Transcript Review - Mr MAC's Physics

Work and
Energy
An Introduction
Work
Work tells us how much a force or combination
of forces changes the energy of a system.
 Work is the product of the force or its
component in the direction of motion and its
displacement.
 W = Fdcos 
 F: force (N)
 d : displacement (m)
 : angle between force and displacement

Units of Work
SI

System: Joule (N m)
1 Joule of work is done when 1 N acts on a body moving it a
distance of 1 meter
 British

(not used in Physics B)
 cgs

System: foot-pound
System: erg (dyne-cm)
(not used in Physics B)
 Atomic
Level: electron-Volt (eV)
Force and direction of motion
both matter in defining work!
 There is no work done by a force if
it causes no displacement.
 Forces
can do positive, negative, or zero
work. When an box is pushed on a flat
floor, for example…



The normal force and gravity do no work, since they are
perpendicular to the direction of motion.
The person pushing the box does positive work, since she is
pushing in the direction of motion.
Friction does negative work, since it points opposite the
direction of motion.
Question

If a man holds a 50 kg box at arms length for 2
hours as he stands still, how much work does he
do on the box?
Question

If a man holds a 50 kg box at arms length for 2
hours as he walks 1 km forward, how much work
does he do on the box?
Question





If a man lifts a 50 kg box 2.0 meters, how much
work does he do on the box?
W =fdcosQ, Q = 0 force & displacement in the
same direction
W = (mg)dcosQ
W=
W=
Work and Energy
 Work
 If
changes mechanical energy!
an applied force does positive work on a
system, it tries to increase mechanical
energy.
 If an applied force does negative work, it
tries to decrease mechanical energy.
 The two forms of mechanical energy are
called potential and kinetic energy.
Sample problem
Jane uses a vine wrapped around a pulley to lift a 70-kg Tarzan to
a tree house 9.0 meters above the ground.
a)How much work does Jane do when she lifts Tarzan?
b)
m = 70 kg, d = 9 m, g = 9.8 m/s2, Q = 0
W=?
W = FdcosQ = (mg)dcosQ
a)
W=
W=
b)How much work does gravity do when Jane lifts Tarzan?
Sample problem
Joe pushes a 10-kg box and slides it across the floor at constant velocity of 3.0 m/s. The
coefficient of kinetic friction between the box and floor is 0.50.
a) How much work does Joe do if he pushes the box for 15 meters?
m = 10 kg, m = 0.5, d = 15 m, Q = 0, v = 15 m/s (constant)
W=?
W = FappdcosQ , since constant velocity Fapp = Fk
W = (mmg)dcosQ
W=
b) How much work does friction do as Joe pushes the box?
Sample problem
A father pulls his child in a little red wagon with constant speed. If
the father pulls with a force of 16 N for 10.0 m, and the handle of
the wagon is inclined at an angle of 60o above the horizontal, how
much work does the father do on the wagon?
F = 16 N, d = 10 m, Q = 60
W=?
W = FdcosQ
W=
W=
Kinetic Energy
Energy
due to motion
K = ½ m v 2
K: Kinetic Energy
m: mass in kg
v: speed in m/s
Unit: Joules
Sample problem
A 10.0 g bullet has a speed of 1.2 km/s.
a) What is the kinetic energy of the bullet?
Do on your own, remember to convert units to proper units.
a) What is the bullet’s kinetic energy if the speed is halved?
b) What is the bullet’s kinetic energy if the speed is doubled?
The Work-Energy Theorem
 The
net work due to all forces equals
the change in the kinetic energy of a
system.
 Wnet = DK
 Wnet:
work due to all forces acting on an
object
 DK: change in kinetic energy (K – K0)
 (FdcosQ = ½ mv2 – ½ mv02)
Constant force and work



The force shown is a
constant force.
W = FDx can be used
to calculate the work
done by this force
when it moves an
object from xa to xb.
The area under the
curve from xa to xb can
also be used to
calculate the work
done by the force
when it moves an
object from xa to xb
F(x)
xa
xb
x
Variable force and work



The force shown is a
variable force.
W = FDx CANNOT be
used to calculate the
work done by this
force!
The area under the
curve from xa to xb can
STILL be used to
calculate the work
done by the force
when it moves an
object from xa to xb
F(x)
xa
xb
x
Springs




When a spring is stretched or compressed from its equilibrium
position, it does negative work, since the spring pulls opposite the
direction of motion.
Ws = - ½ k x2
 Ws: work done by spring (J)
 k: force constant of spring (N/m)
 x: displacement from equilibrium (m)
The force doing the stretching does positive work equal to the
magnitude of the work done by the spring.
Wapp = - Ws = ½ k x2
Springs: stretching
0
F(N)
200
m
100
00
m
x
Fs
-100
-200
1
2
Fs
3
4
5
x (m)
Ws = negative area
= - ½ kx2
Fs = -kx (Hooke’s Law)
Sample problem
A spring with force constant 250 N/m is initially at its
equilibrium length.
a) How much work must you do to stretch the spring 0.050 m?
Do on your own.
b) How much work must you do to compress it 0.050 m?
Sample problem
It takes 1000 J of work to compress a certain spring 0.10 m.
a) What is the force constant of the spring?
Do on your own
a) To compress the spring an additional 0.10 m, does it take 1000 J,
more than 1000 J, or less than 1000 J? Verify your answer with a
calculation.
Sample Problem

How much work is done by the force shown when it acts on an
object and pushes it from x = 0.25 m to x = 0.75 m?
Figure from “Physics”, James S. Walker, Prentice-Hall 2002
Sample Problem
How much work is done by the force shown when it acts on an object
and pushes it from x = 0.25 m to x = 0.75 m?
CALCULATE AREA UNDER CURVE.
Sample Problem

How much work is done by the force shown when it acts on an
object and pushes it from x = 2.0 m to x = 4.0 m?
Figure from “Physics”, James S. Walker, Prentice-Hall 2002
Power
 Power
is the rate of which work is done.
 P = W/t
 W: work in Joules
 t: elapsed time in seconds
 When we run upstairs, t is small so P is
big.
 When we walk upstairs, t is large so P is
small.
Unit of Power
 SI
unit for Power is the Watt.
 1 Watt = 1 Joule/s
 Named after the Scottish engineer
James Watt (1776-1819) who perfected
the steam engine.
 British system
 horsepower
 1 hp = 746 W
How We Buy Energy…
 The
kilowatt-hour is a commonly used
unit by the electrical power company.
 Power companies charge you by the
kilowatt-hour (kWh), but this not
power, it is really energy consumed.
 1 kW = 1000 W
 1 h = 3600 s
 1 kWh = 1000 J/s • 3600s = 3.6 x 106J
Sample problem
A man runs up the 1600 steps of the Empire State Building in 20 minutes. If
the height gain of each step was 0.20 m, and the man’s mass was 80.0 kg,
what was his average power output during the climb? Give your answer in
both watts and horsepower.
Do on your own.
Force types
Forces acting on a system can be
divided into two types according to
how they affect potential energy.
 Conservative forces can be related
to potential energy changes.
 Non-conservative forces cannot be
related to potential energy changes.
 So, how exactly do we distinguish
between these two types of forces?

Conservative forces

Work is path independent.



Work along a closed path is zero.



If the starting and ending points are the same, no work is
done by the force.
Work changes potential energy.
Examples:



Work can be calculated from the starting and ending
points only.
The actual path is ignored in calculations.
Gravity
Spring force
Conservation of mechanical energy holds!
Non-conservative forces

Work is path dependent.




Work along a closed path is NOT zero.
Work changes mechanical energy.
Examples:



Knowing the starting and ending points is not
sufficient to calculate the work.
Friction
Drag (air resistance)
Conservation of mechanical energy does
not hold!
Potential Energy (U)
Energy due to position or configuration
 “Stored” energy
 For gravity: Ug = mgy

m: mass
 g: acceleration due to gravity
 h: height above the “zero” point


For springs: Us = ½ k x2
k: spring force constant
 x: displacement from equilibrium position

Conservative forces and
Potential energy
Wc = -DU
 If a conservative force does positive work on a
system, potential energy is lost.
 If a conservative force does negative work,
potential energy is gained.
 For gravity
 Wg = -DUg = -(mgy – mgyo)
 For springs
 Ws = -DUs = -(½ k x2 – ½ k xo2)

More on paths and
conservative forces.Figure from “Physics”,
Q: Assume a conservative force
moves an object along the
various paths. Which two works
are equal?
A:
W2 = W3
(path independence)
Q: Which two works, when added
together, give a sum of zero?
A:
W1 + W2 = 0
or
W1 + W3 = 0
(work along a closed path is zero)
James S. Walker,
Prentice-Hall 2002
Sample problem
Figure from “Physics”, James S. Walker, Prentice-Hall 2002
A box is moved in the
closed path shown.
a) How much work is
done by gravity
when the box is
moved along the
path A->B->C?
b) How much work is
done by gravity
when the box is
moved along the
path A->B->C->D>A?
Sample problem
Figure from “Physics”, James S. Walker, Prentice-Hall 2002
A box is moved in the closed
path shown.
a)
How much work would
be done by friction if the
box were moved along
the path A->B->C?
b)
How much work is done
by friction when the box
is moved along the path
A->B->C->D->A?
Sample problem
A diver drops to the water from a height of 20.0 m, his
gravitational potential energy decreases by 12,500 J. How
much does the diver weigh?
Do on your own.
Sample problem
If 30.0 J of work are required to stretch a spring from a 2.00
cm elongation to a 4.00 cm elongation, how much work is
needed to stretch it from a 4.00 cm elongation to a 6.00 cm
elongation?
Law of Conservation
of Energy
 In
any isolated system, the total
energy remains constant.
 Energy can neither be created nor
destroyed, but can only be
transformed from one type of
energy to another.
Law of Conservation of
Mechanical Energy

E = K + U = Constant
K: Kinetic Energy (1/2 mv2)
 U: Potential Energy (gravity or spring)


DE = DU + DK = 0
DK: Change in kinetic energy
 DU: Change in gravitational/spring potential energy
 U – Uo + K – Ko = 0
 U + K = Uo + Ko

 ****
mgy + ½ mv2 = mgyo + ½ mvo2 ****
Roller Coaster Simulation
Roller Coaster Physics Simulation
(demonstrates gravitational potential energy
and kinetic energy)
Pendulums and Energy
Conservation
Energy goes back and forth between
K and U.
 At highest point, all energy is U.
 As it drops, U goes to K.
 At the bottom , energy is all K.

Pendulum Energy
½mvmax2 = mgy
h
K1 + U1 = K2 + U2
For any points 1 and 2.
For minimum and
maximum points of
swing
Springs and Energy
Conservation
Transforms energy back and forth
between K and U.
 When fully stretched or extended, all
energy is U.
 When passing through equilibrium, all
its energy is K.
 At other points in its cycle, the energy
is a mixture of U and K.

Spring Energy
0
K1 + U1 = K2 + U2 = E
All U
m
For any two points 1 and 2
-x
All K
m
All U
m
x
½kxmax2 = ½mvmax2
For maximum and minimum
displacements from
equilibrium
Spring Simulation
Spring Physics Simulation
Sample problem
What is the speed of the pendulum bob at point B if it is
released from rest at point A?
40o
1.5 m
A
B
• We’ll do in class.
Sample problem
Problem copyright “Physics”, James S. Walker, Prentice-Hall 2002
A 0.21 kg apple falls from a tree to the ground, 4.0 m below.
Ignoring air resistance, determine the apple’s gravitational potential
energy, U, kinetic energy, K, and total mechanical energy,
E, when its height above the ground is each of the following: 4.0 m,
2.0 m, and 0.0 m. Take ground level to be the point of zero
potential energy.
Sample problem
Problem copyright “Physics”, James S. Walker, Prentice-Hall 2002
A 1.60 kg block slides with a speed of 0.950 m/s on a
frictionless, horizontal surface until it encounters a spring with a
force constant of 902 N/m. The block comes to rest after
compressing the spring 4.00 cm. Find the spring potential
energy, U, the kinetic energy of the block, K, and the total
mechanical energy of the system, E, for the following
compressions: 0 cm, 2.00 cm, 4.00 cm.
Law of Conservation of
Energy

E = U + K + Eint= Constant

Eint is thermal energy.
DU + DK + D Eint = 0
 Mechanical energy may be converted
to and from heat.

Work done by nonconservative forces

Wnet = Wc + Wnc
 Net work is done by conservative and non-conservative
forces
 Wc = -DU
• Potential energy is related to conservative forces only!

Wnet = DK
• Kinetic energy is related to net force (work-energy theorem)

DK = -DU + Wnc
• From substitution

Wnc = DU + DK = DE
 Nonconservative forces change mechanical energy. If
nonconservative work is negative, as it often is, the
mechanical energy of the system will drop.
Sample problem
Problem copyright “Physics”, James S. Walker, Prentice-Hall 2002
Catching a wave, a 72-kg surfer
starts with a speed of 1.3 m/s,
drops through a height of 1.75 m,
and ends with a speed of 8.2 m/s.
How much non-conservative work
was done on the surfer?
Sample problem
Problem copyright “Physics”, James S. Walker, Prentice-Hall 2002
A 1.75-kg rock is released from rest at the
surface of a pond 1.00 m deep. As the
rock falls, a constant upward force of 4.10
N is exerted on it by water resistance.
Calculate the nonconservative work, Wnc,
done by the water resistance on the rock,
the gravitational potential energy of the
system, U, the kinetic energy of the rock,
K, and the total mechanical energy of the
system, E, for the following depths below
the water’s surface: d = 0.00 m, d = 0.500
m, d = 1.00 m. Let potential energy be zero
at the bottom of the pond.
Pendulum lab




Figure out how to demonstrate conservation of energy with a
pendulum using the equipment provided.
The photo-gates must be set up in “gate” mode this time.
The width of the pendulum bob is an important number. To get it
accurately, use the caliper.
Turn in just your data, calculations, and result. Clearly show the
speed you predict for the pendulum bob from conservation of
energy, the speed you measure using the caliper and photo-gate
data, and a % difference for the two.