Transcript Work & Energy - FSU Physics Department

Chapter 7
Conservation of Energy
Recap – Work & Energy
The total work done on a particle
is equal to the change in its
kinetic energy
W  mv  mv
1
2
2
f
1
2
2
i
Potential Energy


The total work done on an object
equals the change in its kinetic
energy
But the total work done on a
system of objects may or may not
change its total kinetic energy.
The energy may be stored as
potential energy.
Potential Energy – A Spring
Both forces do work on the spring. But
the kinetic energy of the spring is
unchanged. The energy is stored as
potential energy
Conservative Forces
If the ski lift takes you
up a displacement h, the
work done on you, by
gravity, is –mgh.
But when you ski downhill
the work done by gravity
is +mgh, independent of the
path you take
Conservative Forces
The work done on
a particle by
a conservative force
is independent of the
path taken
between any two
points
Potential-Energy Function
If a force is conservative, then we can
define a potential-energy function as the
negative of the work done on the particle
s2
U  U 2  U1   F  ds
s1
Potential-Energy Function
potential-energy function associated
with gravity (taking +y to be up)
The value
s
of
s
   (mg ˆj )  (dx iˆ  dy ˆj  dz kˆ) U0 = U(y0)
s
can be set
y
to any
  mg dy
y
convenient
 mg ( y  y0 ) U  U0  mg ( y  y0 ) value
s
U  U  U 0    F  ds
0
0
0
Potential-Energy Function
of a Spring
U  U0  kx
1
2
2
By convention,
one chooses
U0 =U(0) = 0
Force & Potential-Energy
Function
In 1-D, given the potential energy function
associated with a force one can compute
the latter using:
dU
F 
dx
Example:
U  kx
1
2
2
dU
 F 
 kx
dx
7-1
Conservation of
Energy
Conservation of Energy
Energy can be neither
created nor destroyed
Esys  0
Closed System
Open System
Ein
Esys  0
Eout
Conservation of Mechanical
Energy
If the forces acting are conservative
then the mechanical energy is
conserved
Emech  K  U  constant
Example 7-3
(1)
How high does the block go?
Initial mechanical energy
of system
Ei  kx
1
2
2
Final mechanical energy
of system
E f  mgh
Example 7-3
(2)
Forces are conservative, therefore,
mechanical energy is conserved
1
2
kx  mgh
2
Height reached
2
kx
h
2mg
Example 7-4
(1)
How far does the mass drop?
Initial mech. energy
Ei  mgyi  ky  mv
2
i
1
2
2
i
1
2
Final mech. energy
E f  mgy f  ky  mv
1
2
2
f
1
2
2
f
Example 7-4
(2)
Final mech. energy = Initial mech. energy
mgy f  ky  mv 
2
f
1
2
2
f
1
2
mgyi  ky  mv
2
i
1
2
2
i
1
2
mg (d )  (d )  m(0) 
2
1
2
mg (0)  (0)  m(0)
1
2
2
1
2
2
1
2
2
Example 7-4
(3)
Solve for d
( kd  mg )d  0
1
2
Since d ≠ 0
2mg
d
k
2mg
Example 7-4
Note
Espring  kd
1
2
(4)
2
is equal to loss in
gravitational potential
energy
Egrav  mgd
2mg
Conservation of Energy &
Kinetic Friction
Non-conservative forces, such
as kinetic friction, cause
mechanical energy to be
transformed into other forms
of energy, such as thermal
energy.
Work-Energy Theorem
Work done, on a system, by external
forces is equal to the change in energy
of the system
Wext  Esys
The energy in a system can be
distributed in many different ways
Example 7-11
(1)
Find speed of blocks after spring is
released. Consider spring & blocks
as system. Write down initial energy.
Write down final energy.
Subtract initial
from final
Wext  Esys
Wext  Esys
Example 7-11
Initial Energy
Ei  kx
1
2
2
i
Kinetic energy of
system is zero
initially
(2)
Take potential
energy of system
to be zero initially
Wext  Esys
Example 7-11
E f  Es  Em1
(3)
Final Energy
 Em2
 Eother
 0  m1v  m2 v  m2 g s  k m1 g s
1
2
2
1
2
Kinetic and
potential
energies of
system have changed
2
Wext  Esys
Example 7-11
(4)
Subtract initial energy from final energy
Wext  E f  Ei
But since no
external forces
act, Wext = 0, so
Ef = Ei
Wext  Esys
Example 7-11
(5)
And the answer is…
kx  2m2 g s  2k m1 g s
v
m1  m2
2
i
Try to derive
this.
Wext  Esys
E = mc2
In a brief paper in 1905
Albert Einstein wrote
down the most famous
equation in science
E=
2
mc
Sun’s Power Output
 Power
1
Watt = 1 Joule/second
 100 Watt light bulb = 100
Joules/second
 Sun’s
power output
 3.826 x 1026 Watts
Sun’s Power Output
 Mass
to Energy
 Kg/s = 3.826 x 1026 Watts /
(3 x 108 m/s)2
 The
Sun destroys mass at
 ~ 4 billion kg / s
Problems
To go…
Ch. 7, Problem 19
Ch. 7, Problem 29
Ch. 7, Problem 74