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On this page

  • 1. Why talk about energy instead of just forces?
  • 2. Common forms of mechanical energy
  • 3. Systems, boundaries, and the Energy Principle
  • 4. Work WWW: the currency of energy transfer
  • 5. The Work–Kinetic‑Energy Theorem
  • 6. Potential energies in detail
  • 6.1 Gravitational
  • 6.2 Elastic (spring)
  • 7. Dissipative forces & thermal energy
  • 8. Power PPP: how fast energy moves
  • 9. Practice problems
  • Problem 1 Vector work on a catamaran
  • Problem 2 Rocket on a spring
  • Problem 3 Dog-sled power
  • 10 Key take-aways

Mechanics

Chapter 9: Work and Energy

Evan Luo · May 12, 2025

Mechanics

Chapter 9: Work and Energy

Evan LuoMay 12, 2025

5 min read

1. Why talk about energy instead of just forces?

  • Describing motion with Newton's laws is sometimes hard (lots of forces, directions, time-steps).
  • With energy we can treat whole motions in one step: add up the energies before and after.
  • Key idea: Energy can change form (kinetic → potential, etc.) but the total in a closed system stays the same.

2. Common forms of mechanical energy

SymbolNameSimple meaningAlways ≥ 0?Formula
KKKKinetic"How hard it is to stop a moving object"YesK=12mv2K=\tfrac12 m v^2K=21​mv2
UGU_GUG​Gravitational potential"How high you are above (or below) a chosen zero level"Can be ±UG=mgyU_G = m g yUG​=mgy
USpU_{Sp}USp​Elastic (spring) potentialStored in a stretched/compressed springYesUSp=12k(Δs)2U_{Sp}=\tfrac12 k(\Delta s)^2USp​=21​k(Δs)2
EthE_{th}Eth​ThermalMicroscopic jiggling—usually created by frictionN/AΔEth=Fdiss Δs\Delta E_{th}=F_{diss}\,\Delta sΔEth​=Fdiss​Δs

Other forms (chemical, nuclear, electrical…) exist but the three above plus thermal cover most mechanics problems.


3. Systems, boundaries, and the Energy Principle

  • System = the matter you choose to analyze. Draw an imaginary boundary around it.
  • Inside the boundary, energies can transform; across the boundary, energy can transfer by work (mechanical) or heat (thermal, beyond this chapter).
  • Energy Principle (book's wording)
Wext=ΔEsysW_{\text{ext}}=\Delta E_{\text{sys}}Wext​=ΔEsys​

"Work done on the system by the outside equals the change in the system's total energy."


4. Work WWW: the currency of energy transfer

  • Formal definition
W=∫sisfF∥ ds  =  F⃗⋅Δr⃗W=\int_{s_i}^{s_f} F_{\parallel}\,ds \;=\;\vec F\cdot\Delta\vec rW=∫si​sf​​F∥​ds=F⋅Δr

(for a constant force that is the familiar dot product).

  • Only the component of the force parallel to the motion matters.
  • Sign:
    • 0°<θ<90°0°<\theta<90°0°<θ<90°   ⟹  \implies⟹ positive work (adds energy).
    • 90°<θ<180°90°<\theta<180°90°<θ<180°   ⟹  \implies⟹ negative work (removes energy).
    • θ=90°\theta=90°θ=90° or no displacement   ⟹  \implies⟹ zero work.
  • Units: joule (J) = N · m = kg · m²/s².

5. The Work–Kinetic‑Energy Theorem

For a point mass:

Wtot  =  ΔKW_{\text{tot}} \;=\; \Delta KWtot​=ΔK

Total work done on the object equals the change in its kinetic energy. This is just Newton's 2nd law in energy clothing.


6. Potential energies in detail

6.1 Gravitational

Choose a convenient reference level (often y=0y=0y=0 at ground/table). Then

UG=mgyU_G = m g yUG​=mgy

It can be positive (above the reference) or negative (below).

6.2 Elastic (spring)

  • Hooke's law: F⃗Sp=−k Δs s^\vec F_{Sp} = -k\,\Delta s\,\hat sFSp​=−kΔss^ (force points toward equilibrium).
  • Stored energy:
USp=12k (Δs)2U_{Sp} = \tfrac12 k\,(\Delta s)^2USp​=21​k(Δs)2

Zero when the spring is at its relaxed length.


7. Dissipative forces & thermal energy

Kinetic friction, air drag, etc. don't store energy—they convert mechanical energy into thermal:

ΔEth=Fdiss Δs(always +)\Delta E_{th} = F_{diss}\,\Delta s \quad(\text{always }+)ΔEth​=Fdiss​Δs(always +)

So mechanical energy (K+UG+USpK+U_G+U_{Sp}K+UG​+USp​) drops by that same amount.


8. Power PPP: how fast energy moves

  • Instantaneous:
P=dEsysdt  =  F⃗⋅v⃗P = \frac{dE_{\text{sys}}}{dt} \;=\; \vec F\cdot\vec vP=dtdEsys​​=F⋅v
  • Average: Pavg=ΔEΔt=WΔtP_{\text{avg}}=\dfrac{\Delta E}{\Delta t}=\dfrac{W}{\Delta t}Pavg​=ΔtΔE​=ΔtW​
  • Units: watt (W) = J/s. Positive when delivering energy, negative when absorbing it.

9. Practice problems

Try each one first; click Reveal Answer to uncover a full solution.


Problem 1 Vector work on a catamaran

A wind pushes with F⃗1=(−25ı^−15ȷ^)\vec F_1=(-25\hat{\imath}-15\hat{\jmath})F1​=(−25^−15^​) N. Water resists with F⃗2=(13ı^−17ȷ^)\vec F_2=(13\hat{\imath}-17\hat{\jmath})F2​=(13^−17^​) N. During a tack the displacement is Δr⃗=(−260ı^+340ȷ^)\Delta\vec r=(-260\hat{\imath}+340\hat{\jmath})Δr=(−260^+340^​) m. Find the total work done on the craft.

Solution

Net force

F⃗net=F⃗1+F⃗2=(−25+13)ı^+(−15−17)ȷ^=(−12ı^−32ȷ^) N\vec F_{\text{net}}=\vec F_1+\vec F_2 =(-25+13)\hat{\imath}+(-15-17)\hat{\jmath} =(-12\hat{\imath}-32\hat{\jmath})\text{ N}Fnet​=F1​+F2​=(−25+13)^+(−15−17)^​=(−12^−32^​) N

Work = F⃗net⋅Δr⃗\vec F_{\text{net}}\cdot\Delta\vec rFnet​⋅Δr

W=(−12)(−260)+(−32)(+340)=3120−10880=−7760 JW = (-12)(-260) + (-32)(+340) = 3120 - 10880 = -7760\text{ J}W=(−12)(−260)+(−32)(+340)=3120−10880=−7760 J

Negative work   ⟹  \implies⟹ the environment removes 7.76 kJ of mechanical energy.

Click to reveal

Problem 2 Rocket on a spring

A 12 kg rocket sits on a vertical spring k=560k=560k=560 N/m, initially compressed 0.21 m and at rest. Engine fires → rocket moves upward with 1.8 m/s while the spring is now stretched 0.40 m. Neglect friction. How much chemical energy did the engine supply?

Solution

Let "bottom of spring" be y=0y=0y=0.

  1. Initial energies

    • Ki=0K_i = 0Ki​=0 (at rest)
    • USp,i=12k(0.21)2=0.5(560)(0.0441)=12.3 JU_{Sp,i} = \tfrac12 k (0.21)^2 = 0.5(560)(0.0441)=12.3\text{ J}USp,i​=21​k(0.21)2=0.5(560)(0.0441)=12.3 J
    • UG,i=mgyi=12(9.8)(−0.21)=−24.7 JU_{G,i}=mgy_i = 12(9.8)(-0.21) = -24.7\text{ J}UG,i​=mgyi​=12(9.8)(−0.21)=−24.7 J
  2. Final energies

    • Kf=12mv2=0.5(12)(1.8)2=19.4 JK_f = \tfrac12 m v^2 = 0.5(12)(1.8)^2 = 19.4\text{ J}Kf​=21​mv2=0.5(12)(1.8)2=19.4 J
    • USp,f=12k(0.40)2=0.5(560)(0.16)=44.8 JU_{Sp,f} = \tfrac12 k (0.40)^2 = 0.5(560)(0.16)=44.8\text{ J}USp,f​=21​k(0.40)2=0.5(560)(0.16)=44.8 J
    • UG,f=mgyf=12(9.8)(+0.40)=47.0 JU_{G,f}= mgy_f = 12(9.8)(+0.40)=47.0\text{ J}UG,f​=mgyf​=12(9.8)(+0.40)=47.0 J
  3. Change inside system

ΔEsys=(Kf+USp,f+UG,f)−(Ki+USp,i+UG,i)=(19.4+44.8+47.0)−(0+12.3−24.7)=111.2−(−12.4)=123.6 J\Delta E_{\text{sys}} = (K_f+U_{Sp,f}+U_{G,f}) - (K_i+U_{Sp,i}+U_{G,i}) = (19.4+44.8+47.0) - (0+12.3-24.7) = 111.2 - (-12.4) = 123.6\text{ J}ΔEsys​=(Kf​+USp,f​+UG,f​)−(Ki​+USp,i​+UG,i​)=(19.4+44.8+47.0)−(0+12.3−24.7)=111.2−(−12.4)=123.6 J

Engine's chemical energy supplied = +124 J+124\text{ J}+124 J (rounded).

Click to reveal

Problem 3 Dog-sled power

A 220 kg sled starts from rest. A dog team pulls with a constant force giving a=0.75a=0.75a=0.75 m/s² until the sled reaches 3.3 m/s; friction is negligible.

  • (a) Find the team's average power for the whole run.
  • (b) Find their instantaneous power the moment the sled hits 3.3 m/s.

Solution

  1. Time to reach 3.3 m/s: v=atv=a tv=at   ⟹  \implies⟹ t=v/a=3.3/0.75=4.4 st = v/a = 3.3/0.75 = 4.4\,\text{s}t=v/a=3.3/0.75=4.4s.

  2. Distance covered (starting from rest): s=12at2=0.5(0.75)(4.4)2=7.3 ms=\tfrac12 a t^2 = 0.5(0.75)(4.4)^2 = 7.3\text{ m}s=21​at2=0.5(0.75)(4.4)2=7.3 m.

  3. Pulling force: F=ma=220(0.75)=165 NF=ma = 220(0.75)=165\text{ N}F=ma=220(0.75)=165 N.

  4. Work done: W=Fs=165(7.3)=1.2×103 JW = F s = 165(7.3)=1.2\times10^3\text{ J}W=Fs=165(7.3)=1.2×103 J.

  5. Average power: Pavg=W/t=1200/4.4≈2.7×102 WP_{\text{avg}} = W/t = 1200/4.4 ≈ 2.7\times10^2\text{ W}Pavg​=W/t=1200/4.4≈2.7×102 W.

  6. Instantaneous power at 3.3 m/s: P=F⃗⋅v⃗=Fv=165(3.3)=5.45×102 WP = \vec F\cdot\vec v = Fv = 165(3.3)=5.45\times10^2\text{ W}P=F⋅v=Fv=165(3.3)=5.45×102 W.

Answers: Pavg≈272 WP_{\text{avg}}\approx 272\text{ W}Pavg​≈272 W; Pinst≈545 WP_{\text{inst}}\approx 545\text{ W}Pinst​≈545 W.

Click to reveal

10 Key take-aways

  • Think energy for "before vs. after" questions, forces for "what happens right now."
  • Identify your system, list its energy forms, then apply
Wext=Δ(K+UG+USp+Eth+…)W_{\text{ext}} = \Delta(K+U_G+U_{Sp}+E_{th}+…)Wext​=Δ(K+UG​+USp​+Eth​+…)
  • For constant forces: work is easy dot-product geometry.
  • Power tells you "How quickly?"—vital in engineering and physiology.

Source: https://notes.ohevan.com/notes/mechanics/09-work-and-energy

© 2026 Evan Luo. All rights reserved.

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·Last edited May 17, 2025
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© 2026 Evan Luo. All rights reserved.