Download Energy, Work, and Simple Machines

ENERGY, WORK, AND SIMPLE MACHINES
Chapter 10
OBJECTIVES
Chapter 10 Section 1
Chapter 10 Section 2
• Describe the relationship
between work and energy.
• Demonstrate a knowledge of
the usefulness of simple
machines.
• Differentiate between ideal
and real machines in terms of
efficiency.
• Analyze compound machines
in terms of simple machines.
• Calculate efficiencies for
simple and compound
machines.
• Calculate work.
• Calculate the work done by a
variable force.
• Calculate the power used.
10 - 1
ENERGY AND WORK
WORK
W = Fd
Definition:
• Work is the transfer of
energy by mechanical
means.
Where:
• W = work in joules (J)
• Work is done when a
• F = force in newtons (N)
force is exerted on an
object through a distance. • d = displacement in meters (m)
• Work is only done if the
force is applied for a
distance!
 Note: A joule is equal to 1 Nm.
KINETIC ENERGY
Energy Definition:
KE = ½ mv2
• The ability of an object to
produce a change in itself
or the world around it.
Where:
Kinetic Energy
•KE = kinetic energy in
joules (J)
• The energy resulting from
motion.
•m = mass in kg
• Represented by KE.
•v = velocity in m/s
WORK-ENERGY THEOREM
• The work-energy
theorem states that
when work is done
on an object, the
result is a change in
kinetic energy.
• Determined by
English physicist
James Prescott
Joule in the 1800’s.
W = ΔKE
Where:
•W = work in joules (J)
•ΔKE = change in kinetic energy
• KEf – KEi
• Measured in joules
EXAMPLE - CALCULATING WORK
When a club head strikes
a a 46-g golf ball, the ball
picks up 43 J of kinetic
energy. A constant force
of 2300 N is applied to the
ball while the club head
and ball are in contact.
Over what distance is the
club head in contact with
the ball?
Given:
Equations:
m = 46 g
W = ΔKE
ΔKE = 43 J
W = Fd
F = 2300 N
d=?
Calculations:
d = ΔKE / F
= 43 J / 2300 N
= 0.019 m
So, ΔKE = Fd
d = ΔKE / F
HOMEWORK
p. 261
# 2+3
xkcd.com
CALCULATING WORK
Remember:
• A force exerted in the
direction of motion is
given by W=Fd.
• A force exerted
perpendicular to the
direction of motion does
no work.
So, how do you
calculate work
when a constant
force is exerted at
an angle?
CALCULATING WORK
WHEN FORCE IS EXERTED AT AN ANGLE
1.Replace the force by it’s x and y
components.
2.Determine which component of the force is
in the direction of the object’s displacement.
3.Use this component of the force, in W = Fd,
to calculate the work done.
Calculating Work – Force at an Angle
Example:
How much work is
done in pushing a tall
box 15 m with a force
of 4.0 x 102 N that is
applied slightly upward
at an angle of 10.0°
from the horizontal?
F
10.0° incline
Fx
15 m
Given:
F = 400 N
d = 15 m (in x-dir)
W=?
*Need F in x-dir.
Eqn & Calculations:
W = Fd
F = F in x-dir = F cos θ
W = (F cos θ)d
W = (400N)(cos 10)15m
W = 5900 J
Fy
FINDING WORK
WHEN FORCES CHANGE
How can you
calculate work
when forces
change?
• Draw a force vs.
displacement graph.
• The area under the
curve is equal to the
work done on the
object.
CALCULATING WORK
How do you
calculate work when
there is more than
one force?
The work-energy theorem
relates the net work done on
the system to its change in
energy.
If several forces act on
an object:
1. calculate the work
done by each
force
2. then add them
together
* use components
for forces that are
at an angle
HOMEWORK
p. 262 #6-8
xkcd.com
POWER
• Power is the rate at which
the external force
changes the energy of
the system.
• The rate at which work is
done.
• Measured in watts (W)
• 1 W = 1 J/s
• Can also be measured in
kilowatts (kW).
Calculating Power
work
Power =
time
Units:
 Power in watts (W)
 Work in Joules (J)
 Time in seconds (s)
EXAMPLE - CALCULATING POWER
A net force of 2800
N accelerates a
1250-kg vehicle for
8.0 s. The vehicle
travels 80.0 m
during this time.
What power output
does this represent?
Given:
F = 2800 N
m = 1250 kg
t = 8.0 s
d = 80.0 m
P=?
Equations:
P=W/t
W = Fd
so:
P = Fd / t
Calculations:
P = Fd / t
P = (2800 N)(80.0 m) / 8.0 s
P = 28000 W or 28 kW
CLASSWORK
• pg. 264
• This assignment will be checked
before the end of class.
• # 9-13
• QUIZ on Chapter 10 Section 1
TOMORROW!
• QUIZ topics include:
• Work
• Energy
• Work-Energy Theorem
• Power
10 - 2
MACHINES
WHAT ARE THE SIX SIMPLE
MACHINES?
1.Lever
4.Inclined Plane
2.Wheel and Axle
5.Wedge
3.Pulley
6.Screw
LEVER
WHEEL AND AXLE
PULLEY
INCLINED PLANE
WEDGE
SCREW
BENEFITS OF MACHINES
A machine is a
device that
eases the load
by achieving
one or more of
the functions at
the left.
• transferring a force from one
place to another
• changing either the
magnitude of a force
• changing the direction of a
force
• changing the distance or
speed of a force
MECHANICAL ADVANTAGE
Effort Force (Fe)
Mechanical Advantage
• The force exerted by a
person on a machine.
• The mechanical
advantage of a machine
is the ratio of the
resistance force to the
effort force.
• Also called input force.
Resistance Force (F r)
• The force exerted by the
machine.
• Also called output force.
Fr
MA =
Fe
MECHANICAL ADVANTAGE
MA < 1
MA = 1
• The resulting force
from the machine
is less than the
applied force.
• The resulting force
from the machine
is the same as the
applied force.
• So, the machine
changes the
distance or
direction of the
force.
• So, the machine
changes the
distance or
direction of the
force.
• Ex. fishing pole
• Ex. pulley
MA > 1
• The resulting
force from the
machine is more
than the applied
force.
• So, the machine
increases the
force applied by
the person.
• Ex. crowbar
IDEAL MECHANICAL ADVANTAGE
• In an ideal machine, all energy put in would be transferred out,
so the input work would be equal to the output work.
• Therefore, for an ideal machine, the mechanical advantage is
equal to the displacement of the effort force divided by the
displacement of the load – this is called ideal mechanical
advantage (IMA).
de
IMA =
dr
EFFICIENCY
COMPOUND MACHINES