4.5 - Conservation of energy

4.5 - Conservation of energy

Speed can sometimes be found without calculating acceleration or time. You will choose an energy system, conserve mechanical energy when only gravity transfers energy between motion and height, and include the work against resistance when energy is lost. The same balance handles a straight slope or a smooth curved slide.

A conservative system

Two smooth slides start at the same height and finish at the same lower height, but one is longer. Must it give a larger final speed? If the starting speeds match and no energy is lost, only the height change matters.

For a 2 kg2\,\mathrm{kg} body falling 1.8 m1.8\,\mathrm{m}, gravity releases 2(10)(1.8)=36 J2(10)(1.8)=36\,\mathrm{J}. From rest, this becomes kinetic energy. A smooth supporting surface would do no work because its normal force is perpendicular to the motion. Here use g=10 m s−2g=10\,\mathrm{m\,s^{-2}}.

Conservative mechanical-energy model

A force is conservative when its work depends only on the endpoints and can be represented by a potential-energy change. Gravity near Earth's surface is conservative. If the other forces do no net work, the system's mechanical energy K+UK+U stays constant. Friction and air resistance generally transfer mechanical energy into other forms.

An energy balance connects the initial and final states. Time and path shape need not appear when only kinetic and gravitational potential energy change.

Conservation of mechanical energy

When no other force adds or removes mechanical energy,

Kinitial+Uinitial=Kfinal+Ufinal.K_{\text{initial}}+U_{\text{initial}}=K_{\text{final}}+U_{\text{final}}.

Choose one height reference and include the kinetic and potential energies of every body in the selected system. Gravity is already represented through UU.

Not in the formula booklet: learn it.

A useful system can contain several particles and Earth. With a light taut inextensible string over a fixed smooth pulley, the work of its internal tension cancels across the connected particles: tension does equal positive and negative work over their equal movements along the string. The ideal string stores no energy. A stationary pulley support and a normal contact force perpendicular to motion do no work.

Worked example 1: a freely falling body

Question. A 2 kg2\,\mathrm{kg} body falls from rest through 1.8 m1.8\,\mathrm{m}, with air resistance neglected. Find its speed after falling this distance.

Plan. Set the final height as zero and convert the potential-energy loss to kinetic energy.

2(10)(1.8)=12(2)v2⟹v2=36.2(10)(1.8)=\tfrac12(2)v^2\quad\Longrightarrow\quad v^2=36.

Answer. Speed 6 m s−16\,\mathrm{m\,s^{-1}}.

Check. Independent constant-acceleration calculation gives v2=0+2(10)(1.8)=36v^2=0+2(10)(1.8)=36, agreeing with the energy method.

Now let the falling body also set another particle in motion. The potential-energy release is unchanged, but it must supply both kinetic energies.

Worked example 2: energy shared by connected particles

Question. A 2 kg2\,\mathrm{kg} hanging particle is connected to a 3 kg3\,\mathrm{kg} particle on a smooth horizontal table by a light inextensible string over a fixed smooth edge pulley. The table section of string is horizontal. They are released from rest with the string taut. Find their common speed after the hanging particle descends 1.8 m1.8\,\mathrm{m}, before either particle meets an obstruction.

Plan. Include both particles' kinetic energies, but only the hanging particle's potential-energy change.

2(10)(1.8)=12(2)v2+12(3)v236=52v2v2=14.4\begin{aligned} 2(10)(1.8)&=\tfrac12(2)v^2+\tfrac12(3)v^2\\ 36&=\tfrac52v^2\\ v^2&=14.4 \end{aligned}

Answer. Common speed 14.4=3.79 m s−1\sqrt{14.4}=3.79\,\mathrm{m\,s^{-1}}, hanging particle down and table particle towards the pulley.

Check. Newton's law for the pair gives a=20/5=4 m s−2a=20/5=4\,\mathrm{m\,s^{-2}}. Thus v2=2(4)(1.8)=14.4v^2=2(4)(1.8)=14.4. The speed is below free fall because the released energy now moves 5 kg5\,\mathrm{kg}, not just 2 kg2\,\mathrm{kg}.

The string's tension is external to either particle alone, but internal to the combined pair. This is why choosing the system comes before cancelling forces or energy transfers.

Watch out

Conserved mechanical energy is a modelling conclusion, not a rule for every journey. Check what does work. A smooth surface removes friction, but an applied driving force or specified air resistance still changes K+UK+U.

Choose both connected particles as the system in this calculation.

Work against resistance

When resistance acts, total energy is still accounted for, but mechanical energy is reduced by the positive work done against resistance. A longer rough path can lose more energy than a shorter one with the same height change.

Energy with resistance

If no driving force supplies work,

Kinitial+Uinitial=Kfinal+Ufinal+Wagainst resistance.K_{\text{initial}}+U_{\text{initial}} =K_{\text{final}}+U_{\text{final}}+W_{\text{against resistance}}.

For constant opposing resistance rr over travelled distance ss, Wagainst resistance=rsW_{\text{against resistance}}=rs. Add the work over successive sections if their resistances differ.

Not in the formula booklet: learn it.

On a rough straight incline with no extra perpendicular force, R=mgcos⁡αR=mg\cos\alpha and sliding friction is μR\mu R. Its work uses the distance along the incline. The gravity term uses the vertical height change. These distances serve different purposes.

Worked example 3: descending a rough slope

Question. A 2 kg2\,\mathrm{kg} particle starts from rest and slides 5 m5\,\mathrm{m} down a rough plane with sin⁡α=3/5\sin\alpha=3/5, cos⁡α=4/5\cos\alpha=4/5 and μ=1/4\mu=1/4. Find its speed at the bottom.

Plan. Calculate the height loss and friction work, then allocate the remaining energy to kinetic energy.

R=20(4/5)=16 N,F=14(16)=4 Nh=5(3/5)=3 m20(3)=12(2)v2+4(5)v2=40\begin{aligned} R&=20(4/5)=16\,\mathrm{N},\qquad F=\tfrac14(16)=4\,\mathrm{N}\\ h&=5(3/5)=3\,\mathrm{m}\\ 20(3)&=\tfrac12(2)v^2+4(5)\\ v^2&=40 \end{aligned}

The downhill weight component 12 N12\,\mathrm{N} exceeds the 4 N4\,\mathrm{N} friction limit, so the stated release does produce downhill motion.

Answer. Speed 210 m s−12\sqrt{10}\,\mathrm{m\,s^{-1}}, approximately 6.32 m s−16.32\,\mathrm{m\,s^{-1}}.

Check. Force calculation gives a=(12−4)/2=4 m s−2a=(12-4)/2=4\,\mathrm{m\,s^{-2}}. Then v2=2(4)(5)=40v^2=2(4)(5)=40, confirming the energy result.

Reverse the journey: project the same particle up from the bottom with that bottom speed. Friction still removes energy; it does not return the energy lost on the descent.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account