6.02i, 6.02j - Energy conservation and the work-energy principle

6.02i, 6.02j - Energy conservation and the work-energy principle

Build an energy equation from initial and final states, with the right signs for work and energy loss. You will use conservation on smooth tracks, account for resistance through the work-energy principle, and handle spring and elastic-string systems, including a string that starts slack.

Conservation of mechanical energy

Two smooth tracks start and finish at the same heights, but one takes a longer, twisting route. Must a particle arrive more slowly on the longer track? If it reaches the endpoint, starts at the same speed and experiences only gravity and a perpendicular contact force, its final speed is the same. Energy cares about the vertical drop, not the track's length.

You already know

Kinetic energy is K=12mv2K=\tfrac12mv^2; gravitational potential energy is Ug=mghU_g=mgh; elastic potential energy is E=λx2/(2l)E=\lambda x^2/(2l). All use joules. Work done by resistance is negative, while work done against resistance is positive.

A force such as gravity transfers energy between the kinetic and potential stores. On a fixed smooth track, the contact force is perpendicular to the displacement and does no work. If there is no dissipative force or external energy input, the total mechanical energy stays constant.

Conservation of mechanical energy

When the chosen system has no external mechanical-energy input and no mechanical energy is dissipated,

K1+U1=K2+U2K_1+U_1=K_2+U_2

Here UU includes all relevant gravitational and elastic potential-energy stores, measured with consistent reference states. For gravity alone,

12mu2+mgh1=12mv2+mgh2\frac12mu^2+mgh_1=\frac12mv^2+mgh_2

Not in the formula booklet: learn it.

List both states before substituting. A turning point has zero instantaneous speed, but need not be an equilibrium position.

Worked example 1: descent from rest

Question. A particle of mass 2 kg2\,\mathrm{kg} is released from rest on a fixed smooth track. It descends through vertical height 5 m5\,\mathrm{m}. Assuming it reaches the lower point and air resistance is negligible, find its speed there.

Plan. Take zero gravitational potential energy at the lower point and equate initial and final mechanical energies.

0+(2)(9.8)(5)=12(2)v2+0initial energy = final energyv2=98simplifyv=98 m s−1speed is non-negative\begin{aligned} 0+(2)(9.8)(5)&=\frac12(2)v^2+0 && \text{initial energy = final energy}\\ v^2&=98 && \text{simplify}\\ v&=\sqrt{98}\,\mathrm{m\,s^{-1}} && \text{speed is non-negative} \end{aligned}

Why this step? Only the vertical drop determines the gravitational potential-energy decrease; the length of the smooth track is unnecessary.

Answer. The speed is 9.90 m s−19.90\,\mathrm{m\,s^{-1}}, to 3 significant figures.

Check. Final kinetic energy is 98 J98\,\mathrm{J}, exactly the gravitational potential energy lost.

Worked example 2 changes only the initial speed. The same drop now adds to an existing kinetic-energy store.

Worked example 2: an initial kinetic-energy store

Question. The same particle instead starts with speed 6 m s−16\,\mathrm{m\,s^{-1}} and descends through the same vertical height 5 m5\,\mathrm{m} on the smooth track. Find its final speed.

Plan. Include the initial kinetic energy before equating the two states.

12(2)(62)+(2)(9.8)(5)=12(2)v2all initial energy stores36+98=v2simplifyv=134 m s−1positive speed\begin{aligned} \frac12(2)(6^2)+(2)(9.8)(5)&=\frac12(2)v^2 && \text{all initial energy stores}\\ 36+98&=v^2 && \text{simplify}\\ v&=\sqrt{134}\,\mathrm{m\,s^{-1}} && \text{positive speed} \end{aligned}

Why this step? The gravitational contribution has not changed. It is the energy, rather than the speed itself, that adds to the initial value.

Answer. The final speed is 11.6 m s−111.6\,\mathrm{m\,s^{-1}}, to 3 significant figures.

Check. v2−u2=134−36=98=2g(5)v^2-u^2=134-36=98=2g(5).

The mass cancels in these gravity-only equations. That conclusion follows from the model; a mass-dependent resistance could change it.

Watch out

Wrong: v=u+2ghv=u+\sqrt{2gh}. Speeds do not add this way in an energy equation. Right: v2=u2+2ghv^2=u^2+2gh, derived by retaining both kinetic-energy terms for a descent of height hh.

The opening tracks therefore give the same final speed under the stated assumptions, even though their travel times need not be the same.

The work-energy principle and resistance

When forces do net work on a particle, its kinetic energy changes by that signed amount. This statement accounts for all forces, including gravity, driving forces and resistance.

Along a straight line, Newton's second law gives Fresultant=m dv/dt=mv dv/dxF_{\mathrm{resultant}}=m\,dv/dt=mv\,dv/dx. Integrating between two positions gives ∫Fresultant dx=∫uvmv dv=12mv2−12mu2\int F_{\mathrm{resultant}}\,dx=\int_u^v mv\,dv=\tfrac12mv^2-\tfrac12mu^2. The chain-rule acceleration identity used here is not in the formula booklet.

Work-energy principle

For a particle of constant mass,

Wresultant=K2−K1W_{\mathrm{resultant}}=K_2-K_1

Equivalently, if gravity and elastic forces are accounted for through potential-energy stores,

K1+U1+Winput=K2+U2+ElossK_1+U_1+W_{\mathrm{input}}=K_2+U_2+E_{\mathrm{loss}}

Here Eloss≥0E_{\mathrm{loss}}\ge0 is mechanical energy dissipated, and WinputW_{\mathrm{input}} is signed work by other specified driving or applied forces. Do not count the same force both as work and as a potential-energy change or loss.

Not in the formula booklet: learn it.

For constant resistance magnitude RR over distance ss, the work by resistance is −Rs-Rs and the mechanical energy loss is RsRs. A smooth contact force does no work only when it is perpendicular to the displacement; inspect the motion rather than relying on the word “smooth” alone.

Worked example 3: driving against resistance

Question. A particle of mass 2 kg2\,\mathrm{kg} moves along a horizontal line with initial speed 3 m s−13\,\mathrm{m\,s^{-1}}. A constant driving force of 10 N10\,\mathrm{N} acts forwards and a constant resistance of 2 N2\,\mathrm{N} acts backwards. Find its speed after travelling 4 m4\,\mathrm{m}.

Plan. Calculate the signed work of the resultant force and equate it to the kinetic-energy increase.

Wresultant=(10)(4)−(2)(4)=32 Jdriving work minus resistance work magnitude12(2)v2−12(2)(32)=32work-energy principlev2=41include initial kinetic energyv=41 m s−1positive speed\begin{aligned} W_{\mathrm{resultant}}&=(10)(4)-(2)(4)=32\,\mathrm{J} && \text{driving work minus resistance work magnitude}\\ \frac12(2)v^2-\frac12(2)(3^2)&=32 && \text{work-energy principle}\\ v^2&=41 && \text{include initial kinetic energy}\\ v&=\sqrt{41}\,\mathrm{m\,s^{-1}} && \text{positive speed} \end{aligned}

Why this step? The 40 J40\,\mathrm{J} supplied by the driver does not all become kinetic energy: 8 J8\,\mathrm{J} is dissipated against resistance.

Answer. The speed is 6.40 m s−16.40\,\mathrm{m\,s^{-1}}, to 3 significant figures.

Check. The energy balance is 9+40=41+89+40=41+8, in joules.

Worked example 4 changes only the driving force. The same resistance and journey now produce a smaller kinetic-energy gain.

Worked example 4: less driving work

Question. The particle, initial speed, resistance and distance are as in example 3, but the driving force is 4 N4\,\mathrm{N}. Find the final speed.

Plan. Recalculate the net work, keeping the resistance contribution negative.

Wresultant=(4−2)(4)=8 Jnew resultant workv2−9=8same mass and initial energyv=17 m s−1solve for speed\begin{aligned} W_{\mathrm{resultant}}&=(4-2)(4)=8\,\mathrm{J} && \text{new resultant work}\\ v^2-9&=8 && \text{same mass and initial energy}\\ v&=\sqrt{17}\,\mathrm{m\,s^{-1}} && \text{solve for speed} \end{aligned}

Why this step? Changing the driving force changes work over the whole distance. The resistance loss remains 8 J8\,\mathrm{J}.

Answer. The final speed is 4.12 m s−14.12\,\mathrm{m\,s^{-1}}, to 3 significant figures.

Check. 9+16=17+89+16=17+8, so input energy equals final energy plus loss.

For a rough slope, include the vertical potential-energy change and the work against friction along the slope. If a required final kinetic energy comes out negative, that state cannot be reached under the model; the particle must stop earlier.

Watch out

Wrong: add +Rs+Rs to the kinetic-energy gain as “work by resistance”. Right: either use signed work −Rs-Rs on the input side, or positive loss RsRs on the output side. Using both counts the same effect twice.

Choose one of those equivalent accounts and keep it consistent.

Energy transfers involving an elastic spring

A changing spring force makes constant-acceleration formulae unsuitable, but an energy equation can still connect two states directly.

For a particle attached to an ideal light spring, include the spring's elastic potential energy in the system. On a horizontal surface there is no gravitational potential-energy change. Add a positive loss term if friction dissipates energy. The surface is fixed, so the normal contact force is perpendicular to the motion and does no work.

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