Download Chapter 6 - Work and Kinetic Energy

Chapter 6
Work and Kinetic
Energy
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by Wayne Anderson
Copyright © 2012 Pearson Education Inc.
Goals for Chapter 6
• Understand & calculate work done by a force
• Understand meaning of kinetic energy
• Learn how work changes kinetic energy of a
body & how to use this principle
• Relate work and kinetic energy when forces
are not constant or body follows curved path
• To solve problems involving power
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Introduction
• The simple methods we’ve learned using Newton’s
laws are inadequate when the forces are not
constant.
• In this chapter, the introduction of the new concepts
of work, energy, and the conservation of energy will
allow us to deal with such problems.
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Work
• A force on a body does work if the body undergoes a
displacement.
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Work done by a constant force
• The work done by a constant force acting at an angle  to
the displacement is
W = Fs cos .
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Work done by a constant force
• The work done by a constant force acting at an angle  to
the displacement is
W = Fs cos .
• Units of Work = Force x Distance = Newtons x meters
- (Nm) = JOULE of energy!
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Positive, negative, and zero work
• A force can do positive, negative, or zero work depending on
the angle between the force and the displacement.
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Positive, negative, and zero work
• A force can do positive, negative, or zero work depending on
the angle between the force and the displacement.
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Positive, negative, and zero work
• A force can do positive, negative, or zero work depending on
the angle between the force and the displacement.
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Positive, negative, and zero work
• A force can do positive, negative, or zero work depending on
the angle between the force and the displacement.
• Work is done BY an external force, ON an object.
•Positive work done by an external force
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Positive, negative, and zero work
• A force can do positive, negative, or zero work depending on
the angle between the force and the displacement.
• Work is done BY an external force, ON an object.
•Positive work done by an external force
(with no other forces acting in that direction of motion)
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Positive, negative, and zero work
• A force can do positive, negative, or zero work depending on
the angle between the force and the displacement.
• Work is done BY an external force, ON an object.
•Positive work done by an external force
(with no other forces acting in that direction of motion)
will speed up an object!
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Positive, negative, and zero work
• A force can do positive, negative, or zero work depending on
the angle between the force and the displacement.
• Work is done BY an external force, ON an object.
•Positive work done by an external force
(with no other forces acting in that direction of motion)
will speed up an object!
•Negative work done by an external force (wnofaitdom)
will slow down an object!
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Work done by several forces – Example 6.2
• Tractor pulls wood 20 m over level ground;
• Weight = 14,700 N; Tractor exerts 5000 N
force at 36.9 degrees above horizontal.
• 3500 N friction opposes!
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Work done by several forces – Example 6.2
• How much work is done BY the tractor ON the
sled with wood?
• How much work is done BY gravity ON the sled?
• How much work is done BY friction ON the sled?
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Work done by several forces – Example 6.2
• Tractor pulls wood 20 m over
level ground; w = 14,700 N;
Tractor exerts 5000 N force at
36.9 degrees above horizontal.
3500 N friction opposes!
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Work done by several forces – Example 6.2
• Work done BY tractor ON sled:
= +5000N cos(36.9) * 20m = 80 kJ
• Work done BY gravity ON sled?
= 14,700 N sin(90) * 20m = 0
• Work done BY friction ON sled?
= 3500N cos(180) * 20m = - 70 kJ
• NET work: +10 kJ
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Kinetic energy
• The kinetic energy of a particle is KE = 1/2 mv2.
• NET work on body changes its kinetic energy! (& speed!)
Gain KE doing
POSITIVE WORK
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Kinetic energy
• The kinetic energy of a particle is K = 1/2 mv2.
• Net work on body changes its speed & kinetic energy
Lose KE doing
NEGATIVE WORK
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The work-energy theorem
• The work done by the net force on a particle equals
the change in the particle’s kinetic energy.
• Mathematically, the work-energy theorem is
expressed as
Wtotal = KEfinal – KEinitial = KE
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Work done by several forces – Example 6.3
• Tractor pulls wood 20 m over
level ground; w = 14,700 N;
Tractor exerts 5000 N force at
36.9 degrees above horizontal.
3500 N friction opposes!
• Suppose sled moves at 2.0 m/s;
what is speed after 20 m?
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Using work and energy to calculate speed
• Net gain in Work: +10kJ = CHANGE in KE
• KE initial = ½ mvi2 = ½ 1500 kg * 2.0m/s 2 = 3000 J
• KE final = ½ mvf2 = 3000 J + Net work done = 13 kJ
• Vf = 4.2 m/s!
1500
kg
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Net Work done = +10kJ
Forces on a hammerhead – example 6.4
• Hammerhead of pile driver drives a beam into the ground
falling along rails which provide 60N of frictional force.
200 kg head of pile driver lifted
3 m and dropped onto beam
which sinks 7.4 cm into
ground
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Forces on a hammerhead – example 6.4
• Hammerhead of pile driver drives a beam into the ground
falling along rails which provide 60N of frictional force.
Find speed as it hits and
average force on the beam.
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Forces on a hammerhead – example 6.4
• Hammerhead of pile driver drives a beam into the ground
falling along rails which provide 60N of frictional force.
While falling down
along rails…
What does + work?
What does – work?
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Forces on a hammerhead – example 6.4
• Hammerhead of pile driver drives a beam into the ground
falling along rails which provide 60N of frictional force.
How fast is it going
just before hitting
the beam?
KE = Net Work!
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Forces on a hammerhead – example 6.4
• Hammerhead of pile driver drives a beam into the ground
falling along rails which provide 60N of frictional force.
How fast is it going
just before hitting
the beam?
KE = Net Work!
+ 1900 N * 3 m =
5700 J
Vf = 7.55 m/s
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Forces on a hammerhead – example 6.4
• Hammerhead of pile driver drives a beam into the ground,
following guide rails that produce 60 N of frictional force
While hitting the
beam
What does + work?
What does – work?
What distance do
these forces act?
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Forces on a hammerhead – example 6.4
• Hammerhead of pile driver drives a beam into the ground,
following guide rails that produce 60 N of frictional force
While hitting the beam
Know KEi = 5700 J!
Know KEf = 0!
Know distance = 7.4 cm
Get Net Force acting!
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Comparing kinetic energies – example 6.5
• Two iceboats have different masses, m & 2m.
Wind exerts same force; both start from rest.
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Comparing kinetic energies – example 6.5
• Two iceboats have different masses, m & 2m.
Wind exerts same force; both start from rest.
Which boat wins?
Which boat crosses with the most KE?
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Kinetic energy
Kinetic energy does not depend on the direction
of motion.
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Kinetic energy
Kinetic energy increases linearly with the mass of
the object.
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Kinetic energy
Kinetic energy increases with the square of the
speed of the object.
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The work-energy theorem
The work-energy theorem: The work done by
the net force on a particle equals the change in
the particle’s kinetic energy.
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Work and energy with varying forces
• Many forces, such as
force to stretch a spring,
are not constant.
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Work and energy with varying forces
• Many forces, such as
force of a stretched
spring, are not constant.
Suppose a particle moves along
the x-axis from x1 to x2 under a
VARYING force
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Work and energy with varying forces
• Approximate work by
dividing total displacement
into many small segments.
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Work and energy with varying forces
Calculate approximate work done by variable force over
many segments.
Do this for each segment & add results for all segments.
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Work and energy with varying forces—Figure 6.16
• Work = ∫ F∙dx
• An infinite
summation of tiny
rectangles!
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Work and energy with varying forces—Figure 6.16
• Work = ∫ F∙dx
• An infinite
summation of tiny
rectangles!
• Height: F(x)
Width: dx
Area: F(x)dx
• Total area: ∫ F∙dx
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Work and energy with varying forces
The work done by the force in the total
displacement from x1 to x2 is the integral of Fx
from x1 to x2:
On a graph of force as a function of position, the
total work done by the force is represented by
the area under the curve between the initial and
final positions.
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Work done by a constant force
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Stretching a spring
• The force required to stretch a
spring a distance x is proportional
to x:
Fx = kx
• k is the force constant
(or spring constant)
• Units of k = Newtons/meter
– Large k = TIGHT spring
– Small k = L O O S E spring
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Stretching a spring
• The force required to stretch a
spring a distance x is
proportional to x:
Fx = kx
• k is the force constant (or
spring constant) of the spring.
• Area under graph represents
work done on the spring to
stretch it a distance X:
W = ½ kX2
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Work done on a spring scale – example 6.6
• A woman of 600 N weight compresses spring 1.0 cm.
What is k and total work done BY her ON the spring?
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Work done on a spring scale – example 6.6
• A woman of 600 N weight compresses spring 1.0 cm.
What is k and total work done?
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Work done on a spring scale – example 6.6
• A woman of 600 N weight compresses spring 1.0 cm.
What is k and total work done BY her ON the spring?
K = Fs/x = -600N / -0.010 m = + 6.0 x 104 N/m
W = ½ kxf2 – ½ kxi2 = ½ (6.0 x 104N/m) (-0.010 m)2 = 3.0 J
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Motion with a varying force
• An air-track glider mass 0.1 kg is attached to a spring of
force constant 20 N/m,
• Glider starts at rest with spring compressed, then
released. At the point where the spring is no longer
compressed, it is moving with some speed v1.
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Motion with a varying force
• The force on the glider is varying from maximum + to 0
to maximum negative, when it stops momentarily.
• Moving at 1.50 m/s to right when spring is unstretched.
• Find maximum distance moved if no friction, and if
friction was present with mk = 0.47
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Motion with a varying force
• glider mass 0.1 kg
• spring of force constant 20 N/m
• Moving at 1.50 m/s to right when spring is unstretched.
• Final velocity once spring stops glider = 0
• We know 3 things! F/m = a; vi; vf
• Why can’t we use F = ma ?
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Motion with a varying force
• An air-track glider is attached to a spring, so the force on
the glider is varying.
• In general, if the force varies, using ENERGY will be an
easier method than using forces!
• We know:
• Initial KE = ½ mv2
• Final KE = ½ mvf2 = 0
• Get net work done!
• Get ½ kx2
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Motion with a varying force
• An air-track glider is attached to a spring, so the force on
the glider is varying.
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Motion on a curved path—Example 6.8
• A child on a swing moves along a smooth curved
path at constant speed.
• Weight w, Chain length = R, max angle = q0
• You push with force F that varies.
• What is work done by you?
• What is work done by gravity?
• What is work done by chain?
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Motion on a curved path—Example 6.8
• A child on a swing moves along a curved path.
• Weight w, Chain length = R, max angle = q0
• You push with force F that varies.
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Motion on a curved path—Example 6.8
• W = ∫ F∙dl
• F∙dl = F cos q
• dl = ds (distance along the arc)
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Work–energy theorem for motion along a curve
A particle moves along
a curved path from
point P1 to P2, acted
on by a force that
varies in magnitude
and direction.
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Work–energy theorem for motion along a curve
A particle moves along a
curved path from point
P1 to P2, acted on by a
force that varies in
magnitude and
direction.
The work can be found
using a line integral:
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Power
Power is the rate at which work is done.
Average power is:
Instantaneous power is:
The SI unit of power is the watt (1 W = 1 J/s), but another
familiar unit is the horsepower (1 hp = 746 W).
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Power: Lifting a box slowly
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Power: Lifting a box quickly
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Power
• Power is rate work is done.
• Average power is Pav = W/t
• Instantaneous power is P = dW/dt.
• SI unit of power is watt
(1 W = 1 J/s)
• horsepower
(1 hp = 746 W ~ ¾ of kilowatt)
• kilowatt-hour (kwh) is
ENERGY (power x time)
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Power
In mechanics we can also express power in terms
of force and velocity:
Here is a one-horsepower
(746-W) propulsion system.
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A “power climb”
• A person runs up stairs.
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Banked Curve Examples
Ny
N
y
Nx
fsx
fsy
fs
center of curvature
of the banked road
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mg
x
Banked Curve Examples
Maximum Cornering Speed for
Banked Curves
Radius (m)
30
30
30
30
30
30
30
40
50
60
60
40
Coefficient of
Static Friction
0.8
0.8
0.8
0.8
0.6
0.4
0.2
0.2
0.2
0.2
0.2
0.8
Angle
(degrees)
30
20
10
0
30
30
30
30
30
30
0
5
max speed (m/s)
27
22
18
15
23
19
16
19
21
23
11
19
max speed (mph)
60
48
40
34
51
43
35
41
46
50
24
43
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