{top}} = \underbrace{\tfrac{1}{2}mv^{2} + 0}_{\text{bottom}} v=2ghv = \sqrt{2gh}

Three things about this result deserve attention. The mass cancels, so a heavy ball and a light one arrive at the same speed. The shape of the track is irrelevant — only the height drop matters, because the normal force does no work and gravity is conservative. And the answer is identical to the free-fall result v2=u2+2asv^{2} = u^{2}+2as, which it must be, since the same physics is being expressed twice.

Worked example — a pendulum. A pendulum bob on a string of length LL is released from an angle θ\theta to the vertical. The height drop is

h=L−Lcos⁡θ=L(1−cos⁡θ)h = L - L\cos\theta = L(1-\cos\theta)

so at the lowest point

v=2gL(1−cos⁡θ)v = \sqrt{2gL(1-\cos\theta)}

Again the mass cancels, and again the tension in the string does no work, being always perpendicular to the motion. Getting the geometry of hh right — L(1−cos⁡θ)L(1-\cos\theta), not Lcos⁡θL\cos\theta — is the only real difficulty, and it is worth checking at the limits: θ=0\theta = 0 gives h=0h = 0, and θ=90∘\theta = 90^\circ gives h=Lh = L, both correct.

3. Energy Bar Charts and the Shape of an Exchange

A useful way to see a problem before calculating it is to picture the total energy as a fixed quantity being redistributed among stores.

For a body thrown vertically upward: at launch, all the energy is kinetic; at the apex, all of it is gravitational potential; on return, all kinetic again. At any intermediate height the two shares always sum to the same total. Setting 12mv2=mgh\tfrac{1}{2}mv^{2} = mgh at the apex immediately gives h=v22gh = \dfrac{v^{2}}{2g}, the familiar result.

For a mass oscillating on a spring: at the extremes of the motion all the energy is elastic and the speed is zero; passing through the equilibrium position all of it is kinetic and the speed is maximum. The exchange is continuous, and the total is fixed — the structure that defines simple harmonic motion, developed fully in Block C.

For a bouncing ball with losses: each bounce reduces the total mechanical energy, so the bar representing the total gets shorter at every impact, and the ball rises less high each time. The missing energy has become thermal and sound energy.

4. Systems with Friction

When friction acts, mechanical energy is not conserved but total energy still is. The equation from Section 1 becomes an accounting statement:

Ek,i+Ep,i=Ek,f+Ep,f+ΔEthermalΔEthermal=f dE_{k,i} + E_{p,i} = E_{k,f} + E_{p,f} + \Delta E_{\text{thermal}} \qquad \Delta E_{\text{thermal}} = f\,d

where dd is the distance travelled along the path — not the displacement, since friction dissipates energy on every leg of a journey.

Worked example. A block of mass 2.0 kg2.0~\text{kg} slides from rest down a 5.0 m5.0~\text{m} ramp inclined at 30∘30^\circ, arriving at the bottom with a speed of 6.0 ms6.0~\dfrac{\text{m}}{\text{s}}. Find the average friction force.

The height drop is h=5.0sin⁡30∘=2.5 mh = 5.0\sin 30^\circ = 2.5~\text{m}, so

Ep,i=mgh=2.0(9.81)(2.5)=49.1 JE_{p,i} = mgh = 2.0(9.81)(2.5) = 49.1~\text{J} Ek,f=12(2.0)(6.0)2=36.0 JE_{k,f} = \tfrac{1}{2}(2.0)(6.0)^{2} = 36.0~\text{J}

The difference has been dissipated:

f d=49.1−36.0=13.1 J⟹f=13.15.0=2.6 Nf\,d = 49.1 - 36.0 = 13.1~\text{J} \quad\Longrightarrow\quad f = \frac{13.1}{5.0} = 2.6~\text{N}

The method is characteristic: the energy that has gone missing is the work done against friction, and dividing by the distance gives the force.

5. Why the Principle Works: Conservative Forces Again

The reason mechanical energy is conserved is precisely the path-independence established two chapters ago. Because gravity does the same work on every route between two heights, the energy "banked" as potential energy is guaranteed to be returned in full. Friction offers no such guarantee, and it is that broken guarantee — not any failure of energy conservation itself — that makes mechanical energy decrease.

It follows that the normal force and string tension never spoil conservation, even though they are not conservative forces in the technical sense: they do no work at all, being perpendicular to the motion. This is what allows the principle to be applied to rollercoasters, pendulums and beads on wires without knowing anything about the constraining forces.

Two limitations should be stated. First, the principle gives speeds, not velocities: energy is a scalar and cannot tell you the direction of motion. Second, it gives no times: nothing in Ek+Ep=constantE_k + E_p = \text{constant} contains tt. For either of those, return to Newton's laws.

6. A Combined Worked Example

Problem. A 0.30 kg0.30~\text{kg} ball is launched by a spring of constant 400 Nm400~\dfrac{\text{N}}{\text{m}} compressed by 12 cm12~\text{cm}, up a smooth track that rises 0.50 m0.50~\text{m}. Find its speed at the top of the rise.

Reference level: the launch point.

Initial energy — all elastic:

Ei=12kx2=12(400)(0.12)2=2.88 JE_i = \tfrac{1}{2}kx^{2} = \tfrac{1}{2}(400)(0.12)^{2} = 2.88~\text{J}

Final energy — kinetic plus gravitational:

Ef=12mv2+mgh=12(0.30)v2+0.30(9.81)(0.50)E_f = \tfrac{1}{2}mv^{2} + mgh = \tfrac{1}{2}(0.30)v^{2} + 0.30(9.81)(0.50)

Equate (the track is smooth, so mechanical energy is conserved):

2.88=0.15v2+1.472.88 = 0.15v^{2} + 1.47 v2=1.410.15=9.40⟹v=3.1 msv^{2} = \frac{1.41}{0.15} = 9.40 \quad\Longrightarrow\quad v = 3.1~\frac{\text{m}}{\text{s}}

Note the check available at the end: had the rise been higher than 2.880.30(9.81)=0.98 m\dfrac{2.88}{0.30(9.81)} = 0.98~\text{m}, the required v2v^{2} would have come out negative — the algebraic signal that the ball cannot reach that height at all.

7. Common Exam Pitfalls

  • Applying conservation when friction is present. Use Wnon-cons=ΔEk+ΔEpW_{\text{non-cons}} = \Delta E_k + \Delta E_p instead, and identify the missing energy as thermal.
  • Failing to state the reference level, or changing it midway through a problem.
  • Using the distance along a slope as the height. Only the vertical drop h=dsin⁡θh = d\sin\theta enters mghmgh.
  • Getting the pendulum geometry wrong. The drop is L(1−cos⁡θ)L(1-\cos\theta), not Lcos⁡θL\cos\theta — check it at θ=0\theta = 0 and θ=90∘\theta = 90^\circ.
  • Expecting energy conservation to give a direction or a time. It gives speeds only.
  • Forgetting to include every store. Elastic energy, gravitational energy and kinetic energy may all be present in the same problem.
  • Thinking the normal force spoils conservation. It does no work, so it cannot.
  • Assuming a heavier body arrives faster. The mass cancels in every friction-free energy problem of this kind.

Summary

  • Splitting the work–energy theorem into conservative and non-conservative parts gives the master equation Wnon-cons=ΔEk+ΔEpW_{\text{non-cons}} = \Delta E_k + \Delta E_p
  • With no dissipative forces, mechanical energy is conserved: Ek+Ep=constantE_k + E_p = \text{constant}.
  • Solve by choosing and stating a reference level, writing the total energy at two points, and equating them.
  • A body falling through a height hh from rest arrives at v=2ghv = \sqrt{2gh} — independent of mass and of the path, provided the track is smooth.
  • A pendulum released at angle θ\theta drops through h=L(1−cos⁡θ)h = L(1-\cos\theta), giving v=2gL(1−cos⁡θ)v = \sqrt{2gL(1-\cos\theta)} at the lowest point.
  • The normal force and string tension never break conservation, because they do no work at all.
  • With friction, mechanical energy falls by f df\,d, computed along the distance travelled; total energy is still conserved, with the difference appearing as thermal energy.
  • Energy methods give speeds but not velocities, and never times — for those, return to Newton's laws.