Transcript Work-Kinetic Energy

7.4 Work Done by a
Varying Force
Work Done by a Varying Force

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
Assume that during a very
small displacement, Dx, F
is constant
For that displacement, W
~ F Dx
For all of the intervals,
W 
xf
 F Dx
x
xi
Work Done by a Varying
Force, cont

Sum approaches a definite
value:
xf
lim
Dx 0
 F Dx  
x
xi

xi
Fx dx
Therefore:
W 

xf

xf
xi
Fx dx
(7.7)
The work done is equal to
the area under the curve!!
Example 7.7 Total Work Done from a
Graph (Example 7.4 Text Book)

The net work done by this
force is the area under
the curve
W = Area under the Curve
W = AR + AT
W = (B)(h) + (B)(h)/2 =
(4m)(5N) + (2m)(5N)/2
W = 20J + 5J = 25 J
Work Done By Multiple
Forces

If more than one force acts on a system
and the system can be modeled as a
particle, the total work done ON the
system is the work done by the net
force
W  W
net

xf
xi
  F dx
x
(7.8)
Work Done by Multiple
Forces, cont.

If the system cannot be modeled as a
particle, then the total work is equal to
the algebraic sum of the work done by the
individual forces
Wnet  Wby individual forces
Hooke’s Law
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The force exerted BY the spring is
Fs = –kx
(7.9)
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x is the position of the block with respect to the
equilibrium position (x = 0)
k is called the spring constant or force constant and
measures the stiffness of the spring (Units: N/m)
This is called Hooke’s Law
Hooke’s Law, cont.
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When x is positive (spring
is stretched), Fs is
negative
When x is 0 (at the
equilibrium position), Fs is 0
When x is negative (spring
is compressed), Fs is
positive
Hooke’s Law, final
The force exerted by the spring (Fs )
is always directed opposite to the
displacement from equilibrium
 Fs is called the restoring force
 If the block is released it will
oscillate back and forth between
–x and x

Work Done by a Spring
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Identify the block as
the system
The work as the
block moves from:
xi = – xmax to xf = 0
Ws  
xf
xi
1 2
Fx dx  
kx  dx  kxmax (7.10)

 xmax
2
0
Work Done by a Spring,
cont

The work as the
block moves from:
xi = 0 to xf = xmax
Ws  
x max
0
 (kx)dx   kx
1
2
2
max
(7.10) (a)
Work Done by a Spring,
final
Therefore:
Net Work done by the
spring force as the block
moves from –xmax to xmax
is ZERO!!!!
 For any arbitrary
displacement: xi to xf :

xf


Ws    (kx)dx   12 kx2f  12 kxi2  12 kxi2  12 kx2f
xi
(7.11)
Spring with an Applied
Force
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Suppose an external agent,
Fapp, stretches the spring
The applied force is
equal and opposite to
the spring force:
Fapp = –Fs = –(–kx) 
Fapp = kx
Spring with an Applied
Force, final
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Work done by Fapp
when xi = 0 to xf = xmax
is:
WFapp = ½kx2max
For any arbitrary
displacement: xi to xf :
xf
xf
xi
xi
WFapp   Fappdx   kxdx  12 kx2f  12 kxi2
(7.12)
Active Figure 7.10
7.5 Kinetic Energy And the
Work-Kinetic Energy Theorem
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Kinetic Energy is the energy of a particle due to its
motion
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K = ½ mv2
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(7.15)
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Units of K: Joules (J)
1 J = N m = (kg m/s2)m= kg m2/s2 = kg(m/s)2
 A change in kinetic energy is one possible result
of doing work to transfer energy into a system
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•
•
•
Kinetic Energy, cont
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Calculating the work:
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Knowing that:
F = ma = mdv/dt =m(dv/dt)(dx/dx) 
Fdx = m(dv/dx)(dx/dt)dx = mvdv
W 
xf
xi
 F dx  
xf
xi
ma dx
vf
W   mv dv
vi
1 2 1 2
W

 2mv f  2 mvi
(7.14)
Work-Kinetic Energy Theorem
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The Work-Kinetic Energy Principle states
SW = Kf – Ki = DK (7.16)
In the case in which work is done on a
system and the only change in the system is
in its speed, the work done by the net force
equals the change in kinetic energy of the
system.
We can also define the kinetic energy
2
 K = ½ mv
(7.15)
Work-Kinetic Energy Theorem, cont
Summary: Net work done by a constant
force in accelerating an object of mass m
from v1 to v2 is:
Wnet = ½mv22 – ½mv12  DK
“Net work on an object = Change in
Kinetic Energy”
 It’s been shown for a one-dimension
constant force. However, this is valid in
general!!!
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Work-Kinetic Energy REMARKS!!
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Wnet ≡ work done by the net (total) force.
Wnet is a scalar.
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Wnet can be positive or negative since DK can
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be both + or –
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K  ½mv2 is always positive. Mass and v2 are
both positive. (Question 10 Homework)
Units are Joules for both work & kinetic energy.
The work-kinetic theorem: relates work to a
change in speed of an object, not to a change in its
velocity.
Example 7.8 Question #14
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(a). Ki  ½m v2 ≥ 0
K depends on: v2 ≥ 0 & m > 0
If v 2v
Kf = ½m (2v)2 = 4(½mv2 ) = 4Ki
Then: Doubling the speed makes an object’s
kinetic energy four times larger
(b). If SW = 0  v must be the same at
the final point as it was at the initial point
Example 7.9 Work-Kinetic Energy
Theorem (Example 7.7 Text Book)
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m = 6.0kg first at rest is pulled to
the right with a force F = 12N
(frictionless).
Find v after m moves 3.0m
Solution:
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The normal and gravitational forces
do no work since they are
perpendicular to the direction of the
displacement
W = F Dx = (12)(3)J = 36J
W = DK = ½ mvf2 – 0 
36J = ½(6.0kg)vf2 = (3kg)vf2
Vf =(36J/3kg)½ = 3.5m/s
Example 7.10 Work to Stop a Car
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Wnet = Fdcos180°= –Fd = –Fd
Wnet = DK = ½mv22 – ½mv12 = –Fd 
-Fd = 0 – ½m v12  d  v12
If the car’s initial speed doubled, the stopping distance is
4 times greater. Then: d = 80 m
Example 7.11 Moving Hammer
can do Work on Nail
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A moving hammer strikes a nail and
comes to rest. The hammer exerts a
force F on the nail, the nail exerts a
force –F on the hammer (Newton's 3rd
Law) m
Work done on the nail is positive:
Wn = DKn = Fd = ½mnvn2 – 0 > 0

Work done on the hammer is
negative:
Wh = DKh = –Fd = 0 – ½mhvh2 < 0
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Example 7.12 Work on a Car to
Increase Kinetic Energy
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Find Wnet to accelerate the 1000 kg car.
Wnet = DK = K2 – K1 = ½m v22 – ½m v12
Wnet = ½(103kg)(30m/s)2 – ½(103kg)(20m/s)2 
Wnet = 450,000J – 200,000J = 2.50x105J
Example 7.13 Work and Kinetic
Energy on a Baseball
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A 145-g baseball is thrown so that acquires a
speed of 25m/s. ( Remember: v1 = 0)
Find: (a). Its K.
(b). Wnet on the ball by the pitcher.
(a). K  ½mv2 = ½(0.145kg)(25m/s)2 
K  45.0 J
(b). Wnet = DK = K2 – K1 = 45.0J – 0J 
Wnet = 45.0 J
7.6 Non-isolated System
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A nonisolated system is one that interacts
with or is influenced by its environment
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An isolated system would not interact with its
environment
The Work-Kinetic Energy Theorem can be
applied to nonisolated systems
Internal Energy
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The energy associated with
an object’s temperature is
called its internal energy,
Eint
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In this example, the
surface is the system
The friction does work and
increases the internal
energy of the surface
Active Figure 7.16
Potential Energy
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Potential energy is energy related to the
configuration of a system in which the
components of the system interact by
forces
Examples include:
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elastic potential energy – stored in a spring
gravitational potential energy
electrical potential energy
Ways to Transfer Energy Into
or Out of A System
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Work – transfers by applying a force
and causing a displacement of the point
of application of the force
Mechanical Waves – allow a
disturbance to propagate through a
medium
Heat – is driven by a temperature
difference between two regions in space
More Ways to Transfer Energy
Into or Out of A System
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Matter Transfer – matter physically
crosses the boundary of the system,
carrying energy with it
Electrical Transmission – transfer is
by electric current
Electromagnetic Radiation – energy
is transferred by electromagnetic waves
Examples of Ways to Transfer
Energy
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a) Work
b) Mechanical
Waves
c) Heat
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d) Matter transfer
e) Electrical
Transmission
f) Electromagnetic
radiation
Conservation of Energy
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Energy is conserved
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Mathematically: SEsystem  ST
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This means that energy cannot be created or destroyed
If the total amount of energy in a system changes, it can
only be due to the fact that energy has crossed the
boundary of the system by some method of energy transfer
(7.17)
Esystem is the total energy of the system
T is the energy transferred across the system boundary
Established symbols: Twork = W and Theat = Q
The Work-Kinetic Energy theorem is a special
case of Conservation of Energy
Material for the Final
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Examples to Read!!!
Example 7.7
 Example 7.9
 Example 7.12
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(page 195)
(page 201)
(page 204)
Homework to be solved in Class!!!
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Problems: 11, 26