Transcript Work Presentation

Work, Power,
and Energy
Work
• In order for an object to move, a force must
be exerted
• Amount of work done depends on the
amount of force exerted and how far the
object moves
Work (cont.)
Work: the product of the force exerted
and the distance the object moves
W = Fd
ex. A force of 50 newtons is used to drag a crate across
the floor at constant velocity. How much work is done
to drag the crate 8.0 m?
W = Fd
= ( 50 N )( 8.0 m )
= 400 N m
= 400 joules =
400 J
Def’n: 1 joule = 1 N m
James Joule
(1818 – 1889)
ex: (a) How much work is done to lift a bowling ball
of mass 7.0 kg to a height of 1.5 m?
F
W = Fd
To lift an object, must
exert force equal to
the weight
F = Wt. = m g
= ( 7.0 kg )( 9.8 m/s2 )
Wt.
= 68.6 N
W = F d = ( 68.6 N )( 1.5 m ) = W = 103 J
(b) How much work is done to hold it up?
Zero, since there is no displacement
(c) How much work is done to carry it horizontally
a distance of 3.0 m?
Zero
*
The displacement must be in the direction of the
exerted force in order for work to be done
Exerted force is vertical (against gravity); displacement
is horizontal (no vertical displacement)
so Work done = 0
ex: A sled is pulled along the ground by a rope that makes a 25o angle
with the horizontal. The tension in the rope is 20 N. Find the work
necessary to drag the sled 52 m.
20 N
Displacement is horizontal; find
component of force in the
horizontal direction
25o
52 m
ex: A sled is pulled along the ground by a rope that makes a 25o angle
with the horizontal. The tension in the rope is 20 N. Find the work
necessary to drag the sled 52 m.
20 N
Displacement is horizontal; find
component of force in the
horizontal direction
25o
Fh
52 m
W = Fd
Fh
cos 25 =
20
Fh = 20 cos 25
Fh = 18.1 N
= Fh d
= ( 18.1 N )( 52 m )
W = 943 J
ex. A refrigerator weighs 2000 N. How much work would be
done to lift it onto a truck bed 1.5 m above the ground?
W = F d = ( 2000 N )( 1.5 m )
= 3000 J
Problem:
I cannot lift 2000 N
Need to use a machine
I can use a ramp to lift the 2000-N refrigerator onto the truck. The
ramp is 6 m long, and I push it up.
F = 500 N
1.5 m
6m
This is supposed to
be a truck?
I find that I only have exert a force of 500 N to push the
refrigerator up the ramp.
But I have to exert that force over a greater distance.
F = 500 N
1.5 m
6m
This is supposed to
be a truck?
The amount of work done to push the refrigerator
up the ramp is
W = F d = ( 500 N )( 6 m ) = 3000 J
This is the same amount of work as to lift the refrigerator
straight up.
So the ramp did not “save us work”; the same amount was
done in either case
ex. A barbell weighs 800 N. Tom lifts it to a height of
2.0 m and holds it there.
(a) How much work is done to lift it?
W = Fd
= ( 800 N )( 2.0 m )
= 1600 J
(b) How much work is done to hold it up?
Zero; if the barbell does not move,
no work is done
Tom lifts it to
that height in 2.0 s.
Which guy is stronger?
Tom exhibits more power
His friend Pete lifts
it also, but takes
3.0 s to do it.
Power: the time rate of doing work
P =
W
t
• If you do the same amount of work in a
shorter time, you exhibit more power
ex. Sue lifts a 300-N weight to a height of 2.0 m in 3.0 s.
(a) How much work did she do?
W = F d = ( 300 N )( 2.0 m )
W = 600 J
(b) How much power did she exert?
P =
W
t
=
600 J
3.0 s
= 200 J/s
= 200 watts = 200 W
Def’n:
1 J/s = 1 watt
James Watt
( 1736-1819 )
Energy : the ability to do work
Examples:
Potential Energy: energy of position
Forms of PE:
- gravitational
- elastic ( bowstring,
rubber band )
- chemical ( TNT, food )
- nuclear ( stars )
Work-Energy Theorem: the energy acquired by a body
is equal to the work done on it
Apply to Gravitational Potential Energy:
PEg = Work done to lift object
= Fd
Force to lift = weight
= Wt. d
= mgd
PEg = m g h
Gravitational PE of object
lifted to height h
•Work can also be done to cause an object to obtain motion
Kinetic Energy: energy of motion
Depends on:
-speed
-mass
(pitched baseball)
(bowling ball)
Apply Work-Energy Theorem to Kinetic Energy:
KE = Work done on object
= Fd
= ma d
= m(ad)
vf2 = vi2 + 2 a d
start from rest: vi = 0
KE = m ( ½ vf2 )
vf2 = 2 a d
2
2
½ vf2 = a d
KE = ½ m v2
Which object has
potential energy but no
kinetic energy?
Which object has
kinetic energy but no
potential energy?
Which object has
both potential energy and
kinetic energy?
Conservation of Energy: The total amount of energy in an isolated
system remains constant.
Total Energy = Potential Energy + Kinetic Energy
Tot. En. = PE + KE
Rube Goldberg Machine Project
• A Rube Goldberg machine is a series of simple
machines connected to perform a task
The Self-Operating Napkin: As
you raise spoon of soup (A) to
your mouth it pulls string (B),
thereby jerking ladle (C) which
throws cracker (D) past parrot
(E). Parrot jumps after cracker
and perch (F) tilts, upsetting
seeds (G) into pail (H). Extra
weight in pail pulls cord (I),
which opens and lights
automatic cigar lighter (J),
setting off sky-rocket (K) which
causes sickle (L) to cut string
(M) and allow pendulum with
attached napkin to swing back
and forth thereby wiping off
your chin.
•Task: Flip a light switch upward to turn on a bulb
•Must fit on a base 1 m x 1 m
•Must consist of at least five distinct steps
•Must contain at least three different simple machines;
more gets extra credit
•All parts must remain on base
•No live animals
•No explosions, although fire may be allowed
•Must be able to reset and run again in 5 minutes
•Nothing may be plugged in, although battery-powered
objects are ok
Rubric for Rube Goldberg Project
Success in the Objective
At Least Five Steps
Machine Made in Class
On Time/Taken Home
Elegance of Design
Total: __________
4 6 8 10 12 20
1
2
3
4
5
1
2
3
4
5
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
x
1.5
=
__________
Points from individual evaluations (25 max.) = ______
Total = _________