Transcript p211c07

Potential Energy and Energy Conservation
Kinetic Energy: Energy associated with motion K = ½ mv2
Work done by a force on an object
is that forces contribution to DK
may only depend upon initial and final position of object
Conservative Forces
Work expressed as change in Potential Energy
Phys211C7 p1
Gravitational Potential Energy
Work done by gravity:
Dy
force × portion of displacement along force
force is always vertical => work = weight* height lifted
dW = F . dl = -mg dy
accumulated work by force of gravity
W = mg(y1 – y2) = mgy1 – mgy2 = – D(mgy)
dl
Dy
w
Note: work done does not depend upon path!
Identify potential energy: U  mgy
W = – DU
Work comes at the expense of potential energy.
Phys211C7 p2
Conservation of Mechanical Energy (with gravity)
In the absence of any other forces
W = DK
(work energy theorem)
now
W = - DU
so
DK = - DU
or DK + DU = 0
Define total Energy
so
E=K+U
DE = 0 or E1 = E2
In the absence of other forces, total energy is conserved!
With other forces, the other forces still do work:
DE = Wother
Phys211C7 p3
Example: A .145 kg baseball is thrown straight up into the air with an initial speed of 20
m/s. Determine kinetic, potential and total energy at the ball’s initial height as well as when
the ball reaches its maximum height.
Example, Projectile Motion: Show that two balls launched at different angles, but with the
same initial speed, will have the same speed at a given height h.
Use energy methods to relate the maximum height of a projectile to its initial speed and
direction.
Phys211C7 p4
Example, Projectile Motion: Use energy methods to relate the maximum height of a
projectile to its initial speed and direction.
.
Example: An object of mass m slides down a semicircular frictionless ramp of radius R
starting from rest. Find the speed and the normal force at the bottom of the ramp
Phys211C7 p5
Example: A 12 kg crate is slid up a 2.5 m long ramp inclined at 30°. The crate is given an
initial velocity of 5.00 m/s up the ramp, but only makes it 1.6m up the ramp before sliding
downward.
How much work is done by friction on the upward slide?
How fast is the crate moving as it slides back to the bottom of the ramp?
Phys211C7 p6
Elastic Potential Energy
from last chapter: work done on spring W = ½ kx22 - ½ kx12
work done by spring Wel = ½ kx12 - ½ kx22
 elastic potential energy W = - DU
U = ½ kx2
note: x is always measured from equilibrium!
another contribution to total energy
DE = Wother
Phys211C7 p7
Example: A 0.200 kg mass rides on a horizontal frictionless surface, attached to a spring
with a force constant of 5.00 N/m. The mass is pulled .100m from equilibrium.
What is the speed of the mass when it is .080 m from equilibrium?
What is the speed of the mass when it is at equilibrium?
With the same mass-spring system above, a constant force of .610 N is applied to the mass
which is initially at rest at the equilibrium position.
What is the glider’s speed when it has reached x = .100 m?
If the force is turned off at x = .100 m, how much further does the mass go?
Phys211C7 p8
Example: A 2000 Kg elevator is falling at a speed of 25.0 m/s when it “bottoms
out” on a spring at the bottom of the elevator shaft. The spring is supposed to stop
the elevator, compressing 3.00 m to bring it to rest. During this breaking process, a
safety clamp provides a constant 17,000 N frictional force on the elevator.
What is the spring constant?
What is the upward acceleration of the elevator just after it comes to a rest?
Phys211C7 p9
For a Conservative Force
Work can always be expressed as a change in potential energy
Is reversible (Wab = -Wba)
Is independent of the path of the object
Does no work when the initial and final positions of the object
are the same
 
W   F  dl
P2
 
W   F  dl  0
P1
Phys211C7 p10
Example: A 40.0 kg object with a coefficient of kinetic friction of 0.200 may be dragged
across a room by two paths: a direct 2.5m path, and dogleg path of 2.00m on the first leg and
1.50m along the second leg. How much work is done dragging the object across each path?
Phys211C7 p11
In a closed system, energy is conserved!
“loss” (or gain) of energy associated with the change of state
of the materials is due to changes in Internal Energy
change of temperature, melting/freezing, etc.
(Thermodynamics)
can be understood in terms of microscopic kinetic and
potential energy (Statistical Mechanics)
DE = DK + DU + DUint = 0
Phys211C7 p12
Force and Potential Energy
1- d
W  -DU
DU
Fx ,average ( x)Dx  - DU  Fx ( x)  Dx
dU ( x)
Fx ( x)  dx
U
(take limit )
U
U=mgy
U= ½ kx2
y
x
F
F
Fx=-kx
y
y
Fy=-mg
Phys211C7 p13
Force and Potential Energy
2,3 - d
U ( x, y, z )
Fx ( x, y, z )  x
U
U
Fy  , Fz  y
z

 U ˆ U ˆ U ˆ 
F  -
i
j
k
y
z 
 x


F  -U
example : U  mgy

  (mgy ) ˆ  (mgy ) ˆ  (mgy ) ˆ 
F  -
i
j
k
y
z
 x

 - mgˆj
Phys211C7 p14
2-d example
1
U ( x, y )  k ( x 2  y 2 )
2
U
U
Fx   -kx Fy   -ky
x
y

F  -k ( xiˆ  yˆj )
F  Fx  Fy  k x 2  y 2  kr
2
2
1
U  kr 2
2
U
Fr   -kr
r
Works in c ylindrical coordinates, but ...
Phys211C7 p15
U
Interpreting Energy Graphs
force is negative of slope of U
force is “down hill”
F
equilibrium = zero force
for 2-d use “contour maps”
stable
equilibrium
unstable
equilibrium
Phys211C7 p16