work and energy 1.work energy work done by a constant force (scalar product) work done by a...
TRANSCRIPT
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
Chapter 7: Work and Energy
Work-Energy Theorem
Forms of Mechanical Energy
Work and 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 done by several forces
Work and Energy
Work Done by a Constant Force (I)
Work (W) How effective is the force in moving a
body ?
W [Joule] = ( F cos ) d
Both magnitude (F) and directions () must be taken into account.
Work and Energy
Work Done bya 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
= 37.0o) by a person. A friction force Ff =50.0 N is exerted to the crate. Determinethe work done by each force acting on thecrate.
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 )
180o
90o
d
F.B.D.
Work and Energy
Example 1A (cont’d)
WP = 3195 [J]
Wf = -2000 [J] (< 0)
Wg = 0 [J]
WN = 0 [J]
180o
Work and Energy
Example 1A (cont’d)
Wnet = Wi
= 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
Does the Earth do work on the satellite?
Satellite in a circular orbit
Work and Energy
2
B
Work and Energy
Forces
Forces on a hammerhead
Work and Energy
S
S23
Fn
Work and Energy
Spring Force (Hooke’s Law)
FS(x) = - k x
FPFS
Natural Length x > 0
x < 0
Spring Force(Restoring Force):The spring exerts its force in thedirection opposite the displacement.
Work and Energy
Work Done to Stretch a Spring
x2
W = FP(x) dxx1
FS(x) = - k x
W
Natural Length
FPFS
Work and Energy
Work and Energy
lb
W = F|| dl la
Work Done bya Varying Force
l 0
Work and Energy
Example 1A
A person pulls on the spring, stretching it3.0 cm, which requires a maximum force of 75 N. How much work does the person do ? If, instead, theperson compressesthe spring 3.0 cm,how much workdoes the person do ?
Work and Energy
(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 xmax
2 = 1.1 J
(c) x2 = 0.030 mWP = FP(x) d x = 1.1 J x1 = 0
Example 1A (cont’d)
Work and Energy
Example 1B
A person pulls on the spring, stretching it3.0 cm, which requires a maximum force of 75 N. How much work does the spring do ? If, instead, theperson compressesthe spring 3.0 cm,how much workdoes the spring do ?
Work and Energy
(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
(c) x2 = -0.030 m WS = -1.1 J
x2 = -0.030 mWS = FS(x) d x = -1.1 J x1 = 0
Example 1B (cont’d)
Work and Energy
Example 2
A 1.50-kg block is pushed against a spring(k = 250 N/m), compressing it 0.200 m, andreleased. What will be the speed of theblock when it separates from the spring at
x = 0? Assume k =0.300.
(i) F.B.D. first !(ii) x < 0
FS = - k x
Work and Energy
(a) The work done by the spring is
(b) Wf = - kFN (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
x2 = 0 mWS = FS(x) d x = +5.00 J x1 = -0.200 m
Example 2 (cont’d)
Energy Conservation
1. Conservative/Nonconservative Forces Work along a path
(Path integral)
Work around any closed path
(Path integral)
2. Potential Energy
Potential Energy and Energy Conservation
Mechanical Energy Conservation
Energy Conservation
Work Done bythe Gravitational Force (I)
21)(
ˆd ˆ
d
mgymgymgy
y-mg
2y
y1
y2
y1
l2
l1
)jj
lFW
()(
l
y
(Path integral)
Near the Earth’s surface
Energy Conservation
Near the Earth’s surface
Work Done bythe Gravitational Force (II)
21)(
ˆd ˆd ˆ
d
mgymgymgy
y x-mg
2y
y1
y2
y1
l2
l1
)jij
lFW
()(
dl
y
(Path integral)
Energy Conservation
Work Done bythe 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 bythe Gravitational Force (IV)
Wg(ABCA)
= Wg(AB) +
Wg(BC) +
Wg(CA)
= mg(y1 – y2) + 0 +
mg(y2- y1) = 0
dl
A
BC
Energy Conservation
Energy Conservation
Work Done bythe Gravitational Force (V)
Wg = 0 for a closed path
The gravitational force is a conservative force.
Energy Conservation
Work Done by Ff (I)
)(
d
0
)(
)0(
mgl L
Ll
l
2
1
lFW ff
(Path integral)
- μmg L
LB
LA
L depends on the path.
Path APath B
Energy Conservation
The work done by the friction force
depends on the path length.
The friction force:(a) is a non-conservative force;
(b) decreases mechanical energy of the system.
Wf = 0 (any closed path)
Work Done by Ff (II)
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(AC) = Ug(yA) – Ug(yC)
Wg(ABC) = Wg(AB) + Wg(BC)
= mg(yA- yB) + mg(yB - yC)
= mg(yA - yC)
dlA
B
C
y
A
B
Climbing the Sear tower
Work and Energy
Work and Energy
Power
The Burj Khalifa is the largestman made structure in the world and was designed by Adrian Smith class of 1966
thebatt.com Febuary 25th