Download PH 201-4A spring 2007 PH 201 4A spring 2007

PH 201-4A
201 4A spring 2007
Work and Energy
Lectures 16-17
Chapter 6
(Cutnell & Johnson, Physics 7th edition)
1
Work and Energy: Work done by a constant force
• Constant pushing force F pointing in the same direction as the resulting displacement S
• It takes work to push a stalled car.
• Force and displacement
p
are two essential elements of work. W increases when F
increases and S increases
• Work is defined as the magnitude of the force F times the magnitude of the
displacement
W=FxS
• Work does not convey
y directional information, it is a scalar quantity
y
• Unit of measurement for work
Force x Distance = Work
Newton (N) x meter (m) = joule (J)
• important feature: The idea of work is tied up with the idea of motion. If there is no
movement of the object, the work done by the force acting on the object is zero
2
The force is directed at angle θ relative to the displacement
The work done on an object by a constant force F is W = (Fcosθ)s where F is the
magnitude of the force
force, s is the magnitude of the displacement
displacement, θ is the angle between
the force and displacement.
The force component that points along the displacement is Fcosθ and it is used in
defining work.
The force component perpendicular to the displacement does no work. To do work, there
must be a force and a displacement and since there is no displacement in the
perpendicular direction there is no work done by the perpendicular component of the
force.
force
3
Positive and Negative Work. Bench-pressing
During the lifting phase, the force F
does positive work
W = (Fcosθ)s θ = 0° cosθ = 1
If the force component points in the
direction opposite to the
displacement, the work is negative.
During the lowering phase
W = (Fcosθ)s = (Fcos180°)s=
= -Fs
Since cos180° = -1
4
Accelerating a Crate
A 120kg crate on the flatbed of a truck is moving with an acceleration a = +1.5m/s2 along
the positive x axis
axis. The crate does not slip with respect to the truck
truck, as the truck
undergoes a displacement s = 65m. What is the total work done on the crate by all the
forces acting on it?
1.2x104 N
5
Work done by a variable force
The curved plot of Fcosθ vs S has
been approximated by straight line
segments. This is a good
approximation if ΔS→0.
W=(Fcosθ)1⋅ΔS1+…+(Fcosθ)n ⋅ΔSn=
sum of rectangular areas=
area under the graph
Plot of Fcosθ vs S for a force
that varies with position. The
work done by this force as a
particle moves displacement S
equals the area under the plot
plot.
6
Work
Problem 11: A 1200kg car is being driven up a 5.0° hill. The frictional force is directed
opposite
pp
to the motion of the car and has a magnitude
g
of fk = 5.0x102N. The force F is
applied to the car by the road and propels the car forward. In addition to those two
forces, two other forces act on the car: its weight W, and the normal force N directed
perpendicular to the road surface. The length of the hill is 3.0x102m. What should be the
magnitude
g
of F so that the net work done by
y all the forces acting
g on the car is
+150,000 J?
7
The Work-Energy Theorem and Kinetic Energy
• When a net force performs work on an object, there is always a result from the effort –
change in the kinetic energy.
• The relationship that relates work to the change in kinetic energy is the Work-Energy
Theorem.
• A constant net external force ΣF acts on an airplane of mass m
• For
o simplicity
s p c ty itt has
as tthe
e sa
same
ed
direction
ect o as d
displacement
sp ace e t s
Newton’s 2nd Law: net force produces acceleration a = ΣF/m; (ΣF)s = mas
• Speed of the plane changes from v0 to vf
• from kinematics: vf2 = v02 + 2as => as = ½(vf2 – v02) => (ΣF)s = 1/2mvf2 – 1/2mv02
• The kinetic energy KE of an object with mass m and speed v is defined as KE = 1/2mv2
• The work-energy theorem: when a net external force does work W on an object, the
kinetic energy of the object changes from initial value KE0 to the final value KEf, the
difference between the two values being equal to the work
W = KEf – KE0 = 1/2mvf2 – 1/2mv02
8
The Work-Energy Theorem and Kinetic Energy
Problem 12: A 0.075kg arrow is fired horizontally. The bowstring exerts an average force
of 65 N on the arrow over a distance of 0.90 m. With what speed does the arrow leave the
bow?
9
The work-energy theorem deals with the work done by the net external force. The workenergy theorem does not apply to the work done by an individual force. If W>0 then KE
increases; if W<0 then KE decreases; if W=0 then KE remains constant.
Downhill Skiing: A 58kg skier is coasting down a 25° slope. A kinetic friction force fk=70N
opposes her motion. Near the top of the slope, the skier’s speed is v0=3.6m/s. Ignoring air
resistance, determine the speed vf at a point that is displaced 57m downhill.
10
Work and Kinetic Energy
In a circular orbit the gravitational force F is
always perpendicular to the displacement s
of the satellite and does no work
KE = constant
W>0
KE increases
In an elliptical orbit, there can be a
component of the force along the
displacement
Work is done
W<0
KE decreases
11
The Work-Energy Theorem and Kinetic Energy
The speed of a hockey puck decreases from 45.00 to 44.67m/s in coasting 16m across
the ice. Find the coefficient of kinetic friction between the p
puck and the ice.
12
Work done by a force of gravity
• basketball of mass m moving
vertically downward
• mg is the only force acting
• initial height h0, final hf
• the displacement s is downward
s=h0-hf
• Wgravity = (Fcosθ)s = mg(h0-hf)
Valid for any path taken between the
initial and final height.
Since only the difference between h0 and hf appears in the equation, the vertical
distances themselves need not be measured from the earth. For example, they would
be measured
easu ed relative
e at e to a zero
e o level
e e that
t at is
s one
o e meter
ete abo
above
e tthe
eg
ground.
ou d h0-hf will be
the same.
13
Gravitational Potential Energy
• An object in motion has kinetic energy
• An object
j
may
y possess
p
energy
gy by
y virtue of its position
p
relative to the earth – such an
object is said to have gravitational potential energy
Wgravity = mgh0 – mghf
Initial gravit.
potential energy
PE0
Final gravit.
potential
energy PEf
• The gravitational potential energy PE is the energy that an object of mass m has by
virtue of its position relative to the surface of the earth. That position is measured by the
height h of the object relative to an arbitrary zero level.
PE = mgh
SI unit: joule (J)
• The gravitational potential energy belongs to the object and the earth as a system
although one often speaks of the object alone as possessing the gravitational potential
energy.
energy
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Gravitational Potential Energy
ΔPE ≡ ΔU g = −Wg = mgh f − mghi = mgΔy
(1)
The choice of the reference point for the potential energy is completely arbitrary.
It is customary to locate the reference point where the gravitational force is zero.
It is possible to show that for this reference point
Earth
hi
ME
Ug
Ug = −
hf
Δy
y
d
and
m
RE
r
⎛ rf − ri ⎞
⎛ 1 1⎞
⎟ (2)
⎜
⎟
= GM E m⎜
ΔU g = −GM E m
−
⎜
⎟
⎟
⎜
⎝ rf ri ⎠
⎝ ri rf ⎠
Where r is the distance between the
center of the Earth and the object.
0
− GM E m
RE
GM E m
r
Lett us show
L
h
th
thatt ffor a particle
ti l off mass m displaced
di l
d th
through
ha
small vertical distance Δy near the Earth’s surface the general
expression (2) for the change in gravitational energy reduces to a
p ((1).
) If both the initial and final p
positions of the
familiar relationship
particle are close to the Earth’s surface, then rf-ri=Δy and rf⋅ri=RE2
GM E m
ΔU g ≈
Δy = mgΔy
2
RE
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Gravitational Potential Energy
Sign arises from the choice of the
zero reference point (where U=0)
which is r = ∞
Gravitational potential energy of a particle of mass 1kg
gravitationally attracted by another particle of mass
1kg.
g
Total mechanical energy:
E = K + U = 1/2mv2 – GMm/r = constant
If the only
y force is g
gravity.
y
For a circular orbit of planets:
v2 = GMs/r; K = 1/2mv2 = GMsm/2r
E = 1/2mv2 – GMsm/r = - GMsm/2r
GMsm/2r + (-GMsm/r) = - GMsm/2r
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The motor of a Scout rocket uses up all the fuel and stops when the rocket is at an
altitude of 200km above the surface of the Earth and is moving vertically at 8.50km/s.
How high will this rocket rise? Ignore any residual atmospheric friction.
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Example: A meteoroid (a chunk of rock) is initially at rest in interplanetary space at a
large distance from the Sun. Under the influence of gravity, the meteoroid begins to fall
toward the Sun along a straight radial line. With what speed does it strike the Sun?
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Conservative Forces, Nonconservative Forces and the Work-Energy Theorem
• An important property of the gravitational force: when an object is moved from one
place to another, the work done by the gravitational force does not depend on the choice
of path. For this reason it is called a conservative force.
• Definition:
Version 1: A force is conservative when the work it does on a moving object is
independent on the path between the object’s initial and final positions.
Version 2: A force is conservative when it does no net work on an object moving around
a closed path starting and finishing at the same point.
• A force is nonconservative if the work it does on an object moving between two points
depends on the path of the motion between the points.
Kinetic friction force – example of nonconservative force
• In normal conditions, conservative and nonconservative forces (friction, air resistance)
act simultaneously on an object
• Work done by the net force:
W = Wc + Wnc
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The Work-Energy Theorem
Wc + Wnc = 1/2mvf2 – 1/2mv02
Wc = Wgravity = mg(h0 – hf)
mg(h0 – hf) + Wnc = 1/2mvf2 – 1/2mv02
Wnc = (1/2mvf2 – 1/2mv02) + (mghf – mgh0)
Wnc = (KEf – KE0) + (PEf – PE0)
Net work
done by
nonconserv.
forces
Change to kinetic
energy
Change in
potential energy
Wnc = ΔKE + ΔPE
The conservation of mechanical energy:
gy
E = KE + PE
-
total mechanical energy
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Conservation of Mechanical Energy
• E = KE + PE total mechanical energy
• Wnc = (KEf – KE0) + (PEf – PE0) = (KEf + PEf) – (KE0 + PE0)
The net work done by external nonconservative forces changes the total mechanical
energy from an initial value E0 to a final value of Ef
• Suppose Wnc = 0 => Ef = E0
1/2mvf2 + mghf = 1/2mv0 + mgh0
total mechanical energy conserved
Yes
No
The total mechanical energy (E = KE + PE) of an object remains constant as the object
moves provided that the net work done by external nonconservative forces is zero
zero,
Wnc = 0 J
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Conservation of Mechanical Energy
E = KE + PE = constant if Wnc = 0
If friction and wind resistance are ignored, a bobsled run illustrates how kinetic and
potential
t ti l energy can be
b interconverted
i t
t d while
hil the
th total
t t l mechanical
h i l energy remains
i
constant. The total mechanical energy is all potential energy at the top and all kinetic
energy at the bottom.
22
A golf ball of mass 50g released from a height of 1.5m above a concrete floor bounces
back to a height of 1.0m
a) What is the kinetic energy of the ball just before contact with the floor begins? Ignore
air
i friction.
f i ti
b) What is the kinetic energy of the ball just after contact with the floor ends?
c) What is the loss of energy during contact with the floor?
23
A pendulum consists of a mass “m” tied to one end of a string of length “l”. The other
end of the string is attached to a fixed point on the wall. Suppose that the pendulum is
initially held at an angle of 90° with the vertical. If the pendulum is released from this
position what will be the speed of the pendulum at the instant it passes through its
position,
lowest point?
24
In a rollercoaster, a car starts on the top of a 30m high mountain. It rolls down into a
valley and then up a 20m high mountain. What is the speed of the car at the bottom of the
valley, at ground level? What is the speed of the car at the top of the second mountain?
25
A block of mass m slides down an inclined plane into a loop of radius R. a) Neglecting
friction, what is the minimum speed the block must have at the highest point of the loop
to stay in the loop. b) At what vertical height on the inclined plane (in terms of radius of
the loop) must the block be released if it is to have the required minimum speed at the
top of the loop?
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Nonconservative Forces and the Work-Energy Theorem
Problem 17: A 55.0 kg skateboarder starts out with a speed of 1.80 m/s. He does +80.0 J
of work on himself by pushing his foot against the ground. In addition, friction does 265 J of work on him. In both cases, the forces doing the work are nonconservative.
The final speed of the skateboarder is 6.00 m/s.
a) ΔPE = PEf – PE0?
b) How much has the vertical height of the skater changed and is the skater above or
below the starting point?
27
Nonconservative Forces and the Work-Energy Theorem
Problem 54: A 3.00 kg model rocket is launched vertically straight up with sufficient initial
speed to reach a maximum height of 1.00x102 m even though air resistance (a
nonconservative force) performs -8.00X102 J of work on the rocket. How high would the
rocket have gone without air resistance?
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Power
• Car is moving up the hill
• The important characteristic of the engine is not
how much force it can exert, but how much work it
can perform in a given amount of time.
• The force is of secondary
y importance,
p
because by
y shifting
g to a low g
gear we can make
sure that even a “weak” engine exerts enough force on the wheels to propel us uphill.
• Rate of work is crucial – how fast the engine can propel the car up the hill.
• When we move uphill, the gravitational force takes energy from the car (performs
negative
g
work).
) To keep the car moving
g the engine
g
must perform an equal amount of
positive work.
• If the engine performs work at a fast rate, it can propel the car at a fast speed.
• The rate at which a force does work on a body is called power delivered by the force.
• If the force does the amount of work W in time Δt, average power: P = W/Δt
instantaneous power: P = W/Δt, Δt -> 0
SI unit of power: 1 watt = 1 W = 1 J/s
engineering units: 1 horsepower = 1 hp = 746 W
1 kilowatt-hour = 1 KWh = 1000 W x 3600 s = 3.6x106J
Kilowatt-hour is used to measure electric energy delivered to homes and factories
29
Example:
An elevator cage has a mass of 1000kg. How many horsepower must the motor deliver to
the elevator if it is to raise the elevator cage at the rate of 2.0m/s?
30
A car accelerates uniformly from rest to 27 m/s in 7.0s along a level stretch of road.
Ignoring friction, determine the average power required to accelerate the car if
a) The weight of the car is 1.2x104N, and b) weight is 1.6x104N
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Other forms of energy
• If the forces acting on a particle are conservative – mechanical energy is conserved
• If nonconservative forces do work on the particle – mechanical energy is not conserved.
This energy never disappears. It is converted into other forms of energy.
• Chemical and nuclear energy:
Chemical – KE and PE of electrons within atom
Nuclear – KE and PE of protons and neutrons within nuclei of atoms.
• Thermal energy – KE and PE of atoms of the body.
• Electric and magnetic energy – are forms of energy associated with electric charges
and with light and radio waves
• All these forms of energy can be transformed into one another. In any such
transformative process, the sum of all the energies of all the pieces of matter involved in
the process remains constant.
General law of conservation of energy: The form of energy changes, but the
total amount of energy does not change.
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