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W4 Energy and Waves

linear motion summary

  • Fnetย ext=dPdt\boxed{F_{\textmd{net ext}}=\frac{dP}{dt}}
  • net force on a system determines
  • (Fnetย ext)ave=ฮ”Pฮ”t\boxed{(F_{\textmd{net ext}})_{\textmd{ave}}=\frac{\Delta P}{\Delta t}}
  • ฮ”P=Fnetฮ”t\boxed{\Delta P=F_{net}\Delta t}
  • (Version of the above law or a finite time interval.)
  • If Fnetย ext=0F_{\textmd{net ext}}=0, then ฮ”P=0\Delta P=0
  • Linear momentum of the system is conserved
  • Final momentum = initial momentum

Work Done by a Constant Force

  • W=Fโˆฅฮ”x=Fcosโกฮธฮ”x\boxed{W=F_{\parallel}\Delta x=F\cos\theta\Delta x}
  • WW = Work done [JJ, Joules or NmNm, Newton meters]
    • 1J=1Nm1J=1Nm
  • FโˆฅF_{\parallel} = component of constant force (FF) parallel to the displacement (ฮ”x\Delta x)
    • Fโˆฅ=FcosโกฮธF_{\parallel} = F\cos\theta

Net Work

  • Wnet=totalย workย doneย byย eachย forceย onย object=Fnetcosโกฮธฮ”x\boxed{W_{net}=\textmd{total work done by each force on object}=F_{net}\cos\theta\Delta x}

Problem solving

  1. Draw a free-body diagram showing all the forces acting on the object you choose to study
  2. Choose an xyxy coordinate system
  3. Apply Newtonโ€™s laws to determine unknown forces
  4. Find the work done by a specific force on the object by using W=Fโˆฅฮ”x=Fcosโกฮธฮ”x\boxed{W=F_{\parallel}\Delta x=F\cos\theta\Delta x} for a constant force. The work done is negative when a force opposes the displacement.
  5. To find the net work done on the object, either:
    1. find the work done by each force and add the results algebraically
    2. or find the net force on the object, FnetF_{net}, and then use it to find the net work done, which for constant net force is: Wnet=Fnetcosโกฮธฮ”x\boxed{W_{net}=F_{net}\cos\theta\Delta x}

Kinetic Energy, and the Work-Energy Principle

Kinetic Energy

  • KE=12mv2\boxed{KE=\frac{1}{2}mv^2}

Work-Energy Principle

  • the net work done on an object is equal to the change in the objectโ€™s kinetic energy (! only for net work)
  • Wnet=ฮ”KE=12mvf2โˆ’12mvi2\boxed{W_{net}=\Delta KE=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2}

Potential Energy

  • Potential Energy Defined in General
    • change in potential energy associated with a particular force is equal to the negative of the work done by that force when the object is moved from one point to a second point

Gravitational Potential Energy

  • PEG=mgh\boxed{PE_G=mgh}

Potential Energy of Elastic Spring

  • PEel=12kx2\boxed{PE_{el}=\frac{1}{2}kx^2}

[Not in Syllabus] Conservative and Nonconservative Forces

Conservative ForcesNonconservative Forces
Gravitational
Elastic
Electric
Friction
Air resistance
Tension in cord
Motor of rocket propulsion
Push or pull by a person

Work-Energy Extended

Mechanical Energy

  • E=KE+PE=12v2+mgh\boxed{E=KE+PE=\frac{1}{2}v^2+mgh}

Conservation of Mechanical Energy

  • If only conservatice forces do work, the total mechanical energy of a system neither increases nor decreases in any process. It stays constant-it is conserved.
  • E1=E2=constant\boxed{E_1=E_2=constant}
  • KE1+PE1=KE2+PE2KE_1+PE_1=KE_2+PE_2
  • 12mv12+mgh1=12mv22+mgh2\frac{1}{2}mv_1^2+mgh_1=\frac{1}{2}mv_2^2+mgh_2

Law of Conservation of Energy

  • The total energy is neither increased nor decreased in any process. Energy can be transformed from one form to another, and transferred from one object to another, but the total amount remains constant

Elastic/Inelastic collisions

Elastic/Inelasticmomentumkinetic energyObject movement
Perfectly elasticconservedconservedobjects โ€œbounce offโ€ each other and move separately after the collision
Perfectly inelasticconservednot conservedobjects stick togethe rnad move with a common final velocity after the collision
  • Many collisions are neither perfectly elastic nor perfectly inelastic

Simple Harmonic Motion

  • occurs in all systems where the restoring force is proportional to the displacement
    • F=โˆ’kxF=-kx
  • x(t)=Acosโก(ฯ‰t)=Acosโก(2ฯ€Tt)\boxed{x(t)=A\cos(\omega t)=A\cos(\frac{2\pi}{T}t)}
    • AA = amplitude; maximum distance from equilibrium point
    • TT = period
      • T=1fT=\frac{1}{f}
      • T=2ฯ€mk\boxed{T=2\pi\sqrt{\frac{m}{k}}}
    • ff = frequency [HzHz, Hertz]
      • f=1Tf=\frac{1}{T}
      • T=12ฯ€kmT=\frac{1}{2\pi}\sqrt{\frac{k}{m}}
    • ฯ‰\omega = angular frequency (rad/srad/s, radians per second)
    • tt = time (ss, seconds)
    • ff = frequency = ฯ‰2ฯ€\frac{\omega}{2\pi}

Hookes Law

  • F=โˆ’kฮ”x\boxed{F=-k\Delta x}

Velocity

  • vx(t)=dxdt=โˆ’Aฯ‰sinโก(ฯ‰t)\boxed{v_x(t)=\frac{dx}{dt}=-A\omega\sin(\omega t)}
  • vmaxv_{max} when ฯ‰t=2ฯ€\omega t=2\pi?
  • v=0v=0 at xยฑAx\pm A

Acceleration

  • a(t)=dvdt=d2xdt2=โˆ’Aฯ‰2cosโก(ฯ‰t)=โˆ’ฯ‰2x(t)\boxed{a(t)=\frac{dv}{dt}=\frac{d^2x}{dt^2}=-A\omega^2\cos(\omega t)}=-\omega^2x(t)

Dynamics

  • In SHM the restoring force is proportional to the displacement
  • w=km\boxed{w=\sqrt\frac{k}{m}}

TEXTBOOK

11.1 Simple Harmonic Motion - Spring Oscillations

Springs

  • F=โˆ’kx\boxed{F=-kx}
    • FF = force exerted by spring
  • Fext=+kx\boxed{F_{ext}=+kx}
    • FextF_{ext}=external force on spring

Simple Harmonic Motion (SHM)

11.2 Energy in Simple Harmonic Motion

  • total mechanical energy of a simple harmonic oscillator is proportional to the square of the amplitude
  • E=12kA2\boxed{E=\frac{1}{2}kA^2}
  • vmax=kmA\boxed{v_{max}=\sqrt{\frac{k}{m}}A}
  • v=ยฑvmaxaโˆ’x2A2\boxed{v=\pm v_{max}\sqrt{a-\frac{x^2}{A^2}}}

11.3 The Period and Sinusoidal Nature of SHM

  • T=2ฯ€mk\boxed{T=2\pi\sqrt{\frac{m}{k}}}