Transcript E - IBPhysicsLund

Topic 2: Mechanics
2.3 Work, energy and power
2.3.1 Outline what is meant by work.
2.3.2 Determine the work done by a non-constant
force by interpreting a force-displacement
graph.
2.3.3 Solve problems involving the work done by a
force.
Topic 2: Mechanics
2.3 Work, energy and power
Outline what is meant by work.
Dynamics is the study of forces as they are
applied to bodies, and how the bodies respond to
those forces.
Generally, we create free body diagrams to solve
the problems via Newton’s second law.
A weakness of the second law is that we have to
know all of the applied forces in order to solve
the problem.
Sometimes, forces are hard to get a handle on.
In other words, as the following example will
show, Newton’s second law is just too hard to use
to solve some physics problems.
In these cases, the principles of work and energy
are used - the subjects of Topic 2.3.
Topic 2: Mechanics
2.3 Work, energy and power
Outline what is meant by work.
EXAMPLE: Suppose we wish to find the speed of the
ball when it reaches the bottom of the track.
W N
KEY
Discuss the problems one might have in using free
body diagrams to find that final speed.
SOLUTION:
Because the slope of the track is changing, so is
the relative orientation of N and W.
Thus, the acceleration is not constant.
Thus we can’t use the kinematic equations.
Thus we can’t find v at the bottom of the track.
Topic 2: Mechanics
2.3 Work, energy and power
Outline what is meant by work.
As we have stated, the principles of work
and energy need to be mastered in order to
solve this type of problem.
We begin by defining work.
In everyday use, work is usually thought of as
effort expended by a body, you, on homework, or
on a job.
In physics, we define work W as force F times the
displacement s, over which the force acts:
W = Fs
work done by a constant force
The units of work are the units of force
(newtons) times distance (meters). For
convenience, we call a newton-meter (N m) a joule
(J) in honor of the physicist by that name.
Topic 2: Mechanics
2.3 Work, energy and power
Outline what is meant by work.
W = Fs
work done by a constant force
EXAMPLE: Find the work done by the 25-newton
force F in displacing the box s = 15 meters.
F
s
SOLUTION:
W = Fs
W = (25 n)(15 m)
W = 380 n m = 380 J.
FYI
The units for force
can be abbreviated
either (n) or (N).
FYI
The units of (n m) are joules (J). You can just
keep them as (n m) if you prefer.
Topic 2: Mechanics
2.3 Work, energy and power
Outline what is meant by work.
W = Fs
work done by a constant force
If the force is not parallel to the displacement
the formula for work has the minor correction
W = Fs cos 
work done by a constant force
not parallel to displacement
Where  is the angle between F and s.
FYI
If F and s are parallel,  = 0°.
F
s
parallel
F
If F and s are antiparallel,  = 180°. s
antiparallel
Topic 2: Mechanics
2.3 Work, energy and power
Solve problems involving work.
W = Fs cos 
work done by a constant force
not parallel to displacement
Where  is the angle between F and s.
EXAMPLE: Find the work done by the 25-newton
force F in displacing the box s = 15 meters if
the force and displacement are (a) parallel, (b)
antiparallel and (c) at a 30° angle.
SOLUTION:
(a) W = Fs cos  = (25)(15) cos 0° = 380 J.
(b) W = (25)(15) cos 180° = -380 J.
(c) W = (25)(15) cos 30° = 320 J.
FYI
Work can be negative.
The F and the s are the magnitudes of F and s.
Topic 2: Mechanics
2.3 Work, energy and power
Solve problems involving work.
EXAMPLE: Find the work done by the brakes in
bringing a 730-kg Smart Car to a rest in 80.
meters if its starting speed is 32 m/s.
f
s
SOLUTION:
Since f and s are antiparallel then  = 180°.
It is given that s = 80 m.
From v2 = u2 + 2as we get
02 = 322 + 2a(80) so that a = -6.4 m s-2.
Then f = ma = 730(-6.4) = -4672 n. |f| = +4672 n.
Finally, W = Fs cos 
= (4672)(80) cos 180° = -370000 J.
FBD Crate
T
Topic 2: Mechanics
2.3 Work, energy and power
a=0
Solve problems involving work.
EXAMPLE: A pulley system is used to raise a
100-n crate 4 m as shown. Find the work
done by the tension force T if the lift
occurs at constant speed.
T
SOLUTION:
From the FBD since a = 0, T = 100 n.
From the statement of the problem, s = 4 m.
Since the displacement and the tension are
parallel,  = 0°.
s
Thus W = Ts cos 
= (100)(4) cos 0° = 400 J.
FYI
Pulleys are used to redirect tension forces for
convenience.
100
s
T
T
T
FBD Crate
T T
Topic 2: Mechanics
2.3 Work, energy and power
a=0
100
Solve problems involving work.
EXAMPLE: A pulley system is used to raise a
100-n crate 4 m as shown. Find the work
done by the tension force T if the lift
T
T
occurs at constant speed.
SOLUTION:
From the FBD 2T = 100 so that T = 50 n.
From the statement of the problem, s = 4 m.
2s
T
Since the displacement and the tension are
T
parallel,  = 0°.
Thus W = T(2s) cos 
s
T
= (50)(24) cos 0° = 400 J.
FYI
Pulleys are also used gain mechanical advantage.
M.A. = Fout/Fin = 100 n / 50 n = 2.
Topic 2: Mechanics
2.3 Work, energy and power
F F
x
0
Determine the work done by a non-constant force
by interpreting a force-displacement graph.
Consider a spring mounted to a wall as shown.
If we pull the spring to the right, it resists in
direct proportion to the distance it is
stretched.
If we push to the left, it does the same thing.
It turns out that the spring force F is given by
F = -ks
Hooke’s Law (the spring force)
The minus sign gives the force the correct
direction, namely, opposite the displacement s.
Since F is in (n) and s is in (m), the units for
the spring constant k are (n m-1).
Topic 2: Mechanics
2.3 Work, energy and power
Determine the work done by a non-constant force
by interpreting a force-displacement graph.
F = -ks
Hooke’s Law (the spring force)
EXAMPLE: A force vs. displacement plot for a
spring is shown. Find the value of the
spring constant, and find the
F (n)
20
spring force if the displacement
is -65 mm.
s (mm)
SOLUTION:
0
Pick a convenient point on
the plot.
-20
For the point, F = -15 n and
-40
-20
20
0
40
s = 30 mm = .030 m so that
F = -ks or -15 = -k(.030) or k = 500 n m-1.
F = -ks = -(500)(-6510-3) = +32.5 n.
Topic 2: Mechanics
2.3 Work, energy and power
Determine the work done by a non-constant force
by interpreting a force-displacement graph.
F = -ks
Hooke’s Law (the spring force)
EXAMPLE: A force vs. displacement plot for a
spring is shown. Find the work done by you if you
displace the spring from 0 to 40 mm.
F (n)
20
SOLUTION:
The spring itself always does negs (mm)
ative work because the force and 0
the displacement are antiparallel.
The force you apply will always
-20
be parallel. Thus you will do
20
0
40
positive work according to F = ks. 40 20
F = ks is plotted in red.
Topic 2: Mechanics
2.3 Work, energy and power
Determine the work done by a non-constant force
by interpreting a force-displacement graph.
F = -ks
Hooke’s Law (the spring force)
EXAMPLE: A force vs. displacement plot for a
spring is shown. Find the work done by you if you
displace the spring from 0 to 40 mm.
F (n)
20
SOLUTION:
The area under the F vs. s graph
s (mm)
represents the work done by that 0
force.
The area desired is from 0 to
-20
40 mm, shown here:
-40
-20
20
0
40
A = (1/2)bh
= (1/2)(4010-3 m)(20 n)
= 0.4 J
Topic 2: Mechanics
2.3 Work, energy and power
2.3.4 Outline what is meant by kinetic energy.
2.3.5 Outline what is meant by change in
gravitational potential energy.
2.3.6 State the principle of conservation of
energy.
2.3.7 List different forms of energy and describe
examples of the transformation of energy from
one form to another.
Topic 2: Mechanics
2.3 Work, energy and power
Outline what is meant by kinetic energy.
Kinetic energy EK is the energy of motion.
The bigger the speed v, the bigger EK.
The bigger the mass m, the bigger EK.
The formula for EK, justified later, is
EK = (1/2)mv2
kinetic energy
Looking at the units for EK we have
kg(m/s)2 = kg m2 s-2
= (kg m s-2)m
In the parentheses we have a mass times an
acceleration which is a newton.
Thus EK is measured in (n m), which are (J), the
same unit we used for work.
As a final note, many books use the symbol K
instead of EK for kinetic energy.
Topic 2: Mechanics
2.3 Work, energy and power
Outline what is meant by kinetic energy.
EK = (1/2)mv2
kinetic energy
EXAMPLE: What is the kinetic
energy of a 4.0-gram NATO SS
109 bullet traveling at 950
m/s?
SOLUTION:
Convert grams to kg (jump 3
decimal places left) to get
m = .004 kg.
Then K = (1/2)mv2
= (1/2)(.004)(950)2
= 1800 J.
Topic 2: Mechanics
2.3 Work, energy and power
Outline what is meant by kinetic energy.
EK = (1/2)mv2
kinetic energy
EXAMPLE: What is the kinetic
energy of a 220-pound NATO
soldier running at 6 m/s?
SOLUTION:
First convert pounds to kg:
(220 lb)(1 kg/2.2 lb) = 100
kg.
Then K = (1/2)mv2
= (1/2)(100)(6)2
= 1800 J.
FYI
Small and large objects can have the same EK!
Topic 2: Mechanics
2.3 Work, energy and power
Outline what is meant by kinetic energy.
W = Fs
work done by a constant force
EK = (1/2)mv2
kinetic energy
It is no coincidence that work and kinetic energy
have the same units. Observe the following
derivation.
v2 = u2 + 2as
mv2 = m(u2 + 2as)
mv2 = mu2 + 2mas
mv2 = mu2 + 2Fs
(1/2)mv2 = (1/2)mu2 + Fs
EK,f = EK,0 + W
EK,f - EK,0 = W
∆EK = W
This is called the Work-Kinetic Energy theorem.
Topic 2: Mechanics
2.3 Work, energy and power
Outline what is meant by kinetic energy.
∆EK = W
work-kinetic energy theorem
This formula says that the change in kinetic
energy of an object equals the work done on it.
EXAMPLE: Use energy to find the work done by the
brakes in bringing a 730-kg Smart Car to a rest
in 80. meters if its starting speed is 32 m/s.
f
s
SOLUTION:
EK,f = (1/2)mv2 = (1/2)(730)(02) = 0 J.
EK,0 = (1/2)mu2 = (1/2)(730)(322) = 370000 J.
∆EK = EK,f - EK,0 = 0 – 370000 = -370000 J
W = ∆EK = -370000 J.
Topic 2: Mechanics
2.3 Work, energy and power
Outline what is meant by kinetic energy.
EK = (1/2)mv2
EK = p2/(2m)
kinetic energy
One of the many derivations the IBO requires you
to know expresses kinetic energy in terms of
momentum instead of velocity.
p = mv
p2 = m2v2
p2/m = mv2
(1/2)p2/m = (1/2)mv2
EK = p2/(2m)
FBD Ball
F
Topic 2: Mechanics
2.3 Work, energy and power
Outline what is meant by change in gravitational
potential energy.
Consider a bowling ball resting on the floor.
If we let go of it, it just stays put.
If on the other hand we raise it to a height
∆h and then let it go, it will fall and speed
up, gaining kinetic energy as it falls.
Since the lift constitutes work against
gravity (the weight of the ball) we have
W = Fs cos 
W = mg∆h cos 0° = mg∆h.
We call this energy due to the position of a
weight gravitational potential energy.
∆EP = mg∆h
change in gravitational potential energy
a=0
mg
Topic 2: Mechanics
2.3 Work, energy and power
Outline what is meant by change in gravitational
potential energy.
∆EP = mg∆h
change in gravitational potential energy
EXAMPLE: Consider a crane which lifts
a 2000-kg weight 18 m above its
original resting place. What is the
change in gravitational potential
energy of the weight and how much work
did the crane do on the weight?
SOLUTION:
The change in gravitational potential
energy is just
∆EP = mg∆h = 2000(10)(18) = 360000 J.
The crane must have done 360000 J of
work if the acceleration was zero.
Topic 2: Mechanics
2.3 Work, energy and power
State the principle of conservation of energy.
EXAMPLE: Consider a crane which lifts
a 2000-kg weight 18 m above its
original resting place. If the cable
breaks at the top find the speed and
kinetic energy of the mass at the
instant it reaches its starting point.
SOLUTION:
v2 = u2 + 2as
v2 = 02 + 2(-10)(-18) = 360
v = 19 m s-1.
EK = (1/2)mv2
= (1/2)(2000)(192) = 360000 J.
Topic 2: Mechanics
2.3 Work, energy and power
State the principle of conservation of energy.
EXAMPLE: Consider a crane which lifts
a 2000-kg weight 18 m above its
original resting place. If the cable
breaks at the top find its change in
kinetic energy and change in potential
energy the instant it reaches the
ground.
SOLUTION:
EK,f = (1/2)mv2
= (1/2)(2000)(192) = 360000 J.
EK,0 = (1/2)(2000)(02) = 0 J.
∆EK = 360000 – 0 = 360000 J.
∆EP = mg∆h
= (2000)(10)(-18) = -360000 J.
Topic 2: Mechanics
2.3 Work, energy and power
State the principle of conservation of energy.
EXAMPLE: Consider a crane which lifts
a 2000-kg weight 18 m above its
original resting place. If the cable
breaks at the top find the sum of the
change in kinetic and the change in
potential energies the instant it
reaches the ground.
SOLUTION:
From the previous slide
∆EK = 360000 J and
∆EP = -360000 J so
∆EK + ∆EP = 360000 + -360000
∆EK + ∆EP = 0 J.
Topic 2: Mechanics
2.3 Work, energy and power
State the principle of conservation of energy.
We see that ∆EK + ∆EP = 0 from the previous
example.
In general, if there is no friction or drag to
remove energy from a system
∆EK + ∆EP = 0
conservation of energy
In the absence of friction and drag
The above formula is known as the statement of
the conservation of mechanical energy.
Essentially, what it means is that if the kinetic
energy changes (say it decreases), then the
potential energy changes (it will increase) in
such a way that the total change is zero!
Another way to put it is “the total energy of a
system never changes.”
Topic 2: Mechanics
2.3 Work, energy and power
State the principle of conservation of energy.
∆EK + ∆EP = 0
conservation of energy
In the absence of friction and drag
EXAMPLE: Find the speed of the 2-kg ball when it
reaches the bottom of the 20-m tall track.
∆h
SOLUTION:
Use conservation of energy to find EK,f
∆EK + ∆EP =
(1/2)mv2 - (1/2)mu2 + mg∆h =
(1/2)(2)v2 - (1/2)(2)02 + 2(10)(-20) =
v2 =
v =
and v.
0
0
0
400
20 m s-1.
Topic 2: Mechanics
2.3 Work, energy and power
List different forms of energy and describe
examples of the transformation of energy from one
form to another.
We have talked about kinetic energy (motion).
We have talked about potential energy (position).
We have chemical energy and nuclear energy.
We have electrical energy and magnetic energy.
We have sound energy and light energy.
And we also have heat energy.
In mechanics we only have to worry about the
highlighted energy forms.
And we only worry about heat if there is friction
or drag.
Topic 2: Mechanics
2.3 Work, energy and power
State the principle of conservation of energy.
∆EK + ∆EP = 0
conservation of energy
In the absence of friction and drag
EXAMPLE: Suppose the speed of the 2-kg ball is 15
m s-1 when it reaches the bottom of the 20-m tall
track. What is the loss of mechanical energy and
where did it go? The system lost 175 J as drag
and friction heat.
∆h
SOLUTION:

∆EK + ∆EP
(1/2)mv2 - (1/2)mu2 + mg∆h
(1/2)(2)152 - (1/2)(2)02 + 2(10)(-20)
-175 J
=
=
=
=
loss/gain
loss/gain
loss/gain
loss/gain
Topic 2: Mechanics
2.3 Work, energy and power
2.3.8 Distinguish between elastic and inelastic
collisions.
Topic 2: Mechanics
2.3 Work, energy and power
Distinguish between elastic and inelastic
collisions.
In an elastic collision, kinetic energy is
conserved (it does not change). Thus EK,f = EK,0.
In an inelastic collision, kinetic energy is not
conserved (it does change). Thus EK,f ≠ EK,0.
In a completely inelastic collision the colliding
bodies stick together.
Topic 2: Mechanics
2.3 Work, energy and power
Distinguish between elastic and inelastic
collisions.
EXAMPLE: Two train cars having equal masses of
750 kg and velocities u1 = 10. m s-1 and
u2 = 5.0 m s-1 collide and hitch together. What is
their final speed?
u1
v
u2
v
SOLUTION:
Use conservation of linear momentum.
If Fnet = 0 then p = CONST, so that
p1,0 + p2,0 = p1,f + p2,f
mu1 + mu2 = mv + mv
m(u1 + u2) = 2mv
10 + 5 = 2v or v = 7.5 m s-1.
Topic 2: Mechanics
2.3 Work, energy and power
Distinguish between elastic and inelastic
collisions.
EXAMPLE: Two train cars having equal masses of
750 kg and velocities u1 = 10. m s-1 and u2 = 5.0
m s-1 collide and hitch together. Identify the
collision type/s that occur/s.
u1
v
u2
v
SOLUTION:
Since the cars stick together we have a
completely inelastic collision.
From the previous slide we have v = 7.5 so that
∆EK = (1/2)(2m)v2 – [ (1/2)mu12 + (1/2)mu22 ]
= (1/2)(2750)7.52 –(1/2)750102 -(1/2)75052
= -4700 J. Inelastic collision since ∆EK ≠ 0.
Topic 2: Mechanics
2.3 Work, energy and power
2.3.9 Define power.
2.3.10 Define and apply the concept of
efficiency.
Topic 2: Mechanics
2.3 Work, energy and power
Define power.
Power is the rate of energy usage and so has the
equation
P = E/t
power
From the formula we see that power has the units
of energy (J) per time (s) or (J s-1) which are
known as watts (W).
EXAMPLE: How much energy does a 100.-W bulb
consume in one day?
SOLUTION:
From P = E/t we get E = Pt so that
E = (100 J/s)(24 h)(3600 s/h)
E = 8640000 J!
Don’t leave lights on in unoccupied rooms.
Topic 2: Mechanics
2.3 Work, energy and power
Define and apply the concept of efficiency.
Efficiency is the ratio of output power to input
power
Efficiency = Pout/Pin
efficiency
EXAMPLE: Conversion of coal into electricity is
through the following process. Coal burns to heat
up water to steam. Steam turns a turbine. The
turbine turns a generator which produces
electricity. Suppose the useable electricity from
such a power plant is 125 MW, while the chemical
energy of the coal is 690 MW. Find the efficiency
of the plant.
SOLUTION:

Efficiency = Pout/Pin
= 125 MW / 690 MW
= 0.18 or 18%.
Topic 2: Mechanics
2.3 Work, energy and power
2.3.11 Solve problems involving momentum, work,
energy and power.
Topic 2: Mechanics
2.3 Work, energy and power
Solve problems involving momentum, work, energy
and power.
6.0 m
u = 0
EXAMPLE: A 25-kg object resting
on a frictionless incline is
released, as shown. What is its
speed at the bottom?
∆h
v = ?
30°
SOLUTION:
We solved this one long ago using Newton’s
second law. It was difficult!
We will now use energy to solve it.

∆EK + ∆EP = 0
(1/2)mv2 - (1/2)mu2 + mg∆h = 0
(1/2)(25)v2 - (1/2)(25)02 + (25)(10)(-6) = 0
12.5v2 = 1500
v = 11 m s-1.
Topic 2: Mechanics
2.3 Work, energy and power
Solve problems involving momentum, work, energy
and power.
PRACTICE: A 3.0-kg ball thrown at a wall at
9.0 m s-1 (+x-dir). It rebounds at 8.0 m s-1
(–x-dir). Find its change in momentum and change
in kinetic energy. Is this an elastic collision?
p0 = mu = (3)(+9) = 27 kg m s-1.
pf = mv = (3)(-8) = -24 kg m s-1.
Thus ∆p = pf – p0 = -24 – 27 = -51 kg m s-1.
EK,0 = (1/2)mu2 = (1/2)(3)92 = 121.5 J.
EK,f = (1/2)mv2 = (1/2)(3)82 = 96 J.
Thus ∆EK = EK,f – EK,0 = 96 – 121.5 = -26 J.
This is not an elastic collision since ∆EK ≠ 0.
Topic 2: Mechanics
2.3 Work, energy and power
Solve problems involving momentum, work, energy
and power.
EXAMPLE: Suppose a .020-kg bullet traveling
horizontally at 300. m/s strikes a 4.0-kg block
of wood resting on a wood floor. How fast is the
block/bullet combo moving immediately after
collision?
SOLUTION:
If we consider the bullet-block combo as our
system, there are no external forces in the xdirection at collision. Thus pf = p0 so that
 mvf + MVf = mvi + MVi
the bullet and
.02v + 4v = (.02)(300) + 4(0) the block move at
4.02v = 6
the same speed
v = 1.5 m/s
after collision
f
Topic 2: Mechanics
2.3 Work, energy and power
s
Solve problems involving momentum, work, energy
and power.
EXAMPLE: Suppose a .020-kg bullet traveling
horizontally at 300. m/s strikes a 4.0-kg block
of wood resting on a wood floor. The block/bullet
combo slides 6 m before coming to a stop. Find
the friction between the block and the floor.
SOLUTION:
Here we use the work-kinetic energy theorem:

∆EK = W
(1/2)mv2 - (1/2)mu2 = fs cos 
(1/2)(4.02)(0)2 - (1/2)(4.02)(1.5)2 = f(6) cos 180°
-4.5225 = -6f
f = -4.5225/-6
f = 0.75 N.
F s
Topic 2: Mechanics
2.3 Work, energy and power
Solve problems involving momentum, work, energy
and power.
EXAMPLE: Suppose a .020-kg bullet traveling
horizontally at 300. m/s strikes a 4.0-kg block
of wood resting on a wood floor. If the bullet
penetrates .06 m of the block, find the average
force F acting on it during its collision.
SOLUTION:
We again use the work-kinetic energy theorem on
only the bullet:

∆EK = W
(1/2)mv2 - (1/2)mu2 = Fs cos 
(1/2)(.02)(1.5)2 - (1/2)(.02)(300)2 = -F(.06)
-900 = - .06F
F = 15000 n.