Gravitational Potential Energy
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avitational Potential Energy Potential Energy and Principle of Conservation of Energy U U U mgy mgy W grav 2 1 2 1 mgy U the energy that an object of mass m a distance y from the earth has due the gravitational attraction between e object and the earth. Maximum height of projectile K 1 = U 2 +K 2 U 2 =K 1 -K 2 mgh=(1/2)(mv 0 2 -mv 2x 2 )=(1/2) g v g v h y 2 sin 2 0 2 2 0 2 1 nservation of total mechanical energy = K + U =const (if only gravity does work) oof: work-energy theorem → W=ΔK → U 1 –U 2 = K 2 –K 1 → U 1 + K 1 = U 2 +
description
Potential Energy and Principle of Conservation of Energy. Gravitational Potential Energy. is the energy that an object of mass m at a distance y from the earth has due to the gravitational attraction between the object and the earth. Conservation of total mechanical energy - PowerPoint PPT Presentation
Transcript of Gravitational Potential Energy
Gravitational Potential Energy
Potential Energy and Principle of Conservation of Energy
UUUmgymgyW grav 2121
mgyU is the energy that an object of mass mat a distance y from the earth has dueto the gravitational attraction between the object and the earth.
Maximum height of projectileK1 = U2 +K2 U2=K1-K2 mgh=(1/2)(mv0
2-mv2x2)=(1/2)mv1y
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Conservation of total mechanical energyE = K + U =const (if only gravity does work)Proof: work-energy theorem → W=ΔK → U1–U2 = K2–K1 → U1 + K1 = U2 +K2
Exam Example 15: Riding loop-the-loop (problem 7.42)
Data: R= 20 m, v0=0, m=100 kg
Find: (a) min h such that a car does not fall off at point B,(b) kinetic energies for that hmin at the points B, C, and D,(c) if h = 3.5 R, compute velocity and acceleration at C.
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Solution: v
rada
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(a)To avoid falling off, centripetal acceleration v2/R > g → v2 > gR.Conservation of energy: KB+2mgR=mgh → (1/2)mvB
2=mg(h-2R) . Thus, 2g(h-2R) > gR → h > 5R/2 , that is hmin = 5R/2.
(b) Kf+Uf=K0+U0 , K0=0 → KB = mghmin- 2mgR = mgR/2 ,
KC = mghmin- mgR = 3mgR/2 , KD = mghmin = 5mgR/2.
(c) (1/2)mvC2 = KC= mgh – mgR = 2.5 mgR → vC = (5gR)1/2 ;
arad = vC2/R = 5g, atan = g since the only downward force is gravity.
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Elastic Potential Energy
UUUkxkxkxdxWx
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Conservation of total mechanical energyE = K + U =const (if only elastic force does work)
Proof: work-energy theorem → W=ΔK → U1 – U2 = K2 – K1 → U1 + K1 = U2 +K2
Conservation of total mechanical energywhen gravitational and elastic forces do work
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Conservative and Nonconservative Forces
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Two equivalent definitions of conservative forces: A force is conservative (Version 1) when the work it does on a moving object is independent of the path (W2 = W1) , or(Version 2) when it does no net work on an objectmoving around a closed path ( W2 – W1 = 0 ) .
General concept of potential energy: )(0 rUUW cons
Conservative forces: gravitational force, elastic spring force, electric force.
Nonconservative forces are irreversible: static and kinetic friction forces, air resistance, any dissipative forces, tension, normal force, propulsion force of a rocket.
Important: Potential energy is not defined for a nonconservative force !
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Conservative Force is a Gradient of Potential Energy
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Energy Diagrams and Oscillations in a Potential Well
gm Ugrav = mgy
Work-Energy Theorem in Terms of Conservative and Nonconservative Forces
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UUW
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Definition of the total mechanical energy: E = K + U
Work-Energy Theorem: E – E0 = Wnc
Any change in the total mechanical energy of an object is entirely due to the work done by external nonconservative forces.
Typical examples:1. Friction and resistance irreversibly decrease the total mechanical energy Wnc = - fk·s < 0 → E < E0 2. Propulsion force increases the total mechanical energy Wnc = F·s > 0 → E > E0
skf
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The Principle of Conservation of Mechanical Energy:K + U = const if Wnc = 0
The Principle of Conservation of EnergyWork-Energy Theorem → ΔK + ΔU = Wother Let us introduce an internal energy so that ΔEint = - Wother , then ΔK + ΔU + ΔEint = 0 , i.e., there is no missing energy !
K + U + Eint = Etotal =constEnergy can neither be created nor destroyed, butcan only be converted from one form to another.
Forms of Energy: kinetic energy, energy of gravity, energy of e.m. fields (electric and magnetic energies), heat (thermal energy), chemical and nuclear energies, …Nuclear fission: U235 + n → A1 +A2 + 2n + 200 MeV , 1 eV = 1.6·10-19 J
Nuclear fusion: ITER (tokomak reactor) 1D2 + 1T3 → 2He4 + n + 17 MeV;p-p cycle in stars 4p → 2He4 + 25 MeV; 2He3 + 1D2 → 2He4 + p + 18 MeV; p + 3Li6 → 2He4 + 2He3 + 4 MeV; 2He3 + 3Li6 → 22He4 + p + 17 MeV
Li-fission: n + 3Li6→ 1T3+2He4+5 MeV & n+3Li7→ 1T3+2He4+n–2.5 MeV Castle Bravo fusion bomb test in 1956 yield 15Mt instead of predicted 6MtThermonuclear bomb: Teller-Ulam and Sakharov-Ginzburg-Zeldovich designs Tsar Bomb (Russia, 1961) 50 Mt yield = 3000 Hiroshima nuclear bombsEinstein’s equation: E0 = mc2 → Mass is equivalent to energy!
mec2 = 9·10-31 kg (3·108 m/s)2 = 10-13 J = 0.5 MeV; annihilation e- + e+ → 2γ Energy of the sun is finite! Esun = Msun c2 = 2·1030 kg (3·108 m/s)2 = 2·1047 J
Exam Example 16: Spring on the Incline (Fig. 7.25, p.227)Data: m = 2 kg, θ = 53.1o, y0 = 4 m, k = 120 N/m, μk = 0.2, v0 =0.
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Solution: work-energy theorem Wnc=ΔK+ΔUgrav+ΔUel
(a)1st passage: Wnc= -y0μkmg cosθ since fk=μkFN==μkmg cosθ, ΔK=K1 , ΔUgrav= - mgy0 sinθ, ΔUel=0 → K1=mgy0(sinθ-μkcosθ), v1=(2K1/m)1/2 =[2gy0(sinθ–μkcosθ)]1/2 2nd passage: Wnc= - (y0+2|ys|) μkmg cosθ, ΔK=K2, ΔUgrav= -mgy0sinθ, ΔUel=0 →K2=mgy0sinθ-(y0+2|ys|) μkmgcosθ, v2=(2K2/m)1/2 (b) (1/2)kys
2 = Uel = ΔUel = Wnc – ΔUgrav = mg (y0+|ys|) (sinθ-μkcosθ) →αys
2 +ys –y0 =0, where α=k/[2mg (sinθ-μkcosθ)], → ys =[-1 - (1+4αy0)1/2]/(2α) Wnc = - (y0+|ys|) mgμkcosθ(c) Kf =0, ΔUel=0, ΔUgrav= -(y0–yf) mg sinθ, Wnc= -(y0+yf+2|ys|) μkmg cosθ →
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Find: (a) kinetic energy and speed at the 1st and 2nd passages of y=0, (b) the lowest position ys and friction energy losses on a way to ys, (c) the highest position yf after rebound. m
Exam Example 17: Proton Bombardment (problem 6.72)
Data: mass m, potential energy U=α/x,initial position x0>0 and velocity v0x<0.
Find: (a) Speed v(x) at point x. α=QUQp=92e2, e=4.8∙10-10 (CGS system)
(b) How close to the repulsive uranium nucleus 238U does the proton get?(c) What is the speed of the proton when it is again at initial position x0?
Solution: Proton is repelled by 238U with a force
Newton’s 2nd law, ax=Fx/m, allows one to find trajectory x(t) as a solution
of the second order differential equation: (a)Easier way: conservation of energy
(b)Turning point: v(xmin)=0
(c)It is the same since the force is conservative: U(x)=U(x0) v(x)=v(x0)
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