Transcript Lecture09-09

Chapter 7
Work and Kinetic Energy
Reading and Review
Vertical circular motion
Centripetal acceleration must be
C
vertical
(down)
B
horizontal
A
vertical
(up)
Condition for falling: N=0
at C:
So, as long as:
at the top, then N>0 and pointing down.
(now apparent weight is in the
opposite direction to true weight!)
Barrel of Fun
A rider in a “barrel of fun”
finds herself stuck with
her back to the wall.
Which diagram correctly
shows the forces acting
on her?
a
b
c
d
e
Barrel of Fun
A rider in a “barrel of fun”
finds herself stuck with
her back to the wall.
Which diagram correctly
shows the forces acting
on her?
a
b
c
d
The normal force of the wall on the
rider provides the centripetal force
needed to keep her going around in a
circle. The downward force of gravity
is balanced by the upward frictional
force on her, so she does not slip
vertically.
Follow-up: What happens if the rotation of the ride slows down?
e
Around the Curve
You are a passenger in a car,
not wearing a seat belt. The car
makes a sharp left turn, and you
find yourself hitting the
passenger door. What is the
correct description of what is
actually happening?
a) centrifugal force is pushing you
into the door
b) the door is exerting a leftward
force on you
c) both of the above
d) neither of the above
Around the Curve
You are a passenger in a car,
not wearing a seat belt. The car
makes a sharp left turn, and you
find yourself hitting the
passenger door. What is the
correct description of what is
actually happening?
a) centrifugal force is pushing you
into the door
b) the door is exerting a leftward
force on you
c) both of the above
d) neither of the above
The passenger has the tendency to
continue moving in a straight line.
There is a net centripetal force,
provided by the door, that forces
the passenger into a circular path.
Working Hard... or Hardly Working
Atlas holds up the world.
Sisyphus pushes his rock
up a hill.
(b)
(a)
Who does more work?
Working Hard... or Hardly Working
Atlas holds up the world.
Sisyphus pushes his rock
up a hill.
(b)
(a)
With no displacement,
Atlas does no work
Work Done by a Constant Force
The definition of work, when the force is parallel to
the displacement:
SI unit: newton-meter (N·m) = joule, J
Friction and Work I
A box is being pulled
across a rough floor
a) friction does no work at all
at a constant speed.
b) friction does negative work
What can you say
c) friction does positive work
about the work done
by friction?
Friction and Work I
A box is being pulled
across a rough floor
a) friction does no work at all
at a constant speed.
b) friction does negative work
What can you say
c) friction does positive work
about the work done
by friction?
Friction acts in the opposite direction to
N Displacement
the displacement, so the work is negative.
Or using the definition of work (W = F
Pull
f
(Δr)cos  ), because  = 180º, then W < 0.
mg
Friction and Work II
Can friction ever do
positive work?
a) yes
b) no
Friction and Work II
Can friction ever do
positive work?
a) yes
b) no
Consider the case of a box on the back of a pickup truck. If
the box moves along with the truck, then it is actually the
force of friction that is making the box move.
Forces not along displacement
If the force is at an angle to the displacement:
Convenient notation: the dot product
The work can also be written as the dot product
of the force and the displacement:
Force and displacement
The work done may be positive, zero, or negative,
depending on the angle between the force and the
displacement:
Sum of work by forces = work by sum of forces
If there is more than one force acting on an object,
we can find the work done by each force, and also
the work done by the net force:
Units of Work
1 kcal = 1 Cal = 4.186 kJ
Lifting 0.5 L H2O up 20 cm = 1 J
Play Ball!
In a baseball game, the
catcher stops a 90-mph
a) catcher has done positive work
pitch. What can you say
b) catcher has done negative work
about the work done by the
c) catcher has done zero work
catcher on the ball?
Play Ball!
In a baseball game, the
catcher stops a 90-mph
a) catcher has done positive work
pitch. What can you say
b) catcher has done negative work
about the work done by the
c) catcher has done zero work
catcher on the ball?
The force exerted by the catcher is opposite in direction to the
displacement of the ball, so the work is negative. Or using the
definition of work (W = F (Δr)cos ), because  = 180º, then W < 0.
Note that the work done on the ball is negative, and its speed
decreases.
Follow-up: What about the work done by the ball on the catcher?
Tension and Work
A ball tied to a string is
being whirled around in
a circle. What can you
say about the work done
by tension?
a) tension does no work at all
b) tension does negative work
c) tension does positive work
Tension and Work
A ball tied to a string is
being whirled around in
a circle. What can you
say about the work done
a) tension does no work at all
b) tension does negative work
c) tension does positive work
by tension?
No work is done because the force
acts in a perpendicular direction to the
displacement.
Or using the
definition of work (W = F (Δr)cos  ),
because  = 90º, then W = 0.
T
v
Follow-up: Is there a force in the direction of the velocity?
Work by gravity
Fg

a
A ball of mass m drops a distance h. What is the
total work done on the ball by gravity?
W = Fd = Fg x h
h
W = mgh
Path doesn’t matter when asking “how
much work did gravity do?” Only the
change in height!
A ball of mass m rolls down a ramp of height h at an
angle of 45o. What is the total work done on the ball by gravity?
a
Fgx = Fg sinθ
N
h = L sinθ
Fg
W = Fd = Fgx x L = (Fg sinθ) (h / sinθ)
W = Fg h = mgh
θ
h
Path independence
• If a force depends on POSITION only then
the work done by it on an object moving from
r1 to r2 will NOT depend upon the path.
• Such a force is called a Conservative Force
Motion and energy
When positive work is done on an object, its speed
increases; when negative work is done, its speed
decreases.
Kinetic Energy
After algebraic manipulations of the equations of
motion, we find:
Therefore, we define the kinetic energy:
Work-Energy Theorem
Work-Energy Theorem: The total work done on an
object is equal to its change in kinetic energy.
(True for rigid bodies that remain intact)
Lifting a Book
You lift a book with your hand in
a) mg  ∆ r
such a way that it moves up at
b) FHAND  ∆ r
constant speed. While it is
c) [FHAND + mg]  ∆ r
moving, what is the total work
d) zero
done on the book?
e) none of the above
∆r
FHAND
mg
v = const
a=0
Lifting a Book
You lift a book with your hand in
a) mg  ∆ r
such a way that it moves up at
b) FHAND  ∆ r
constant speed. While it is
c) (FHAND + mg)  ∆ r
moving, what is the total work
d) zero
done on the book?
e) none of the above
The total work is zero because the net force
acting on the book is zero. The work done by
∆r
FHAND
the hand is positive, and the work done by
gravity is negative. The sum of the two is
v = const
a=0
zero. Note that the kinetic energy of the book
does not change either!
mg
Follow-up: What would happen if FHAND were greater than mg?
Kinetic Energy I
By what factor does the
a) no change at all
kinetic energy of a car
b) factor of 3
change when its speed is
c) factor of 6
tripled?
d) factor of 9
e) factor of 12
Kinetic Energy I
By what factor does the
a) no change at all
kinetic energy of a car
b) factor of 3
change when its speed is
c) factor of 6
tripled?
d) factor of 9
e) factor of 12
Because the kinetic energy is
mv2, if the speed increases
by a factor of 3, then the KE will increase by a factor of 9.
Slowing Down
If a car traveling 60 km/hr can
brake to a stop within 20 m, what
is its stopping distance if it is
traveling 120 km/hr? Assume
that the braking force is the same
in both cases.
a) 20 m
b) 30 m
c) 40 m
d) 60 m
e) 80 m
Slowing Down
If a car traveling 60 km/hr can
brake to a stop within 20 m, what
is its stopping distance if it is
traveling 120 km/hr? Assume
that the braking force is the same
in both cases.
F d = Wnet = ∆KE = 0 –
and thus, |F| d =
mv2,
mv2.
Therefore, if the speed doubles,
the stopping distance gets four
times larger.
a) 20 m
b) 30 m
c) 40 m
d) 60 m
e) 80 m
Application: ball on a track
how high must I place the ball so
that it can complete a loop?
Condition: Fcp > mg at top of loop
Fcp = mv2/r = mg
v2 = gr
KE = mv2 / 2 = mgr/2
Gravity must provide
this energy
Wg = mg∆h = KE
∆h = r/2 above the top of the loop!
Work Done by a Variable Force
If the force is constant, we can interpret the
work done graphically:
Work Done by a Variable Force
If the force takes on several successive constant
values:
Work Done by a Variable Force
We can then approximate a continuously varying
force by a succession of constant values.
Work Done by a Variable Force
The force needed to stretch a spring an amount x is
F = kx.
Therefore, the work
done in stretching the
spring is
Application: work by a spring
Hooke’s Law: F = - kx
k = (3kg)(9.8 m/s2) / (3.9 cm)
k = 760 N/m
Loaded spring: W = kx2/2 = (760 N/m) (0.04m)2/ 2
W = 0.61 J
How fast?: v = d/t = (0.020 m) (0.020 s) = 1 m/s
Kinetic Energy: KE = mv2/2 = (1kg)(1m/s)2 / 2
KE = 0.55 J
Power
Power is a measure of the rate at which work is
done:
Pave
W

t
SI unit: J/s = watt, W
1 horsepower = 1 hp = 746 W
Power
Power
If an object is moving at a constant speed in the
face of friction, gravity, air resistance, and so forth,
the power exerted by the driving force can be
written:
F x
 x 
P
 F

Fv

t
 t 
Question: what is the total work per unit
time done on the object?
Electric Bill
When you pay the electric company
by the kilowatt-hour, what are you
actually paying for?
a) energy
b) power
c) current
d) voltage
e) none of the above
Electric Bill
When you pay the electric company
by the kilowatt-hour, what are you
actually paying for?
a) energy
b) power
c) current
d) voltage
e) none of the above
We have defined: Power = energy / time
So we see that: Energy = power × time
This means that the unit of power × time
(watt-hour) is a unit of energy !!
A block rests on a horizontal frictionless surface. A string is attached
to the block, and is pulled with a force of 45.0 N at an angle above the
horizontal, as shown in the figure. After the block is pulled through a
distance of 1.50 m, its speed is 2.60 m/s, and 50.0 J of work has been
done on it. (a) What is the angle (b) What is the mass of the block?
The pulley system shown is used
to lift a 52 kg crate. Note that one
chain connects the upper pulley
to the ceiling and a second chain
connects the lower pulley to the
crate. Assuming the masses of
the chains, pulleys, and ropes
are negligible, determine
(a) the force F required to lift the
crate with constant speed, and
(b) the tension in two chains
(a) the force F required to lift the crate
with constant speed, and
(b) the tension in two chains
(a) constant velocity, a=0, so net
force =0.
2T - (52kg)(9.8m/s2) = 0
T = 250 N
F = -250 Ny
(b) upper pulley doesn’t move:
Tch - 2Trope = 0
Tch = 500 N
Mechanical
Advantage!
lower pulley has constant
acceleration
Tch -2Trope =0 Tch = 500 N
What about work?
(a) how much power is applied to the
box by the chain?
(b) how much power is applied on the
rope by the applied force?
Trope = 250 N
Tchain = 500 N
F = -250 Ny
(a) P = Fv = 500 N * vbox
(b) P = Fv = 250 N * vhand
 hand moves twice as fast
 hand moves twice as far