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Lecture 12

Goals
 Analyze situations with Work and a Constant Force
 Analyze situations with Work and a Varying Force
 Introduce Hooke’s Law springs
 Relate external or “net” work to speed
 Introduce concepts of Kinetic and Potential energy
Physics 201: Lecture 12, Pg 1
Definition of Work, The basics
Ingredients: Force ( F ), displacement (  r )
Work, W, of a constant force F
acts through a displacement  r :
W ≡ F · r
F

r
(Work is a scalar)
Work tells you something about what happened on the path!
Did something do work on you?
Did you do work on something?
If only one force acting: Did your speed change?
Physics 201: Lecture 12, Pg 2
Net Work: 1-D Example
(constant force)

A force F = 10 N pushes a box across a frictionless floor
for a distance x = 5 m.
Finish
Start
F
 = 0°
x
Net Work is F x = 10 x 5 N m = 50 J
 1 Nm ≡ 1 Joule and this is a unit of energy

Physics 201: Lecture 12, Pg 3
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 x = 5 m.
Finish
Start
F
 = 180°
x

Net Work is F x = -10 x 5 N m = -50 J
Physics 201: Lecture 12, Pg 4
Work: “2-D” Example
(constant force)

An angled force, F = 10 N, pushes a box across a
frictionless floor for a distance x = 5 m and y = 0 m
Start
F
Finish
 = -45°
Fx
x

(Net) Work is Fx x = F cos(-45°) x = 50 x 0.71 Nm = 35 J
Physics 201: Lecture 12, Pg 5
Infinitesimal Work vs speed along a linear path
Let dW  Fx dx
F
Start
F Finish
Fx
 ma x dx
m
dvx
dt
 m dv
x
dx
dx
x dt
 m dvx vx
Physics 201: Lecture 12, Pg 6
Work vs speed along a linear path
dW  Fx dx  m vx dvx
xf
vxf
xi
 If F is constant
vxi
Wext   Fx dx   mvx dvx
xf
vxf
xi
vxi
Wext  Fx  dx m  vx dvx
Fx ( x f  xi )  Fx x 

1
2
2
mv xf

1
2
2
mv xi
 K
K is defined to be the change in the kinetic energy
Physics 201: Lecture 12, Pg 7
Work in 3D….

x, y and z with constant F:
Fx ( x f  xi )  Fx x 
Fy ( y f  yi )  Fy y 
1
2
mv
xf
2
1
2
mv yf
2


1
2
mv
xi
2
1
2
mv yi
2
1
2
mv zi
2
1
2
Fz ( z f  zi )  Fz z  2 mv zf 
1
2
1
2
Fx x  Fy y  Fz z  2 mv f  2 mvi 
with v
2
2
 vx

2
vy

2
vz
K
 
 v v
Kinetic energy :
KE  mv
1
2
Physics 201: Lecture 12, Pg 8
2
Examples of External Work
K = Wext

Pushing a box on a smooth floor with a constant force;
there is an increase in the kinetic energy
Examples of No External or Net Work
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 “potentially”
involves changes in the “potential” energy. The answer
depends on what we call the system. )

Physics 201: Lecture 12, Pg 10
Exercise
Work in the presence of friction and non-contact forces

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
A.
B.
C.
D.
2
3
4
5
Physics 201: Lecture 12, Pg 11
Work and Varying Forces (1D)

Area = Fx x
F is increasing
Here W = F · r
becomes dW = F dx
Consider a varying force F(x)
Fx
xf
x
x
Wext   F ( x ) dx
Finish
Start
F
F
xi
 = 0°
x
Work has units of energy and is a scalar!
Physics 201: Lecture 12, Pg 12
A Hooke’s Law spring
Most springs have a force which increases linearly with
displacement from the equilibrium

F ( x  xeq )  k ( x  xeq )
spring at an equilibrium position
50
m
Force (N)
40
x
30
20
F
spring
compressed
10
1.0
x
2.0 3.0 4.0
Displacement (m)
F
Slope: spring constant k
Physics 201: Lecture 12, Pg 13
•
Example: Work Kinetic-Energy Theorem with variable force
How much will the spring compress (i.e. x = xf - 0) to bring
the box to a stop (i.e., v = 0 ) if the object is moving initially at a
constant velocity (vo) on frictionless surface as shown below ?
Wext

f
 
 F  r dr
i
x x

f
Fx dx
x x
i
t o vo

eq
xf
eq
Wext
m
   kx dx
0
spring at an equilibrium position
Wext
x
 - 2 kx
1
2 xf
0
 - 2 k x  K
2
f
1
V=0
t
|
F
- k x  m0  mv
m
1
2
spring compressed
2
1
2
2
1
2
Physics 201: Lecture 12, Pg 14
2
0
Example: Work Kinetic-Energy Theorem with variable force
•
How much will the spring compress (i.e. x) to bring the box
to a stop (i.e., v = 0 ) if the object is moving initially at a
constant velocity (vo) on frictionless surface as shown below ?
to vo
Notice that the spring force is
opposite the displacement
m
spring at an equilibrium position
x
For the mass m,
work is negative, Ws
V=0
t
F
m
For the spring,
spring compressed
work is positive, Wapp
Physics 201: Lecture 12, Pg 15
What happens when I release the spring?
•
How fast will the block be moving when it loses contact with
the spring?
1
2
1
2
1
2

f
0
2
2
2
W  k x  mv  mv
V=0
F
m
spring compressed
vo
m
spring now at an equilibrium position
Physics 201: Lecture 12, Pg 16
Work done by friction

A small block slides inside a hoop of radius r. The walls of the
hoop are frictionless but the horizontal floor has a coefficient
of sliding friction of m. The block’s initial velocity is v and is
entirely tangential. How far, in angle around the circle does
the block travel before coming to rest ?
SFy = 0 = -mg + N
 SFx = m ar = m v2 /R
 Ff = m N = m mg
 W = Ff s = K = ½ mvf2 – ½ mvi2
= – ½ mv2
 - m mg s = – ½ mv2
 s = v2 / 2mg =  r
  = v2 / (2mgr)

Finish
v
Start
R
Physics 201: Lecture 12, Pg 17
Friction….again
Looking down on an air-hockey table with no air
flowing (m > 0).
 WPath 1 friction = Ff xpath 1
 WPath 2 friction = Ff xpath 2


xpath 2 > xpath 1 so W2 > W1
Path 2
Path 1
Physics 201: Lecture 12, Pg 18
The spring and friction have an important difference
For the spring I can reverse the process and
recover the kinetic energy
 The compressed spring has the ability to do work
 For the spring the work done is independent of
path
 The spring is said to be a conservative force

In the case of friction there is no immediate way to
back transfer the energy of motion
 In this case the work done is dependent on path
 Friction is said to be a non-conservative force

Physics 201: Lecture 12, Pg 19
Potential Energy (U)
For the compressed spring the energy is “hidden”
but still has the ability to do work (i.e., allow for
energy transfer)
 This kind of “energy” is called “Potential Energy”

U ( x) spring  k ( x  xeq )
1
2

2
The gravitation force, if constant, has the same
properties.
U ( y ) gravity  mg ( y  y0 )
Physics 201: Lecture 12, Pg 20
Mechanical Energy
If only “conservative” forces, then total mechanical energy
(potential U plus kinetic K energy) of a system is conserved
For an object in a gravitational “field”

½ m vyi2 + mgyi = ½ m vyf2 + mgyf = constant
K ≡ ½ mv2
U ≡ mgy
Emech = K + U
Emech = K + U = constant

K and U may change, but Emech = K + U remains a fixed value.
Emech is called “mechanical energy”
Physics 201: Lecture 12, Pg 21
For Tuesday


Read Chapter 8
Start HW6
Physics 201: Lecture 12, Pg 22