Transcript potential energy

Gravitational Potential Energy
p. 191 1-6
extra p. 194 1-5,7
Conservation of Energy
p. 197 1-6, 7(tricky)
extra p. 201 1-11
Chapter Review for above
p. 226 14-21,37-39
Elastic Potential Energy
p. 206 1-5
extra p. 219 5-14
p. 211 8-11, 12,13 tough
Chapter Review for Elastic Potential
p. 227 22,23,35
GRAVITATIONAL POTENTIAL ENERGY
Conservative and Non-Conservative Forces
A ball is thrown upwards and returns to the
thrower with the same speed it departed with.
A block slides into a spring, compresses it and
leaves the spring with the same speed it first
contacted it with.
A force is conservative if the kinetic energy of a
particle returns to its initial value after a round trip
(during the trip the Ek may vary). A force is nonconservative if the kinetic energy of the particle
changes after the round trip (Assume only one
force does work on the object). Gravitational,
electrostatic and spring forces are conservative
forces.
Friction is an example of a non-conservative
force. For a round trip the frictional force
generally opposes motion and only leads to a
decrease in kinetic energy.
We must introduce the concept of potential
energy. This is energy of configuration or position.
As kinetic energy decreases the energy of
configuration increases and vice versa.
DEk  DE p  0
DEp
Change in potential energy
DEg
Change in gravitational potential
energy.
DEe
Change in elastic potential
energy.
DEk  DE g
 DE g  W
 DE g  FDd cos
 DE g  mg Dd ( 1)
or
 DE g  mg Dd ( 1)
DE g  mgDh
define Eg as mgh
h is height relative to
a reference point
Gravity does work on an object as its height
changes. As an object increases its height
gravity does negative work on the object and
the object’s kinetic energy decreases. This
loss of kinetic energy is a gain of potential
energy.
(
2
mv2

2
mv1
)  (mgh2  mgh1 )  0
2
2
2
2
mv1
mv2
 mgh1 
 mgh2
2
2
Em1  Em2
define Em as the mechanical energy
Mechanical Energy is conserved when an
object is acted upon by conservative forces.
LAW OF CONSERVATION OF ENERGY
Energy may be transformed from one kind to
another, but it cannot be created or destroyed:
the total energy is constant.
There are many forms of energy such as
electromagnetic, electrical, chemical, nuclear,
and thermal.
DEa  DEb  DEc  DEd ....  0
A ball is launched from a height of 2 m with
an initial velocity of 25 m/s [35o ath]. What
is the speed of the ball when its height is 7.5
m?
2
mv1
2
2
v1
2
 mgh1 
 gh1 
2
v2
2
2
mv2
2
 gh2
 mgh2
v22  v12  2 gh1  2 gh2
m 2
m
m
 (25 )  2(9.81 2 )( 2m)  2(9.81 2 )( 7.5m)
s
s
s
m
v2  22 .73
s
2
v2
The speed of the ball is 22.73 m/s
The energy approach doesn’t calculate the
velocity but it is a quicker method. The kinematic approach is longer but more precise.
let’s try it for exam review!!!
ELASTIC POTENTIAL ENERGY
Hooke’s Law (Robert Hooke 1678)
The magnitude of the force exerted by a spring
is directly proportional to the distance the spring
has moved from its equilibrium position.
An ideal spring obeys Hooke’s Law because it
experiences no internal or external friction.
+Fx= force exerted by
hand on spring
breaking point
F (N)
slope = k
non-elastic region
elastic limit
x (m)
The linear region is sometimes called Hooke’s
Law region. It applies to many elastic devices.
Hooke’s Law
F  kx
F = force exerted on the spring (N)
k = force constant of spring (N/m)
x = position of spring relative to the
equilibrium (deformation) (m)
The direction of compression on the spring is
negative while the direction of elongation is
positive (for F and x).
The spring exerts an equal and opposite force on
the object.
Derivation of Elastic Potential Energy
A spring exerts a conservative force on a object.
An object will have the same kinetic energy
after a round trip with a spring. The spring will
begin at its equilibrium position with zero
potential energy.
DEk  DE p  0
 DEe  DEk
 DEe  W


 ( Ee 2  Ee1 )  F D d cos 
kx
 Ee 2  x(1)
2
2
kx
Ee 2 
2
zero since at equilibrium
the force on the object
and its Dd have opposite directions
For an object interacting with an ideal
spring.
F  kx
1 2
Ee  kx
2
The potential energy of this object must
be considered in the mechanical energy.
2
2
1
2
2
2
mv1
mv2
kx
kx
 mgh1 

 mgh2 
2
2
2
2
Remember a reference height is needed for
height. The direction of x is not important
unless solving for x. If it is known that the
answer is compression then –x is correct. If
the answer is elongation then +x is correct.
If one form of energy is not present then it need
not be included in the equation.
try p.206 1-5
p.211 8-10 (they are quick)
A 2 kg ball is dropped from a height of 10 m onto a spring
that is 0.75 m in length and has a spring constant of 1000
N/m. How far will the ball compress the spring? What
force is exerted on the ball at its lowest point?
10 m
initial
v1=0
h1=10 m
x1=0
final
v2=0
h2=0.75+x2
0.75 m
x2
0.75+x2
2
2
1
2
2
2
mv1
mv2
kx
kx
 mgh1 

 mgh2 
2
2
2
2
2
kx2
mgh1  mgh2 
2
2
kx2
mgh1  mg (0.75  x2 ) 
2
Use the quadratic formula to solve.
x2   0.5832 m or x2   0.6224 m
 the spring compresses 0.6224 m
The force exerted by the spring on the
ball at this is calculated by. . .
F  kx
N
F  (1000 )(0.6224 m)
m
F  622.4 N [up ]