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Potential Energy and
Principle of Conservation of Energy
W grav  mgy 1  mgy 2  U 1  U 2    U
Gravitational Potential Energy
U  mgy
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
E = K + U =const (if only gravity does work)
Proof: work-energy theorem → W=ΔK → U1–U2 = K2–K1 → U1 + K1 = U2 +K2
Maximum height of projectile
K1 = U2 +K2
U2=K1-K2
mgh=(1/2)(mv02-mv2x2)=(1/2)mv1y2
v 02 sin 2  0
h

2g
2g
v12y
Exam Example 15: Riding loop-the-loop (problem 7.42)

g

a rad

a tan D 
a
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.

Solution:
v
(a)To avoid falling off, centripetal acceleration v /R > g → v
2
2
> gR.
Conservation of energy: KB+2mgR=mgh → (1/2)mvB2=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.
Elastic Potential Energy
1 2
U  kx
2 1
   kxdx  kx
2
x2
2
1
Welastic
x1
Fx = -kx
1 2
 kx2  U1  U 2  U
2
Conservation of total mechanical energy
E = K + U =const (if only elastic force does work)
Proof: work-energy theorem → W=ΔK →
U1 – U2 = K2 – K1 → U1 + K1 = U2 +K2

g

Fspring

w
0
y
 Conservation of total mechanical energy
v when gravitational and elastic forces do work
1 2
1 2
K  U grav  U elastic  const  mv  mgy  ky  const
2
2
Conservative and Nonconservative Forces

r0
 
F (r )
Two equivalent definitions of conservative forces:
A force is conservative
  1
(Version 1) when the work it does on a moving
F (r )
object is independent of the path (W2 = W1) , or
(Version 2) when it does no net work on an object
moving around a closed path ( W2 – W1 = 0 ) .
General concept of potential energy:
2

r

W cons  U 0  U ( r )
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.

fk

v
Important: Potential energy is not defined for a nonconservative force !
 
 F  dr  U 
Conservative Force is a Gradient of Potential Energy
W  U
dU ( x )
1  D motion : Fx ( x )x  U  Fx ( x )  
dx


3  D motion : Fx dx  F y dy  Fz dz  U  F  U
 
 U  U  U  
U
U
U
F (r )  
ix 
iy 
i z   Fx  
, Fy  
, Fz  
y
z 
x
y
z
 x
Energy Diagrams and Oscillations in a Potential Well

mg
Ugrav = mgy
Work-Energy Theorem in Terms of
Conservative and Nonconservative Forces
W  K  K 0



W  Wnc  Wcons   Wnc  K  U  ( K 0  U 0 ) or Wnc  K  U
W  U  U 
0
 cons
 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
fk
2. Propulsion force increases the total mechanical energy
Wnc = F·s > 0 → E > E0

F

s
s
The Principle of Conservation of Mechanical Energy:
K + U = const if Wnc = 0
The Principle of Conservation of Energy
Work-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 =const
Energy can neither be created nor destroyed, but
can 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 6Mt
Thermonuclear bomb: Teller-Ulam and Sakharov-Ginzburg-Zeldovich designs
Tsar Bomb (Russia, 1961) 50 Mt yield = 3000 Hiroshima nuclear bombs
Einstein’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.
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,
m
(c) the highest position yf after rebound.
Solution: work-energy theorem Wnc=ΔK+ΔUgrav+ΔUel

fk
y
y0
(a)1st passage: Wnc= -y0μkmg cosθ since fk=μkFN=
yf
=μkmg cosθ, ΔK=K1 , ΔUgrav= - mgy0 sinθ, ΔUel=0
0
N
→ K1=mgy0(sinθ-μkcosθ),
ys
v1=(2K1/m)1/2 =[2gy0(sinθ–μkcosθ)]1/2
2nd passage: Wnc= - (y0+2|ys|) μkmg cosθ, ΔK=K2,
θ
ΔUgrav= -mgy0sinθ, ΔUel=0 →
K2=mgy0sinθ-(y0+2|ys|) μkmgcosθ, v2=(2K2/m)1/2
(b) (1/2)kys2 = Uel = ΔUel = Wnc – ΔUgrav = mg (y0+|ys|) (sinθ-μkcosθ) →
αys2 +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 →

F

mg
2( y0  | y s |)  k cos 
2( y0  | y s |)  k
y f  y0 
 y0 
sin    k cos 
tan    k
Exam Example 17: Proton Bombardment (problem 6.72)
Data: mass m, potential energy U=α/x,
238U
initial position x0>0 and velocity v0x<0.
0
 proton
v0
m

F
x0
xmin
x
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?
dU 
238
 2 0
Solution: Proton is repelled by U with a force Fx  
dx x
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
d 2x


2
dt
mx2
1 2 1 2
2
2
mv  mv0  U ( x0 )  U ( x)  v( x)  v02  U ( x0 )  U ( x)  v02 
2
2
m
m
(b) Turning point: v(xmin)=0 
1
xmin
 1 1
  
 x0 x 
2 x0
1 mv02
 
 xmin 
x0 2
2  mv02 x0
(c) It is the same since the force is conservative: U(x)=U(x0) v(x)=v(x0)