Mechanics

Chapter 9: Work and Energy

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
KKKinetic"How hard it is to stop a moving object"YesK=12mv2K=\tfrac12 m v^2
UGU_GGravitational potential"How high you are above (or below) a chosen zero level"Can be ±UG=mgyU_G = m g y
USpU_{Sp}Elastic (spring) potentialStored in a stretched/compressed springYesUSp=12k(Δs)2U_{Sp}=\tfrac12 k(\Delta s)^2
EthE_{th}ThermalMicroscopic jiggling—usually created by frictionN/AΔEth=FdissΔs\Delta E_{th}=F_{diss}\,\Delta s

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}}

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


4. Work WW: the currency of energy transfer

  • Formal definition
W=sisfFds  =  FΔrW=\int_{s_i}^{s_f} F_{\parallel}\,ds \;=\;\vec F\cdot\Delta\vec r

(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°     \implies positive work (adds energy).
    • 90°<θ<180°90°<\theta<180°     \implies negative work (removes energy).
    • θ=90°\theta=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 K

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=0 at ground/table). Then

UG=mgyU_G = m g y

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

6.2 Elastic (spring)

  • Hooke's law: FSp=kΔss^\vec F_{Sp} = -k\,\Delta s\,\hat s (force points toward equilibrium).
  • Stored energy:
USp=12k(Δs)2U_{Sp} = \tfrac12 k\,(\Delta s)^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=FdissΔs(always +)\Delta E_{th} = F_{diss}\,\Delta s \quad(\text{always }+)

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


8. Power PP: how fast energy moves

  • Instantaneous:
P=dEsysdt  =  FvP = \frac{dE_{\text{sys}}}{dt} \;=\; \vec F\cdot\vec v
  • Average: Pavg=ΔEΔt=WΔtP_{\text{avg}}=\dfrac{\Delta E}{\Delta t}=\dfrac{W}{\Delta t}
  • 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 F1=(25ı^15ȷ^)\vec F_1=(-25\hat{\imath}-15\hat{\jmath}) N. Water resists with F2=(13ı^17ȷ^)\vec F_2=(13\hat{\imath}-17\hat{\jmath}) N. During a tack the displacement is Δr=(260ı^+340ȷ^)\Delta\vec r=(-260\hat{\imath}+340\hat{\jmath}) m. Find the total work done on the craft.

Solution

Net force

Fnet=F1+F2=(25+13)ı^+(1517)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}

Work = FnetΔr\vec F_{\text{net}}\cdot\Delta\vec r

W=(12)(260)+(32)(+340)=312010880=7760 JW = (-12)(-260) + (-32)(+340) = 3120 - 10880 = -7760\text{ J}

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

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Problem 2 Rocket on a spring

A 12 kg rocket sits on a vertical spring k=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=0.

  1. Initial energies

    • Ki=0K_i = 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}
    • UG,i=mgyi=12(9.8)(0.21)=24.7 JU_{G,i}=mgy_i = 12(9.8)(-0.21) = -24.7\text{ 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}
    • 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}
    • UG,f=mgyf=12(9.8)(+0.40)=47.0 JU_{G,f}= mgy_f = 12(9.8)(+0.40)=47.0\text{ J}
  3. Change inside system

ΔEsys=(Kf+USp,f+UG,f)(Ki+USp,i+UG,i)=(19.4+44.8+47.0)(0+12.324.7)=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}

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

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Problem 3 Dog-sled power

A 220 kg sled starts from rest. A dog team pulls with a constant force giving a=0.75a=0.75 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 t     \implies t=v/a=3.3/0.75=4.4st = v/a = 3.3/0.75 = 4.4\,\text{s}.

  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}.

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

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

  5. Average power: Pavg=W/t=1200/4.42.7×102 WP_{\text{avg}} = W/t = 1200/4.4 ≈ 2.7\times10^2\text{ W}.

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

Answers: Pavg272 WP_{\text{avg}}\approx 272\text{ W}; Pinst545 WP_{\text{inst}}\approx 545\text{ W}.

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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}+…)
  • For constant forces: work is easy dot-product geometry.
  • Power tells you "How quickly?"—vital in engineering and physiology.
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