Chapter 7: Work and Energy
1. Work Energy
 Work done by a constant force
(scalar product)
 Work done by a varying force
(scalar product & integrals)
2. Kinetic Energy
Work-Energy Theorem
Work and Energy
Forms of Mechanical Energy
Work and Energy
CONSERVATION
OF ENERGY
Work and Energy
Work and Energy
Work by a Baseball Pitcher
A baseball pitcher is doing work on
the ball as he exerts the force over
a displacement.
v1 = 0
v2 = 44 m/s
Work and Energy
Work done by several forces
Work Done by
a Constant Force (I)
Work (W)
 How effective is the force in moving a
body ?
 Both magnitude (F) and
directions (q ) must be
taken into account.
W [Joule] = ( F cos q ) d
Work and Energy
Work Done by
a Constant Force (II)
Example: Work done on
the bag by the person..
 Special case: W = 0 J
a) WP = FP d cos ( 90o )
b) Wg = m g d cos ( 90o )
 Nothing to do with the
motion
Work and Energy
Example 1A
A 50.0-kg crate is pulled 40.0 m by a
constant force exerted (FP = 100 N and
q = 37.0o) by a person. A friction force Ff =
50.0 N is exerted to the crate. Determine
the work done by each force acting on the
crate.
Work and Energy
Example 1A (cont’d)
WP = FP d cos ( 37o )
Wf = Ff d cos ( 180o )
Wg = m g d cos ( 90o )
WN = FN d cos ( 90o )
F.B.D.
180o
d
90o
Work and Energy
Example 1A (cont’d)
WP = 3195 [J]
180o
Wf = -2000 [J] (< 0)
Wg = 0 [J]
WN = 0 [J]
Work and Energy
Example 1A (cont’d)
Wnet = SWi
= 1195 [J] (> 0)
The body’s speed
increases.
Work and Energy
Work-Energy Theorem
Wnet = Fnet d = ( m a ) d
= m [ (v2 2 – v1 2 ) / 2d ] d
= (1/2) m v2 2 – (1/2) m v1 2
= K2 – K1
Work and Energy
Example 2
A car traveling 60.0 km/h to can brake to
a stop within a distance of 20.0 m. If the car
is going twice as fast, 120 km/h, what is its
stopping distance ?
(a)
(b)
Work and Energy
Example 2 (cont’d)
(1) Wnet = F d(a) cos 180o
= - F d(a) = 0 – m v(a)2 / 2
 - F x (20.0 m) = - m (16.7 m/s)2 / 2
(2) Wnet = F d(b) cos 180o
= - F d(b) = 0 – m v(b)2 / 2
 - F x (? m) = - m (33.3 m/s)2 / 2
(3) F & m are common. Thus, ? = 80.0 m
Work and Energy
Work and Energy
Satellite in a circular orbit
Does the Earth do work on the satellite?
Work and Energy
B
2
Work and Energy
Forces on a hammerhead
F
o
r
c
e
s
Work and Energy
S
S23
Fn
Work and Energy
Spring Force (Hooke’s Law)
Spring Force
(Restoring Force):
The spring exerts
its force in the
direction opposite
the displacement.
FS
Natural Length
FP
x>0
x<0
FS(x) = - k x
Work and Energy
Work Done to Stretch a Spring
FS
FS(x) = - k x
Natural Length
W =
x2
FP(x) dx
x1
Work and Energy
W
FP
Work and Energy
Work Done by
a Varying Force
lb
W =
F|| dl
la
Work and Energy
Example 1A
A person pulls on the spring, stretching it
3.0 cm, which requires a maximum force
of 75 N. How much work does the person
do ? If, instead, the
person compresses
the spring 3.0 cm,
how much work
does the person do ?
Work and Energy
Example 1A (cont’d)
(a) Find the spring constant k
k = Fmax / xmax
= (75 N) / (0.030 m) = 2.5 x 103 N/m
(b) Then, the work done by the person is
WP = (1/2) k xmax2 = 1.1 J
(c)
x2 = 0.030 m
WP =
FP(x) d x = 1.1 J
x1 = 0
Work and Energy
Example 1B
A person pulls on the spring, stretching it
3.0 cm, which requires a maximum force
of 75 N. How much work does the spring
do ? If, instead, the
person compresses
the spring 3.0 cm,
how much work
does the spring do ?
Work and Energy
Example 1B (cont’d)
(a) Find the spring constant k
k = Fmax / xmax
= (75 N) / (0.030 m) = 2.5 x 103 N/m
(b) Then, the work done by the spring is
x2 = -0.030 m
WS =
FS(x) d x = -1.1 J
x1 = 0
(c) x2 = -0.030 m  WS = -1.1 J
Work and Energy
Example 2
A 1.50-kg block is pushed against a spring
(k = 250 N/m), compressing it 0.200 m, and
released. What will be the speed of the
block when it separates from the spring at
x = 0? Assume mk =
FS = - k x
0.300.
(i) F.B.D. first !
(ii) x < 0
Work and Energy
Example 2 (cont’d)
(a) The work done by the spring is
x2 = 0 m
WS =
FS(x) d x = +5.00 J
x1 = -0.200 m
(b) Wf = - mk FN (x2 – x1) = -4.41 (0 + 0.200)
(c) Wnet = WS + Wf = 5.00 - 4.41 x 0.200
(d) Work-Energy Theorem: Wnet = K2 – K1
 4.12 = (1/2) m v2 – 0
 v = 2.34 m/s
Work and Energy
Potential Energy and Energy
Conservation
1. Conservative/Nonconservative Forces
 Work along a path
(Path integral)
 Work around any closed path
(Path integral)
2. Potential Energy
Mechanical Energy Conservation
Energy Conservation
Work Done by
the Gravitational Force (I)
Near the Earth’s surface
 
W   F  dl (Path integral)
l2
l1
y2
  (-mg  ˆj )  (dy  ˆj )
y1
 ( mgy ) y1  mgy1  mgy 2
y2
Energy Conservation
y
l
Work Done by
the Gravitational Force (II)
Near the Earth’s surface
 
W   F  dl
y
l2
(Path integral)
l1
dl
y2
  (-mg  ˆj )  (dx iˆ  dy  ˆj )
y1
 ( mgy ) yy12  mgy1  mgy 2
Energy Conservation
Work Done by
the Gravitational Force (III)
Wg < 0 if y2 > y1
Wg > 0 if y2 < y1
The work done by the gravitational
force depends only on the initial and
final positions..
Energy Conservation
Work Done by
the Gravitational Force (IV)
Wg(ABCA)
= Wg(AB) +
Wg(BC) +
Wg(CA)
= mg(y1 – y2) +
0+
mg(y2- y1)
=0
Energy Conservation
C
B
dl
A
Energy Conservation
Work Done by
the Gravitational Force (V)
Wg = 0 for a closed path
The gravitational force is a
conservative force.
Energy Conservation
Work Done by Ff (I)
l2 (  L )


W f   F f  dl (Path integral)
l1 (  0 )
 ( mmgl ) L - μmg L
0
L depends on the path.
LB
LA
Energy Conservation
Work Done by Ff (II)
The work done by the friction force
depends on the path length.
Wf = 0 (any closed path)
The friction force:
(a) is a non-conservative force;
(b) decreases mechanical energy of the system.
Energy Conservation
Example 1
A 1000-kg roller-coaster car moves from
point A, to point B and then to point C.
What is its gravitational potential energy
at B and C
relative to
point A?
Energy Conservation
Wg(AC) = Ug(yA) – Ug(yC)
Wg(ABC) = Wg(AB) + Wg(BC)
= mg(yA- yB) + mg(yB - yC)
= mg(yA - yC)
y
B
A
B
dl
C
A
Energy Conservation
Climbing the Sear tower
Work and Energy
Power
Work and Energy
The Burj Khalifa is the largest
man made structure in the world
and was designed by Adrian
Smith class of 1966
thebatt.com Febuary 25th