Physics 1403-001
General Physics
Lecture 18
– Chapter 6 –
Feb 28, 2014
Sung-Won Lee
Sungwon.Lee@ttu.edu
Announcements
•  Course webpage: http://highenergy.phys.ttu.edu/~slee/1403
Syllabus, lecture note, etc…
•  Online homework: http://webassign.net/login.html
HW.5 is placed on WebAssign and is due by March 3.
HW.5: Ch.5 1, 6, 16, 18, 20, 35, 44, 50
HW.6 is placed on WebAssign and is due by March 10.
HW.6: Ch.6 3, 5, 11, 13, 16, 19, 25, 31, 32, 34, 45, 54,
58,70, 72
Reminder
•  Work:
Chapter 6
•  Kinetic energy is energy of motion:
Work & Energy
•  Work-energy principle: The net work done
on an object equals the change in its kinetic
energy.
Potential Energy, final
Mechanical Energy
•  Relation between work and potential energy
•  ΔPE = PEf – PEi = - W
•  The sum of the potential and kinetic energies is
called the mechanical energy
•  The potential energy of an object when it is at a
height h is PE = mgh
•  Conservation of Mechanical Energy
•  KEi + PEi = KEf + PEf
•  A very powerful tool for understanding, analyzing,
•  Potential energy is stored energy
•  It can be converted to kinetic energy
and predicting motion (we will see this today)
Section 6.3
Conservation of Mechanical Energy
•  Look at work done by the book as it falls
from some height to a lower height
•  From work-kinetic energy theorem,
Won book = ΔKbook
•  Also, W = FΔr = mg(yb)– mg(ya ) = ΔKbook
•  mgyb – mgya = (Ui - Uf) = -(Uf- Ui) = -ΔUg
Section 6.3
Charting the Energy
•  A convenient way of
illustrating conservation of
energy is with a bar chart
•  The sum of the energies
is the same at the start
and end
•  So, ΔK = -ΔUg => ΔK + ΔUg = 0 "
•  We define the sum of kinetic and potential energies as
mechanical energy in the system: Emech = K + Ug
•  The statement of Conservation of Mechanical Energy for
an isolated system: Kf + Uf = Ki+ Ui
Section 6.3
Problem Solving Using Conservation of
Mechanical Energy
In the image on the left,
the total mechanical
energy at any point is:
Problem Solving Using Conservation of
Mechanical Energy
If the original height of the
rock is y1 = h = 3.0 m, calculate
the rock s speed when it has
fallen to 1.0 m above the
ground.
The rock s potential energy changes to kinetic energy as it falls.
Note: bar graphs representing potential energy U and kinetic
energy K for the three different positions.
Problem Solving Using Conservation of
Mechanical Energy
Assuming the height of the hill is 40 m, and the
roller-coaster car starts from rest at the top,
calculate (a) the speed of the roller-coaster car at
the bottom of the hill
1
1
mv12 + mgy1 = mv22 + mgy2
2
2
8-4 Problem Solving Using Conservation of
Mechanical Energy
Assuming the height of the hill is 40 m, and the
roller-coaster car starts from rest at the top,
calculate (b) at what height it will have half this
speed. Take y = 0 at the bottom of the hill.
Solving Using Conservation of Mechanical
Energy
Estimate the kinetic energy and the speed
required for a 70-kg pole vaulter to just
pass over a bar 5.0 m high. Assume the
vaulter s center of mass is initially 0.90 m
off the ground and reaches its maximum
height at the level of the bar itself.
1
1
mv12 + mgy1 = mv22 + mgy2
2
2
A re-look at some problems
Let s say that we want to know the velocity of a block sliding on
a frictionless inclined plane after it has slid down from a height h.
h
We determined the acceleration down
the plane before using
F = ma
s
θ!
(m)adown plane = (m)g sinθ
2
v2 = v2 + 2 a s = v + 2 g sinθ (h / sinθ)
0
0
v2 = v2 + 2 g h
0
v = √2 g h
if it starts from rest (i.e v0=0)
Using Energy Conservation
Let s say that we want to know the velocity of a block sliding on
a frictionless inclined plane after it has slid down from a height h.
h
s
θ!
Using Energy Conservation
Let s say that we want to know the velocity of a block sliding on
a frictionless inclined plane after it has slid down from a height h.
Kf + Uf = Ki + Ui
h
s
θ!
Kf + Uf = Ki + Ui
2
1
– mv
2
f
2
+0=
1
–
2
mv2 + mgh
2
i
vf = vi + 2 g h
if it starts from rest
vf = √2 g h
A Swinging Pendulum
Now what about a block sliding
down an incline like this
Here the acceleration down the
plane is continually changing
since the angle of plane with
the horizontal changes.
h
Kf + Uf = Ki + Ui
2
1
– mv
f
2
2
2
+ 0 =1– mvi + mgh
2
2
vf = vi + 2 g h
if it starts from rest
vf = √2 g h
Total Energy
is same everywhere.
Top of swing: v=0, K=0
(A,C)
all U
v = √2 g h
Bottom of swing: U=0
(B) max v, All K
starts from rest at height h
solve for velocity
at bottom of swing
h
Kf + Uf = Ki + Ui
2
1
– mv
f
2
+ 0 =1– mv2 + mgh
v2f = v2i + 2 g h
2
i
v f= √2 g h