Download Kinetic Energy & Work

Energy, Work, & Power!
• Why does studying physics take so much energy…. but
• studying physics at your desk is NOT work? 
Copyright © 2012 Pearson Education Inc.
Goals for This Chapter
• 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
Copyright © 2012 Pearson Education Inc.
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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Energy, Work, & Power!
• Energy: A new tool in your belt for solving physics
problems
• Sometimes, sawing a branch is MUCH easier than
hammering it off!
Copyright © 2012 Pearson Education Inc.
7-1 Kinetic Energy

Energy is required for any sort of motion

Energy:
o
o
o

A scalar quantity assigned to object or system
Can be changed from one form to another
Is conserved in a closed system, (total amount of
energy of all types is always the same)
Chapter 7: one type of energy (kinetic energy) &
one method of transferring energy (work)
© 2014 John Wiley & Sons, Inc. All rights reserved.
7-1 Kinetic Energy
o
Faster an object moves, greater its kinetic energy
o
Kinetic energy is zero for a stationary object

For an object with v well below the speed of light:

The unit of kinetic energy is a joule (J)
© 2014 John Wiley & Sons, Inc. All rights reserved.
7-1 Kinetic Energy
Example Energy of 2200 lb car moving 65 mph?
Mass = 1000 kg
Velocity = ~ 30 m/s
KE = 4.4 x 105 J
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7-1 Kinetic Energy
Example Energy of 18-wheel semi moving 65 mph?
Mass = ~ 67000 kg
Velocity = ~ 30 m/s
KE = 6.0 x 107 J
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7-1 Kinetic Energy
Example Energy dissipated in collision?
4.4 x 105 J + 6.0 x 107 J = ~ 6.1 x 107 J
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7-1 Kinetic Energy
Example
•
Force of Truck on Car?
• Force of Car on Truck?
• ACCELERATION of car? Of Truck?
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7-2 Work and Kinetic Energy


Account for changes in kinetic energy by saying energy
has been transferred to or from the object
In a transfer of energy via an external force, work is:
o
Done on the object by the force
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7-2 Work and Kinetic Energy

This is not the common meaning of the word “work”
o
To do work on an object, energy must be transferred
o
Throwing a baseball does work
o
Pushing an immovable wall does not do work
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7-2 Work and Kinetic Energy

Start from force equation and 1-dimensional velocity:

Rearrange into kinetic energies:

Left side is change in energy

Work is:
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Work
• A force on a body does work only 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
= (N m) = (kg m/s2) m = JOULE of energy!
• Work has the SI unit of joules (J), the same as energy
• In the British system, the unit is foot-pound (ft lb)
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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.
Copyright © 2012 Pearson Education Inc.
7-2 Work and Kinetic Energy

For an angle φ between force and displacement:

As vectors we can write:

Notes on these equations:
o
Force is constant
o
Object is particle-like (rigid)
o
Work can be positive or negative
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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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7-2 Work and Kinetic Energy
© 2014 John Wiley & Sons, Inc. All rights reserved.
7-2 Work and Kinetic Energy


For two or more forces, net work is sum of work done
by all individual external forces
Two methods to calculate net work:

Find all work and sum individual terms
OR….

Take vector sum of forces (Fnet) and calculate net work
…..once 
© 2014 John Wiley & Sons, Inc. All rights reserved.
Work done by several forces – Example
• 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
• 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
• 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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Kinetic energy
• The kinetic energy of a particle is KE = 1/2 mv2.
• Net work on body changes its speed & kinetic energy
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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
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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
Copyright © 2012 Pearson Education Inc.
7-2 Work and Kinetic Energy

The work-kinetic energy theorem states:

(change in kinetic energy) = (net work done)

Or we can write it as:
(final KE) = (initial KE) + (net work)
© 2014 John Wiley & Sons, Inc. All rights reserved.
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
• 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 20 m/s;
what is speed after 20 m?
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Using work and energy to calculate speed
Copyright © 2012 Pearson Education Inc.
7-2 Work and Kinetic Energy

The work-kinetic energy theorem holds for positive and
negative work
Example If the kinetic energy of a particle is initially 5 J:
o
A net transfer of 2 J to the particle (positive work)
• Final KE = 7 J
o
A net transfer of 2 J from the particle (negative work)
• Final KE = 3 J
© 2014 John Wiley & Sons, Inc. All rights reserved.
7-2 Work and Kinetic Energy
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7-2 Work and Kinetic Energy
Answer: (a) energy decreases (b) energy remains the same
(c) work is negative for (a), and work is zero for (b)
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Comparing kinetic energies – example
• 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?
Copyright © 2012 Pearson Education Inc.
7-3 Work Done by the Gravitational Force

We calculate the work as we would for any force

Our equation is:
Eq. (7-12)

For a rising object:
Eq. (7-13)

For a falling object:
Eq. (7-14)
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Forces on a hammerhead – example
• Hammerhead of pile driver drives a beam into the ground.
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,
following guide rails that produce 60 N of frictional force
Copyright © 2012 Pearson Education Inc.
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
Copyright © 2012 Pearson Education Inc.
7-3 Work Done by the Gravitational Force



Orientations of forces and
associated works for
upward displacement
Note works need not be
equal, they are only equal if
initial and final kinetic
energies are equal
If works are unequal, you
need to know difference
between initial & final kinetic
energy to solve for work
© 2014 John Wiley & Sons, Inc. All rights reserved.
7-3 Work Done by the Gravitational Force

Orientations of forces and
their associated works for
downward displacement
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7-3 Work Done by the Gravitational Force
Examples You are a passenger:
o
Being pulled up a ski-slope
• Tension does positive work, gravity does negative
work
Figure 7-8
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7-3 Work Done by the Gravitational Force
Examples You are a
passenger:
o
Being lowered down in
an elevator
• Tension does negative
work,
• Gravity does positive
work
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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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7-4 Work Done by a Spring Force

Spring force is a variable force from a spring
o
Particular mathematical form
o
Many forces in nature have this form


Figure (a) shows spring in a
relaxed state:
neither compressed nor
extended, no force is applied
Stretch or extend spring?
It resists & exerts a
restoring force that attempts
to return the spring to its
relaxed state
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7-4 Work Done by a Spring Force





Spring force is given by Hooke's law:
Negative sign means force always opposes
displacement
Spring constant
spring
k
measures stiffness of the
This is a variable force (function of position) and it
exhibits a linear relationship between F and d
For a spring along the x-axis we can write:
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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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Work and energy with varying forces
• Suppose an applied force
(like that of a stretched
spring) is not constant.
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Work and energy with varying forces—Figure 6.16
• Approximate work by
dividing total displacement
into many small 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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7-4 Work Done by a Spring Force

We can find the work by integrating:

Plug kx in for Fx:

Work:
o
Can be positive or negative
o
Depends on the net energy transfer
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7-4 Work Done by a Spring Force


The work:
o
Can be positive or negative
© 2014 John Wiley & Sons, Inc. All rights reserved.
7-4 Work Done by a Spring Force

For an initial position of x = 0:
Area under graph
represents work done on
the spring to stretch it a
distance X:
W = ½ kX2
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7-4 Work Done by a Spring Force

For an applied force where the initial and final kinetic
energies are zero:
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7-4 Work Done by a Spring Force
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7-4 Work Done by a Spring Force
Answer: (a)
positive
(b) negative
(c) zero
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7-5 Work Done by a General Variable Force



We take a one-dimensional
example
We need to integrate the
work equation (which
normally applies only for a
constant force) over the
change in position
We can show this process
by an approximation with
rectangles under the curve
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Figure 7-12
7-5 Work Done by a General Variable Force

Our sum of rectangles would be:
Eq. (7-31)

As an integral this is:
Eq. (7-32)

In three dimensions, we integrate each separately:
Eq. (7-36)

The work-kinetic energy theorem still applies!
© 2014 John Wiley & Sons, Inc. All rights reserved.
7-6 Power
Learning Objectives
7.18 Apply the relationship
between average power,
the work done by a force,
and the time interval in
which that work is done.
7.19 Given the work as a
function of time, find the
instantaneous power.
7.20 Determine the
instantaneous power by
taking a dot product of the
force vector and an
object's velocity vector, in
magnitude-angle and unitvector notations.
© 2014 John Wiley & Sons, Inc. All rights reserved.
7-6 Power


Power is the time rate at which a force does work
A force does W work in a time Δt; the average power
due to the force is:
Eq. (7-42)

The instantaneous power at a particular time is:
Eq. (7-43)


The SI unit for power is the watt (W): 1 W = 1 J/s
Therefore work-energy can be written as (power) x
(time) e.g. kWh, the kilowatt-hour
© 2014 John Wiley & Sons, Inc. All rights reserved.
7-6 Power

Solve for the instantaneous power using the definition
of work:
Eq. (7-47)

Or:
Eq. (7-48)
Answer: zero (consider P = Fv cos ɸ, and note that ɸ = 90°)
© 2014 John Wiley & Sons, Inc. All rights reserved.
7
Summary
Kinetic Energy

Work
The energy associated with
motion
Eq. (7-1)
Work Done by a Constant
Force

Energy transferred to or from an
object via a force

Can be positive or negative
Work and Kinetic Energy
Eq. (7-7)
Eq. (7-8)

The net work is the sum of
individual works
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Eq. (7-10)
Eq. (7-11)
7
Summary
Work Done by the
Gravitational Force
Work Done in Lifting and
Lowering an Object
Eq. (7-12)
Eq. (7-16)
Spring Force
Spring Force

Relaxed state: applies no force

Spring constant k measures
stiffness

For an initial position x = 0:
Eq. (7-20)
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Eq. (7-26)
7
Summary
Work Done by a Variable
Force

Found by integrating the
constant-force work equation
Power

The rate at which a force does
work on an object

Average power:
Eq. (7-42)
Eq. (7-32)

Instantaneous power:
Eq. (7-43)

For a force acting on a moving
object:
Eq. (7-47)
Eq. (7-48)
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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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Motion with a varying force
• An air-track glider mass 0.1 kg is attached to a spring of
force constant 20 N/m, so the force on the glider is
varying.
• 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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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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Force and power
• Jet engines develop power to fly the plane.
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A “power climb”
• A person runs up stairs
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