Download Physics 207: Lecture 2 Notes

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Lecture 14
Goals:
• Chapter 10
• Understand spring potential energies & use energy diagrams
• Chapter 11
 Understand the relationship between force, displacement
and work
 Recognize transformations between kinetic, potential, and
thermal energies
 Define work and use the work-kinetic energy theorem
 Use the concept of power (i.e., energy per time)
Assignment:
 HW7 due Wednesday, Mar. 10
 For Tuesday: Read through Ch. 12, Sections 1-3, 5 & 6
Do not concern yourself with the integration process in
regards to “center of mass” or “moment of inertia”
Physics 207: Lecture 14, Pg 1
Energy for a Hooke’s Law spring
dvx
dvx dx
dvx
Fx  m
m
m
vx  k ( x  xeq )
dt
dx dt
dx
Fx dx  mvx dvx  k ( x  xeq )dx
uf
vxf
ui
1
2
vxi
1
2
u  x  xeq  dx  du
 Fu du   mvx dvx

ku  mv  ku  mv
2
i
2
i
1
2
2
f
1
2
2
f
Associate ½ ku2 with the
“potential energy” of the spring
m
Physics 207: Lecture 14, Pg 2
Energy for a Hooke’s Law spring
1
2
ku  mv  ku  mv
2
i
1
2
2
i
1
2
2
f
1
2
2
f
U si  K i  U sf  K f  constant

Ideal Hooke’s Law springs are
conservative so the mechanical energy
is constant
m
Physics 207: Lecture 14, Pg 3
Energy diagrams

In general:
Ball falling
Emech
Spring/Mass system
Emech
U
y
K
Energy
Energy
K
U
x
Physics 207: Lecture 14, Pg 4
Equilibrium

Example
 Spring: Fx = 0 => dU / dx = 0 for x=xeq
The spring is in equilibrium position

In general: dU / dx = 0  for ANY function establishes
equilibrium
U
stable equilibrium
U
unstable equilibrium
Physics 207: Lecture 14, Pg 5
Comment on Energy Conservation

We have seen that the total kinetic energy of a system
undergoing an inelastic collision is not conserved.
 Mechanical energy is lost:
 Heat (friction)
 Deformation (bending of metal)

Mechanical energy is not conserved when non-conservative
forces are present !

Momentum along a specific direction is conserved when there
are no external forces acting in this direction.
 Conservation of momentum is a more general result than
mechanical energy conservation.
Physics 207: Lecture 14, Pg 6
Mechanical Energy

Potential Energy (U)
Kinetic Energy (K)
 If “conservative” forces
(e.g, gravity, spring) then

Before
Emech = constant = K + U
During
1
2
During  Uspring+K1+K2 = constant = Emech
After

Mechanical Energy conserved
Physics 207: Lecture 14, Pg 7
Energy (with spring & gravity)
1
h
0
-x







2
3
mass: m
Given m, g, h & k, how much does the
spring compress?
Emech = constant (only conservative forces)
At 1: y1 = h ; v1y = 0 At 2: y2 = 0 ; v2y = ? At 3: y3 = -x ; v3 = 0
Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0
Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2
Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0
Given m, g, h & k, how much does the spring compress?
Em1 = Em3 = mgh = -mgx + ½ kx2  Solve ½ kx2 – mgx +mgh = 0
Physics 207: Lecture 14, Pg 8
Energy (with spring & gravity)
1
h
mass: m
2
3
0
-x

When is the child’s speed greatest?
(A) At y1 (top of jump)
(B) Between y1 & y2
(C) At y2 (child first contacts spring)
(D) Between y2 & y3
(E) At y3 (maximum spring compression)
Physics 207: Lecture 14, Pg 9
Inelastic Processes

If non-conservative” forces (e.g, deformation, friction)
Before
then
Emech is NOT constant

After  K1+2 < Emech (before)


Accounting for this loss we introduce
Thermal Energy (Eth , new)
During
1
2
After
where Esys = Emech + Eth = K + U + Eth
Physics 207: Lecture 14, Pg 10
Energy & Work



Impulse (Force vs time) gives us momentum transfer
Work (Force vs distance) tracks energy transfer
Any process which changes the potential or kinetic energy
of a system is said to have done work W on that system
DEsys = W

W can be positive or negative depending on the direction
of energy transfer
Net work reflects changes in the kinetic energy
Wnet = DK
This is called the “Net” Work-Kinetic Energy Theorem
Physics 207: Lecture 14, Pg 11
Circular Motion


I swing a sling shot over my head. The tension in the rope
keeps the shot moving at constant speed in a circle.
How much work is done after the ball makes one full revolution?
(A) W > 0
v
(B) W = 0
(C) W < 0
(D) need more info
Physics 207: Lecture 14, Pg 12
Examples of “Net” Work (Wnet)
DK = Wnet

Pushing a box on a smooth floor with a constant force;
there is an increase in the kinetic energy
Examples of No “Net” Work
DK = Wnet
Pushing a box on a rough floor at constant speed
 Driving at constant speed in a horizontal circle
 Holding a book at constant height
This last statement reflects what we call the “system”
( Dropping a book is more complicated because it involves
changes in U and K, U is transferred to K )

Physics 207: Lecture 14, Pg 13
Changes in K with a constant F
xf
vxf
xi
vxi
 Fx dx   mv x dvx

If F is constant
xf
vxf
xi
vxi
Fx  dx   mvx dvx
Fx ( x f  xi )  Fx Dx 
1
2
2
mv xf

1
2
2
mv xi
 DK
Physics 207: Lecture 14, Pg 14
Net Work: 1-D Example
(constant force)

A force F = 10 N pushes a box across a frictionless floor
for a distance Dx = 5 m.
Finish
Start
F
q = 0°
Dx



Net Work is F Dx = 10 x 5 N m = 50 J
1 Nm ≡ 1 Joule and this is a unit of energy
Work reflects energy transfer
Physics 207: Lecture 14, Pg 15
Units:
Force x Distance = Work
Newton x Meter = Joule
[M][L] / [T]2 [L]
[M][L]2 / [T]2
mks
N-m (Joule)
cgs
Dyne-cm (erg)
= 10-7 J
Other
BTU = 1054 J
calorie= 4.184 J
foot-lb = 1.356 J
eV
= 1.6x10-19 J
Physics 207: Lecture 14, Pg 16
Net Work: 1-D 2nd Example
(constant force)

A force F = 10 N is opposite the motion of a box across a
frictionless floor for a distance Dx = 5 m.
Finish
Start
F
q = 180°
Dx

Net Work is F Dx = -10 x 5 N m = -50 J

Work reflects energy transfer
Physics 207: Lecture 14, Pg 17
Work in 3D….

x, y and z with constant F:

Fy ( y f  yi )  Fy Dy 
1
2
mv xf
2
1
2
mv yf
2
Fx ( x f  xi )  Fx Dx 
Fz ( z f  zi )  Fz Dz 
1
2

Fx Dx  Fy Dy  Fz Dz  
with v
2
2
 vx

2
vy

1
2
2
mv zf
2
mv f


1
2
1
2
mv xi
2
1
2
mv yi
2
1
2
mv zi
2
2
mvi
 DK
2
vz
Physics 207: Lecture 14, Pg 18
Work: “2-D” Example
(constant force)

An angled force, F = 10 N, pushes a box across a
frictionless floor for a distance Dx = 5 m and Dy = 0 m
Start
F
Finish
q = -45°
Fx
Dx
Fx Dx = F cos(-45°) Dx = 50 x 0.71 Nm = 35 J

(Net) Work is

Work reflects energy transfer
Physics 207: Lecture 14, Pg 19
Scalar Product (or Dot Product)
A · B ≡ |A| |B| cos(q)

Useful for performing projections.
A  î = Ax
îî=1
îj=0

A
q Ay
î Ax
Calculation can be made in terms of
components.
A  B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz )
Calculation also in terms of magnitudes and relative angles.
A  B ≡ | A | | B | cos q
You choose the way that works best for you!
Physics 207: Lecture 14, Pg 20
Scalar Product (or Dot Product)
Compare:
A  B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz )
with A as force F, B as displacement Dr
and apply the Work-Kinetic Energy theorem
Notice:
F  Dr = (Fx )(Dx) + (Fy )(Dz ) + (Fz )(Dz)
Fx Dx +Fy Dy + Fz Dz = DK
So here
F  Dr = DK = Wnet
More generally a Force acting over a Distance does Work
Physics 207: Lecture 14, Pg 21
Definition of Work, The basics
Ingredients: Force ( F ), displacement ( D r )
Work, W, of a constant force F
acts through a displacement D r :
W = F ·D r
F
q
Dr
(Work is a scalar)
“Scalar or Dot Product”
 
F  dr
If we know the angle the force makes with the
path, the dot product gives us F cos q and Dr
If the path is curved

r
f
and

W

dW

 
F  dr
at each point

 F  dr
ri
Physics 207: Lecture 14, Pg 22
Remember that a real trajectory
implies forces acting on an object
v
path
and time
a
a = atang + aradial
a
a
a = a + a
a =0
F = Ftang + Fradial
Two possible options:

Change in the magnitude of
v
a
=0
Change in the direction of
v
a
=0
Only tangential forces yield work!
 The distance over which FTang is applied: Work
Physics 207: Lecture 14, Pg 23
Exercise
Work in the presence of friction and non-contact forces

A.
B.
C.
D.
A box is pulled up a rough (m > 0) incline by a rope-pulleyweight arrangement as shown below.
 How many forces (including non-contact ones) are
doing work on the box ?
 Of these which are positive and which are negative?
 State the system (here, just the box)
 Use a Free Body Diagram
 Compare force and path
v
2
3
4
5
Physics 207: Lecture 14, Pg 25
Lecture 14
Assignment:
 HW7 due Wednesday, March 10
 For Tuesday: Read Chapter 12, Sections 1-3, 5 & 6
do not concern yourself with the integration process
Physics 207: Lecture 14, Pg 26
Related documents