Transcript Lecture5_PotentialEnergy

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 a 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
A horizontal
surface (if strong
enough to hold up!)
can give full support
to an object’s weight
Normal Force
Weight
If tipped completely
of course the ramp
gives NO SUPPORT!
Weight
What about any of the
positions in between?
Normal Force
We can “add”
these forces
in a diagram
Normal
by showing
Force Weight
the arrows
that represent
them, flowing
from head
into tail from
one to the other
(in either order
in fact!)
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.
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
Some Answers
Question 1
3. exactly the same work
This won’t be obvious
until we work through
the next several questions!
But already this much should be clear: lifting straight up requires a large force.
Pushing along the ramp a smaller force, but the push must be applied over a
much larger distance (the full length of the ramp). As you’ll see by the end of
this lecture the proportions in the triangles guarantee the two compensate exactly!
The large force time small distance = the smaller force times longer distance.
The small triangle drawn to display the forces acting
other box, and the large triangle formed by the ramp,
wall and floor are similar. d is the hypotenuse of the
large triangle, W the hypotenuse of the small one.
Question 2
1. h
Question 3
2. no 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).
Question 4
3. the same final speed by either route.
The small triangle drawn to display the forces acting
other box, and the large triangle formed by the ramp,
wall and floor are similar. d is the hypotenuse of the
large triangle, W the hypotenuse of the small one.