Mw6, Mw7 - Energy conservation and the work-energy principle

Mw6, Mw7 - Energy conservation and the work-energy principle

Energy methods link the beginning and end of a motion without finding every intermediate acceleration. In Mw6 you identify when mechanical energy is conserved; in Mw7 you include the work of driving and resistive forces when it is not. The main habit is to account for each energy transfer exactly once.

When mechanical energy is conserved

Two smooth slides start at the same height and finish 2.5 m2.5\,\mathrm{m} lower. One is direct and the other winds around. Which gives a particle released from rest the greater final speed?

Provided each track is fixed, the particle stays in contact and resistance is negligible, neither wins. Gravity does the same work for the same vertical drop, and each normal contact force acts perpendicular to the motion. Energy connects the endpoints even when the acceleration changes along a curved route.

You already know

K=12mv2K=\frac12mv^2, U=mghU=mgh, and the work done by gravity is −ΔU-\Delta U. A force perpendicular to the displacement does zero work. Use g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}.

Mechanical energy is conserved when conservative forces account for the energy transfer and no external driving or dissipative work changes the mechanical total. A smooth surface can still exert a force: what matters here is that its normal force does no work.

Conservation of mechanical energy

When gravity is the only force doing work:

Kinitial+Uinitial=Kfinal+UfinalK_{\mathrm{initial}}+U_{\mathrm{initial}} =K_{\mathrm{final}}+U_{\mathrm{final}}

Equivalently,

12mu2+mghinitial=12mv2+mghfinal\frac12mu^2+mgh_{\mathrm{initial}} =\frac12mv^2+mgh_{\mathrm{final}}

This requires negligible dissipative work and no additional energy input.

Not in the formula booklet: learn it.

Choose a convenient zero height, often the final point. The difference in potential energy will be the same whichever reference you choose.

[DIAGRAM: asset_name: Different smooth routes with the same vertical drop; asset_slug: smooth_routes_energy; caption: A shared vertical drop gives the same final speed when the particle starts with the same speed and resistance is absent; description: Two schematic smooth tracks connect points at a common starting height to a common finishing height 2.5 m lower. One route is direct and the other curved. Both have a blue particle released from rest at the top and a sage final arrow labelled speed 7 m per second. A vertical height marker is separate from the path lengths. A side note says fixed smooth track, negligible resistance and contact maintained.]
A shared vertical drop gives the same final speed when the particle starts with the same speed and resistance is absent

Worked example 1: release from rest on a smooth track

Question. A particle is released from rest on a fixed smooth track. It descends through a vertical height of 2.5 m2.5\,\mathrm{m}. Neglect resistance and assume contact is maintained. Use energy to find its final speed.

Plan. Set the final height to zero and equate the lost gravitational potential energy to the gained kinetic energy.

mg(2.5)=12mv2conserve mechanical energyv2=2g(2.5)=49cancel the massv=7 m s−1speed is non-negative\begin{aligned} mg(2.5)&=\frac12mv^2&&\text{conserve mechanical energy}\\ v^2&=2g(2.5)=49&&\text{cancel the mass}\\ v&=7\,\mathrm{m\,s^{-1}}&&\text{speed is non-negative} \end{aligned}

Why this step? The normal force does no work, so the route's shape does not enter the endpoint energy equation. The mass cancels from both energy terms.

Answer. The final speed is 7 m s−17\,\mathrm{m\,s^{-1}}, for either slide in the opening question.

Check. Per kilogram, the lost potential energy is 9.8(2.5)=24.5 J kg−19.8(2.5)=24.5\,\mathrm{J\,kg^{-1}}, equal to 12(72)\frac12(7^2).

The next example changes only the initial speed. Its initial kinetic energy must now remain in the equation.

Worked example 2: an initial speed is already present

Question. A particle starts at 4 m s−14\,\mathrm{m\,s^{-1}} on the same fixed smooth track and descends through 2.5 m2.5\,\mathrm{m}. With the same assumptions, use energy to find its final speed.

Plan. Add the initial kinetic energy to the gravitational potential energy released.

12m(42)+mg(2.5)=12mv2initial total equals final totalv2=16+2(9.8)(2.5)divide by 12m=65add the two contributionsv=65=8.06 m s−1to 3 significant figures\begin{aligned} \frac12m(4^2)+mg(2.5)&=\frac12mv^2&&\text{initial total equals final total}\\ v^2&=16+2(9.8)(2.5)&&\text{divide by }\tfrac12m\\ &=65&&\text{add the two contributions}\\ v&=\sqrt{65}=8.06\,\mathrm{m\,s^{-1}}&&\text{to }3\text{ significant figures} \end{aligned}

Why this step? The descent adds energy to motion that already exists. Dropping the 424^2 term would model a release from rest instead.

Answer. 8.06 m s−18.06\,\mathrm{m\,s^{-1}}.

Check. The final speed exceeds both 4 m s−14\,\mathrm{m\,s^{-1}} and the 7 m s−17\,\mathrm{m\,s^{-1}} rest-release result, as the extra initial energy requires.

This gives a useful mathematical surprise: in the ideal gravity-only model, neither mass nor route shape affects the final speed for given initial speed and vertical drop. They can affect the motion and travel time, but not this endpoint energy result.

Watch out

Wrong: conserve K+UK+U merely because the surface is smooth. A separate resistive force or a motor can still change mechanical energy. Right: list which forces do work and check the model's assumptions before choosing conservation.

Change the initial speed once more, retaining the same height drop.

Maximum height and motion at the endpoint

Choose endpoints where you know the relevant speed, and distinguish being momentarily highest from being completely at rest.

A vertically projected particle is at rest at its highest point. An oblique projectile still has horizontal velocity there: only its vertical component has become zero. Energy must include that surviving motion.

Energy for an ideal projectile

With no air resistance, and initial height chosen as zero:

12mu2=12mv2+mgh\frac12mu^2=\frac12mv^2+mgh

At maximum height of a projectile launched at angle θ\theta to the horizontal, the remaining speed is ucos⁡θu\cos\theta. Substitution gives

hmax⁡=u2sin⁡2θ2gh_{\max}=\frac{u^2\sin^2\theta}{2g}

This follows from conservation and constant horizontal velocity; derive it when using an energy method.

Not in the formula booklet: learn it.

For other constrained motions, use the geometry to find the vertical height change. A light inextensible rope of length ℓ\ell, fixed at its upper end, gives a vertical drop ℓ−ℓcos⁡θ\ell-\ell\cos\theta from an angle θ\theta to the downward vertical to its lowest point. The tension is perpendicular to the motion and does no work.

[DIAGRAM: asset_name: Heights for a projectile and a swing; asset_slug: projectile_and_swing_heights; caption: A projectile retains horizontal speed at the top; a swing loses height ell minus ell cos theta; description: The left panel shows a projectile launched at angle theta, its highest point marked with a horizontal velocity u cos theta and zero vertical velocity, and a vertical height marker h. The right panel shows a rope of length ell from a fixed pivot, once at angle theta to the downward vertical and once vertically down. The vertical projection ell cos theta and the remaining drop ell minus ell cos theta are labelled. Tension at the inclined position points towards the pivot and the velocity arrow is perpendicular to the rope.]
A projectile retains horizontal speed at the top

Worked example 3: a vertical launch

Question. A particle is projected vertically upwards at 14 m s−114\,\mathrm{m\,s^{-1}}. Neglect air resistance. Use energy to find its maximum height above the launch point.

Plan. Use zero final speed at the top and transfer all the initial kinetic energy to gravitational potential energy.

12m(142)=mghtop speed is zeroh=1962(9.8)cancel the mass=10 mvertical rise\begin{aligned} \frac12m(14^2)&=mgh&&\text{top speed is zero}\\ h&=\frac{196}{2(9.8)}&&\text{cancel the mass}\\ &=10\,\mathrm{m}&&\text{vertical rise} \end{aligned}

Why this step? For a vertical launch, there is no horizontal motion left at the top. All of the initial kinetic energy can become potential energy.

Answer. 10 m10\,\mathrm{m}.

Check. Per kilogram, the initial kinetic energy is 98 J kg−198\,\mathrm{J\,kg^{-1}}, equal to 9.8(10)9.8(10).

The next example changes only the launch direction, keeping the same initial speed. Some kinetic energy now survives at the top.

Worked example 4: an oblique launch

Question. A particle is projected at 14 m s−114\,\mathrm{m\,s^{-1}}, at 30∘30^\circ above the horizontal. Neglect air resistance. Use energy to find its maximum height above the launch point.

Plan. Retain the horizontal speed 14cos⁡30∘14\cos30^\circ in the final kinetic energy.

vtop=14cos⁡30∘=73 m s−1horizontal component remains12m(142)=12m(73)2+mghconserve mechanical energygh=12(196−147)=24.5remove the remaining kinetic energyh=24.59.8=2.5 mvertical rise\begin{aligned} v_{\mathrm{top}}&=14\cos30^\circ=7\sqrt3\,\mathrm{m\,s^{-1}}&&\text{horizontal component remains}\\ \frac12m(14^2)&=\frac12m(7\sqrt3)^2+mgh&&\text{conserve mechanical energy}\\ gh&=\frac12(196-147)=24.5&&\text{remove the remaining kinetic energy}\\ h&=\frac{24.5}{9.8}=2.5\,\mathrm{m}&&\text{vertical rise} \end{aligned}

Why this step? A highest point has zero vertical velocity, not necessarily zero speed. Omitting the horizontal kinetic energy would incorrectly give the vertical-launch height.

Answer. 2.5 m2.5\,\mathrm{m}.

Check. The initial vertical speed is 14sin⁡30∘=7 m s−114\sin30^\circ=7\,\mathrm{m\,s^{-1}}, whose kinetic energy per kilogram is 24.5 J kg−124.5\,\mathrm{J\,kg^{-1}}, exactly the potential-energy gain per kilogram.

The same endpoint reasoning works for a swing: identify the starting and finishing heights, then include whatever speeds are actually present at those points.

Watch out

Wrong: put v=0v=0 at the highest point of every projectile. Right: retain ucos⁡θu\cos\theta unless the launch is vertical. The energy that becomes GPE comes from the vertical motion.

The next launch changes only the angle again.

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