M2 3.1 - Work and mechanical energy

M2 3.1 - Work and mechanical energy

Calculate work, kinetic energy and gravitational potential energy, then balance them to find speeds and distances. Decide when mechanical energy is conserved and when resistance removes it. Use g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}} unless another value is stated.

Work by a constant force

Two smooth tracks start at the same height and finish at the same lower height: must a particle released from rest reach the same speed on both? We will answer this by tracking energy rather than time.

First consider pulling a box 6 m6\,\mathrm m horizontally with a 40 N40\,\mathrm N force at 60∘60^\circ above the horizontal. Only the component 40cos⁡60∘=20 N40\cos60^\circ=20\,\mathrm N acts along the displacement. The vertical component does no work because there is no vertical displacement.

Work

The work done by a constant force is its component along a displacement multiplied by the displacement. Work is a scalar, measured in joules: 1 J=1 N m1\,\mathrm J=1\,\mathrm{N\,m}.

[DIAGRAM: asset_name: Resolving a force for work; asset_slug: m2-work-force-component; caption: Only the force component along the displacement contributes to work.; description: A box moves six metres horizontally right. A forty newton force points sixty degrees above the horizontal, with its twenty newton horizontal component shown by a dashed projection.]
Only the force component along the displacement contributes to work.

Work by a constant force

W=Fscos⁡θW=Fs\cos\theta

Here θ\theta is the angle between the force and displacement. Positive work transfers energy to the particle; negative work transfers energy away. A perpendicular force does zero work.

Not in the formula booklet: learn it.

Worked example 1: pulling forwards

Question. A constant 40 N40\,\mathrm N force makes 60∘60^\circ with a horizontal displacement of 6 m6\,\mathrm m. Find its work.

Plan. Multiply the displacement by the force component along it.

W=(40cos⁡60∘)×6=20×6=120 J.W=(40\cos60^\circ)\times6=20\times6=120\,\mathrm J.

Answer. 120 J120\,\mathrm J.

Check. The component is half the whole force, so the work is half of 40×6=240 J40\times6=240\,\mathrm J.

Change only the force angle. A component opposing motion must produce negative work.

Worked example 2: a resisting component

Question. The same force acts at 120∘120^\circ to the same displacement. Find its work.

Plan. Keep the angle with the displacement, including its obtuse sign.

W=40×6cos⁡120∘=−120 J.W=40\times6\cos120^\circ=-120\,\mathrm J.

Answer. −120 J-120\,\mathrm J.

Check. Its component along the displacement is −20 N-20\,\mathrm N, so (−20)×6=−120 J(-20)\times6=-120\,\mathrm J.

Watch out

The work done by a resistance of magnitude RR over distance ss is −Rs-Rs. The work done against it is the positive amount RsRs. State which you mean.

Kinetic and gravitational potential energy

A moving particle stores energy because of its speed, and a raised particle stores energy because of its height in a gravitational field.

Mechanical energy

Kinetic energy is energy associated with motion. Gravitational potential energy is energy associated with height. Their sum is the mechanical energy considered here.

For a constant resultant force along the motion, Fs=masFs=mas. Using v2−u2=2asv^2-u^2=2as gives Fs=12mv2−12mu2Fs=\tfrac12mv^2-\tfrac12mu^2. This identifies kinetic energy as 12mv2\tfrac12mv^2. Raising a mass vertically through hh against weight requires work mghmgh.

Kinetic and gravitational potential energy

K=12mv2,U=mgh,Δ(K+U)=12m(v2−u2)+mgΔh.K=\frac12mv^2,\qquad U=mgh,\qquad \Delta(K+U)=\frac12m(v^2-u^2)+mg\Delta h.

Take height above any fixed chosen level. Only changes in height matter. Mass is in kilograms, speed in metres per second and height in metres, giving energy in joules.

Not in the formula booklet: learn them.

Worked example 3: gaining speed and height

Question. A 2 kg2\,\mathrm{kg} particle speeds up from 33 to 7 m s−17\,\mathrm{m\,s^{-1}} while rising 1 m1\,\mathrm m. Find its increase in mechanical energy.

Plan. Calculate the kinetic and potential changes separately, then add.

ΔK=12(2)(72−32)=40 J,ΔU=2(9.8)(1)=19.6 J.\Delta K=\tfrac12(2)(7^2-3^2)=40\,\mathrm J, \qquad \Delta U=2(9.8)(1)=19.6\,\mathrm J. Δ(K+U)=40+19.6=59.6 J.\Delta(K+U)=40+19.6=59.6\,\mathrm J.

Answer. An increase of 59.6 J59.6\,\mathrm J, to three significant figures.

Check. Choosing initial height zero gives energies 9 J9\,\mathrm J initially and 49+19.6=68.6 J49+19.6=68.6\,\mathrm J finally; their difference is 59.6 J59.6\,\mathrm J.

Change only the sign of the height change: the particle now descends.

Worked example 4: gaining speed while descending

Question. The same particle has the same speed change while descending 1 m1\,\mathrm m. Find its increase in mechanical energy.

Plan. Keep the kinetic change and use a negative potential change.

Δ(K+U)=12(2)(72−32)+2(9.8)(−1)=40−19.6=20.4 J.\Delta(K+U)=\tfrac12(2)(7^2-3^2)+2(9.8)(-1) =40-19.6=20.4\,\mathrm J.

Answer. An increase of 20.4 J20.4\,\mathrm J.

Check. Set final height zero. Initial energy is 9+19.6=28.6 J9+19.6=28.6\,\mathrm J; final energy is 49 J49\,\mathrm J. The difference is again 20.4 J20.4\,\mathrm J.

Watch out

Doubling speed multiplies kinetic energy by four. Kinetic energy uses speed squared, so it is never negative; a negative velocity still gives positive kinetic energy.

Work-energy balance with resistance

Work done by all forces changes kinetic energy, so resistance and gravity must both appear in the balance.

Work-energy principle

total work done by all forces=Kfinal−Kinitial.\text{total work done by all forces}=K_{\rm final}-K_{\rm initial}.

An equivalent arrangement, with gravity accounted for as potential energy, is

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

Use one arrangement consistently: include either work by gravity or the potential energy change, never both.

Not in the formula booklet: learn it.

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