Transcript Lecture3_Work

A rocket test projectile is launched
skyward at 88 m/sec (198 mi/hr).
How high does it go?
vo = 88
? m/sec
vfinal = ?
a = 9.8
? m/sec2
time to reach peak, t = ?
height reached,
x=?
When does it peak? What has happened by that point?
v = 0  88
m
sec
 ( 9.8
m
 88 sec
 ( 9.8
m
 88 sec
m
 9.8
sec
m
sec
2
m
sec
2
)t
 t = 9 sec
2
2
m 1
1
So x  x  (v88t 
at
)(
9
sec
)

(9.8
2
sec 2
0
)t
0
= 792m – 396.9m
= 395 m (1/4 mile)
m
sec
2
)(
81
sec
)
2
A ball is thrown straight upward and
caught when it returns to the height
from which it was released.
1. At its peak position, the ball’s
A. instantaneous velocity is maximum.
B. instantaneous velocity is zero.
C. instantaneous acceleration is zero.
D. both B & C are true.
2. The time to fall back from its peak position
is
A. greater than B. equal to C. less than
the time it took to climb that high.
3. The speed it builds up to downward
by the moment it is caught is
A. greater than B. equal to C. less than
the speed it was thrown upward with.
For objects for which air friction
is negligible,
time up = time down
speed down = speed up
Two spheres of identical mass are
Released when the mechanism above
Is triggered. One sphere is launched
Horizontally by a spring, the other is
simultaneously dropped from rest.
A. The launched sphere will reach
the ground first.
B. Both spheres touch ground at
the same time.
C. The dropped sphere reaches the
ground first.
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Galileo’s critics argued that if the earth moved:
A stone is flung horizontally at 8 m/sec
from a point 1100 meters above the base
of a cliff.
How far away from the cliff does it land?
y  1100m
or +1100m
2
+9.80m/s
a  g  9.80m / s
Careful!!!
t ?
Actually ay = 9.8m/s2
x  ?
ax = ?0
v0  8m / s
x
v0  0
y
Its y-component of motion
is like dropping from rest!
VERTICALLY
 1100m  0 
m 2
1
( 9.80 2 )t
2
s
t = 15 sec
HORIZONTALLY
x  (8m / s)(15s) x = 120 m
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Weight
Support (floor)
Adding all these supporting forces together
some pull left,
some pull right,
some pull forward,
some pull back
(a tug-of-war
balancing)
all tend to
pull UP!
Styrofoam bridge
weighted at center
Pressure applied to rigid glass bar
on
IfWhat
this forces
trunk isact
moved
this storage
trunk?
across
the floor
at
constant velocity,
Push
Friction
Support of floor
(distributed among
the 4 casters it sits on)
Weight
1. the man’s PUSH must exceed the trunk’s
WEIGHT.
2. the man’s PUSH must exceed the total
friction forces.
3. all the above forces must exactly balance.
Forces all balance for constant velocity.
Imagine the work involved in sliding
this crate from the loading dock to
the center of the machine shop floor.
Compare that to the task of
pushing against the far wall
(TWICE AS FAR)…
TWICE
AS MUCH
…or to the task of pushing WORK?
two identical crates together
to the center of the room
(TWICE AS MUCH).
Consider the work
involved in lifting
this heavy box to the
bottom storage shelf.
2h
compare this to
h
doing it twice (once for each box)
doing it once with BOTH boxes together
TWICE
lifting a single box to the
AS MUCH
upper shelf (TWICE AS FAR) WORK?
HALF
What about: half as high?
AS MUCH
half the weight? WORK?
Work  Force used in performing it.
Work  distance over which work is done.
Lifting weights is definitely work
even by this physics definition!
What about lowering weights back down?
How much work
is being done
balancing this
tray in place?
Work was done
picking this
briefcase up
from the floor.
How much work
is being done in
holding it still?
How much work
does the cable do
in supporting
the bowling ball?
T
How much work does a crane do in
holding its load in
place above ground?
The crane lifts
its load up at
constant speed.
1.It lifts with a force > the load’s weight.
2.It lifts with a force = the load’s weight.
3.It lifts with a force < the load’s weight.
When holding this load steady in place
how much energy must its motor consume?
B
C
A
D
Work is done on the box during which stage(s)?
A. Lifting box up from floor.
B. Holding box above floor.
C. Carrying box forward across floor.
D. Setting box down gently to floor.
E. At all stages: A,B,C,D.
F. At stages A and C.
G. At stages A and D
H. At none of the stages illustrated.
Normal Force
Weight
Weight
Normal Force
Normal
Force
Weight
The unbalanced (net) force is parallel
to the inclined surface of the road!
Exactly the direction we know a car
left in neutral will start to coast!
When work is
done by a completely
unbalanced force
it can set objects in
motion…or increase
their speed.
You push and it
simply coasts away
beyond your reach.
F
d
F
d
F
d
When an object
moves in the
direction of an
unbalanced push
it accelerates
over the entire
distance d.
W=Fd
produces the
acceleration
a = F/m
accelerating
through the
distance d
in a time t:
d = ½ at2
Work is done on a wagon, initially at
rest, by pushing it with an unbalanced
force F over a distance D. With what
final speed does it coast away?
v  v0  at
Final speed?
With what acceleration
a does it build speed?
a  F /m
But for how long does it accelerate?
d  at
1
2
2
2d  at
2
2d / a  t
t  2d / a
2d
So
va

finally
a
2
2ad
But there’s another way to think about this
W=Fd
F = ma
d
1 2
= 2at
So the amount of work done is also equal to:
W = ma  12at2
W = 12 ma2t2
Then, just recall that the
acceleration builds up,
over t, to a final velocity:
v  v0  at
0, if starting
from rest!
W=
1
2
mv
2
We consider this an expression of
the energy possessed by a mass m
when moving with a velocity v
Moving objects carry energy
to scatter pins
to knock down walls
push snow aside
In fact, energy is defined as
the ability to perform work!
Raised weights also
possess energy
…potential energy…
the potential to fall and
build kinetic energy.
A piledriver uses a raised weight
to drive piles into the ground.
Raising a mass against gravity is doing work,
even if there is no resulting final velocity.
Work done in lifting a mass to a height h:
W=Fd
average force
lifted with
mg
W = mgh
In falling through a distance h an mass
builds speed to what final kinetic energy?
d  at
1
2
h  gt
2
1
2
2
t  2h / g
And in that time reaches
v  at  g 2h / g  2 gh
So:
KE  mv  m(2 gh)
1
2
2
1
2
KE = mgh
Normal Force
Normal
Force
Weight
The unbalanced (net) force is parallel
to the inclined surface of the road!
Exactly the direction we know a car
left in neutral will start to coast!
h
d
N
W
F
x
Pushing this box up along the ramp is
1. more work
2. less work
3. exactly the same work
as simply lifting it straight up h.
W
F
=
d
?
1. h
2. N
3. x
W
F
=
d
h
h
d
N
W
F
x
Work in simply lifting the box h
Force in lifting  distance lifted
Wh
Work pushing box along the ramp is
Fd
Wh  Fd
F
d
A man, his briefcase in hand as shown,
walks down the corridor from elevator
to office door. He does
1. positive work on the briefcase.
2. no work on the briefcase.
3. negative work on the briefcase.
The force F does is not along (nor back against) the
direction of motion. Instead F and d are perpendicular!
The force of support neither raises nor lowers the briefcase
(neither increasing or decreasing its potential energy). It
Neither speeds up nor slows down the briefcase (neither
increasing nor decreasing the briefcases kinetic energy).
W
F
=
d
h
h
d
N
W
assume
frictionless
rollers on
ramp
F
x
The box may slip down from its position
on the shelf by sliding back down the
ramp or falling straight down to the floor.
By the time it reaches the floor it will reach
1. a higher final speed by falling.
2. a higher final speed by sliding.
3. the same final speed by either route.
W
F
=
d
h
h
d
N
W
F
Falling: h  gt
1
2
2
t  2h / g
v  v0  gt  0  g 2h / g  2 gh
h
h
Sliding: a  F / m
F  W  mg
d
d
h mg hg
a

d m
d
1  hg  2
2
d   t
t  2d / hg
2 d 
2
hg  2d

then v  
 2hg

 d  hg
The merry-go-round is easier
to push if all the riders
1. hang on at the outer edges.
2. huddle together at the center.
3. It makes little difference
where they ride since its
total mass that matters.