Download Mechanical Energy

Energy Transformations
and Conservation of
Mechanical Energy
8.01
W05D2
Today’s Reading Assignment:
W05D2
Young and Freedman: 7.1-7.5,13.3, 13.6
Review: Potential Energy
Difference
Definition: Potential Energy Difference between the
points A and B associated with a conservative force Fc
is the negative of the work done by the conservative
force in moving the body along any path connecting
the points A and B.
B
U    Fc  dr   Wc
A
Review: Examples of Potential
Energy with Choice of Zero
Point
(1) Constant Gravity:
U ( y)  mgy
U ( y  0)  0
(2) Inverse Square Gravity
Gm1m2
U (r) = 
r
U (r0  )  0
(3) Spring Force
U (x)  (1 / 2)kx 2
U (x  0)  0
Review: Work-Energy Theorem:
Conservative Forces
The work done by the total force in moving an
object from A to B is equal to the change in kinetic
energy
W
total
zf
 F
z0
total
1 2 1 2
 dr  mv f  mv0  K
2
2
When the only forces acting on the object are
conservative forces
F total  Fc
then the change in potential energy is
U  W  K
Therefore
U  K  0
Forms of Energy
• kinetic energy
• gravitational potential energy
• elastic potential energy
• thermal energy
• electrical energy
• chemical energy
• electromagnetic energy
• nuclear energy
• mass energy
Energy Transformations
 Falling water releases stored ‘gravitational potential energy’
turning into a ‘kinetic energy’ of motion.
 Human beings transform the stored chemical energy of food
into catabolic energy
 Burning gasoline in car engines converts ‘chemical energy’
stored in the atomic bonds of the constituent atoms of
gasoline into heat
 Stretching or compressing a spring stores ‘elastic potential
energy’ that can be released as kinetic energy
Energy Conservation
Energy is always conserved
N
 E
i 1
i
 E1  E2  ...  0
It is converted from one form into another, as the system
transforms from an “initial state” to a “final state”, each form
of energy can undergo a change
E  Efinal  Einitial
Energy can also be transferred from a system to its
surroundings
Esystem  Esurroundings  0
Concept Question: Energy
Transformations
You lift a ball at constant velocity from a height hi to
a greater height hf. Considering the ball and the
earth together as the system, which of the following
statements is true? You may neglect any change in
the motion of the earth.
1.
2.
3.
4.
5.
6.
The potential energy of the system increases.
The kinetic energy of the system decreases.
The earth does negative work on the system.
You do negative work on the system.
Two of the above.
None of the above.
Mechanical Energy
When a sum of conservative forces are acting on an object,
the potential energy function is the sum of the individual
potential energy functions with an appropriate choice of
zero point potential energy for each function
U  U1  U 2    
Definition: Mechanical Energy
The mechanical energy function is the sum of the kinetic
and potential energy function
E
mech
 K U
Conservation of Mechanical Energy
When the only forces acting on an object are conservative
K  U  0
(K f  Ki )  (U f  U i )  0
Equivalently, the mechanical energy remains constant in time
mech
E mech

K

U

K

U

E
f
f
f
i
i
i
mech
E mech  E mech

E
0
f
i
Non-Conservative Forces
Definition: Non-conservative force Whenever the
work done by a force in moving an object from an
initial point to a final point depends on the path, r
then the force is called a non-conservative force Fnc
and the work done is called non-conservative work
B
r
r
Wnc   Fnc  dr
A
Non-Conservative Forces
Work done on the object by the force depends
on the path taken by the object
Example: friction on an object moving on a
level surface
Ffriction  k N
Wfriction   Ffriction x   k N x  0
Change in Energy for Conservative
and Non-conservative Forces
Force decomposition:
r r
r
F  Fc  Fnc
Work done is change in kinetic energy:
B
r r B r
r
r
W   F  d r   (Fc  Fnc )  dr  U  Wnc  K
A
A
Mechanical energy change:
K  U  E mech  Wnc
Concept Question: Energy and
Choice of System
A block of mass m is attached to a relaxed spring on
an inclined plane. The block is allowed to slide down
the incline, and comes to rest. The coefficient of kinetic
friction of the block on the incline is µk. For which
definition of the system is the change in energy of the
system (after the block is released) zero?
1.
2.
3.
4.
block
block + spring
block + spring + incline
block + spring + incline +
Earth
Worked Example: Block Sliding
off Hemisphere
A small point like object of
mass m rests on top of a
sphere of radius R. The object
is released from the top of the
sphere with a negligible speed
and it slowly starts to slide. Find
an expression for the angle θf
with respect to the vertical at
which the object just loses
contact with the sphere.
Strategy: Using Multiple Ideas
Energy principle: No non-conservative work
K  U  E mech  Wnc  0
For circular motion, you will also need to
Newton’s Second Law in the radial direction
because no work is done in that direction hence
the energy law does not completely reproduce
the equations you would get from Newton’s
Second Law
2
rφ: Fr   m
Constraint Condition:
vf
R
N  0 at    f
Worked Example: Energy
Changes
Wnc  0
Kf 
Ki ; 0
U f  mgR(1  cos f )
Ui  0
mech
i
E
1 2
mv f
2
E mech

f
; 0
1 2
mv  mgR(1 cos f )
2 f
mech
Wnc  0  E mech

E

f
i
1

0  0   mv 2f  mgR(1  cos f ) 
2

1 2
mv f  mgR(1  cos f )
2
Worked Example: Free Body
Force Diagram
Newton’s Second Law
v2
rˆ : N  mg cos   m
R
2
d

φ
 : mg sin  mR 2
dt
Constraint condition:
N  0 at    f
Radial Equation becomes
mg cos f  m
v 2f
R
Worked Example: Combining
Concepts
Newton’s Second Law Radial Equation
mg cos f  m
v 2f
R

R
1
mg cos f  mv 2f
2
2
Energy Condition:
1 2
mv f  mgR(1  cos f )
2
Combine Concepts:
2
R
1  2 
  f  cos  
mgR(1  cos f )  mg cos f  cos f 
3
2
 3
Modeling the Motion: Newton’s
Second Law
 Define system, choose coordinate system.
 Draw free body force diagrams.
 Newton’s Second Law for each direction.
 Example: x-direction
 Example: Circular motion
ˆi : F total
x
rˆ : F
total
r
d 2x
m 2 .
dt
v2
 m .
R
Modeling the Motion Energy
Concepts
Change in Mechanical Energy:
Identify non-conservative forces.
final
Calculate non-conservative work
Wnc   Fnc  dr .
initial
Choose initial and final states and draw energy diagrams.
Choose zero point P for potential energy for each
interaction in which potential energy difference is welldefined.
Identify initial and final mechanical energy
Apply Energy Law.
Wnc  K  U  E mech
Table Problem: Loop-the-Loop
An object of mass m is released from rest at a height h
above the surface of a table. The object slides along the
inside of the loop-the-loop track consisting of a ramp and
a circular loop of radius R shown in the figure. Assume
that the track is frictionless. When the object is at the top
of the track (point a) it just loses contact with the track.
What height was the object dropped from?
Demo slide: Loop-the-Loop B95
http://scripts.mit.edu/~tsg/www/index.php
?page=demo.php?letnum=B
95&show=0
A ball rolls down an inclined track and
around a vertical circle. This
demonstration offers opportunity for
the discussion of dynamic equilibrium
and the minimum speed for safe
passage of the top point of the circle.
Demo slide: potential to kinetic
energy B97
http://scripts.mit.edu/~tsg/www/index.ph
p?page=demo.php?letnum=B
97&show=0
This demonstration consists of dropping a ball and a
pendulum released from the same height. Both balls are
identical. The vertical velocity of the ball is shown to be
equal to the horizontal velocity of the pendulum when
they both pass through the same height.
Table Problem: Experiment 3
Cart-Spring on an Inclined Plane
A cart of mass m slides down a plane that is inclined at an angle θ from
the horizontal. The cart starts out at rest. The center of mass of the cart
is a distance d from an unstretched spring with spring constant k that
lies at the bottom of the plane. Assume that the inclined plane has a
coefficient of kinetic friction μ. Find an equation whose solution
describes how far the spring will compress when the cart first comes to
rest.
Experiment 3
Energy Transformation
Potential Energy and Force
In one dimension, the potential difference is
B
U (x)  U (x0 )    Fx dx
A
Force is the derivative of the potential energy
dU
Fx  
dx
Examples: (1) Spring Potential Energy:
Fx , spring
dU
d 1 2

   kx   kx
dx
dx  2

(2) Gravitational Potential Energy:
Fr , gravity
Gm1m2
dU
d  Gm1m2 

  

dr
dr 
r 
r2
1 2
U spring ( x)  kx
2
Gm1m2
U grav (r ) = 
r
Energy Diagram
Choose zero point for potential energy:
U (x  0)  0
Potential energy function:
1 2
U (x)  kx , U (x  0)  0
2
Mechanical energy is represented by a
horizontal line since it is a constant
E
mechanical
1 2 1 2
 K(x)  U(x)  mvx  kx
2
2
Kinetic energy is difference between
mechanical energy and potential energy
(independent of choice of zero point)
K  E mechanical  U
Graph of Potential energy function
U(x) vs. x
Table Problem: Energy
Diagram
The figure above shows a graph of potential energy U(x) verses position for a
particle executing one dimensional motion along the x-axis. The total mechanical
energy of the system is indicated by the dashed line. At t =0 the particle is
somewhere between points A and G. For later times, answer the following
questions.
a)At which point will the magnitude of the force be a maximum?
b)At which point will the kinetic energy be a maximum?
c)At how many of the labeled points will the velocity be zero?
d)At how many of the labeled points will the force be zero?
Next Reading Assignment:
W05D3
Young and Freedman 7.1-7.5, 12.3, 12.6
31