Transcript Work Power Energy PPT

Work demo
EWPM
The relationship between force
and distance has a name.
 WORK
WORK - the product of the net force and distance
through which an object moves in the direction of
the net force.
W=F∆d where
W=work
joules
F=force
kg*m/s2 (Newtons)
d=distance m (through which the force moves)
The SI unit for work is the joule (J)
aka a Newton-Meter
The American Unit is the Ft-lb
Ex 1: You discover a box of rocks in middle
of your room. They have a mass of 20 kg.
You apply an effort force of 65N to push
them 2 meters so they are against the wall.
How much work have you done?



W = Fd
W = 65N x 2m
W = 130 J
20 kg
You now have a box of rocks against the wall
of your room. You decide to lift them up to a
height of 1.5 m. Now how much work have
you done? What is your total work since
discovering this box of rocks?






W = Fd
F in this case is equal and
opposite the Fw
Fw= mg
W = (20 kg)(9.8m/s2) x 1.5m
W = 294 J
WT = 130J + 294J =424J
20 kg
Ex 2 Artie pulls a 500-kg boulder a distance of 4m at a
constant velocity along a horizontal surface by applying
a 250-N force along a rope that is parallel with the
horizontal. He then stoops down and lifts the boulder
above his head to a height of 2 m. Artie carries the
boulder at a constant velocity for a distance of 5 m.
How much work does he do to the boulder?
BONUS: Where does Artie get his strength from?
Answer: 10800 J



One of the oddest characters in The Adventures of Pete & Pete is Artie, the Strongest
Man in the World. In trying to figure out the mystery of Artie, the Strongest Man in the
World, we have put together this list of known facts:
Artie, the Strongest Man in the World, is the strongest man in the world. He is used in
several episodes for his super-human strength. [Various episodes]
Artie, the Strongest Man in the World, derives his super powers from the red and blue
underwear that is his costume, a 40/60 blend of titanium and cotton. [One of the early
shorts]
Ex 2
Part I: Drags
W = Fd
W = (250N)(4m) = 1000J
Part II: Lifts
W = F wd
W = (500kg)(9.8 m/s2)(2m) = 9800J
Part III: Carrries
No work
CHALLENGE
Example 3 A reindeer pulls a sled a distance
of 3.0 m across a frictionless surface. He
applies a force of 50.0 N on the rope to the
horizontal. How much work was done on the
sled?
50 N
W = FD Work = (50N)(3m) = 150 J
What happens if force is
applied at an angle?

We have to consider the Horizontal Component
of the applied force. How do we calculate this?
F
Fvert
θ
Fhor

We have to consider the Horizontal Component
of the applied force. How do we calculate this?
50N
Fvert
30°
Fhor
Fcosθ
Example 3 cont. A reindeer pulls a sled a
distance of 3.0 m across a frictionless
surface. He applies a force of 50.0 N on the
rope at an angle of 30.0° with the horizontal.
Calculate the work done on the sled.
50N
30°
Fcosθ
Example 3 A reindeer pulls a sled a distance
of 3.0 m across a frictionless surface. He
applies a force of 50.0 N on the rope at an
angle of 30.0° with the horizontal. Calculate
the work done on the sled.
50N
30°


W=Fdcos.
130J
Fcosθ
Example 3 The work on the sled decreased
by applying the force at an angle. The Force
of effort was the same.
50N
30°
Fcosθ
Who is doing the most work?
A.
B.
C.
Mary holds a 40 N bag of groceries for 15
minutes
Veronica slides a bag a 40 N bag of
groceries from the edge of a counter and
then carries it 10 meters across the room
Phillip lifts a 40 N bag from the floor to a
shelf 2 meters high
Who is doing the most work?
A.
B.
C.
Briton holds a 40 N bag of groceries for 15
minutes
Teagan slides a bag a 40 N bag of
groceries from the edge of a counter and
then carries it 10 meters across the room
Anna lifts a 40 N bag from the floor to a
shelf 2 meters high
KEY IDEA Work Force &
Direction



Phillip is the only one doing work
Why?
When calculating work, the applied force
of effort is parallel and in the direction of
motion.
What is the difference?


Ricky pushes a notebook 2 m across a table
with a force of 10 N in 4 seconds.
Christian pushes his notebook 5 meters
across the table with a with a force of 4 N in
10 seconds.
Power

The rate at which work is done

P = W/t

Power is measured in Watts. (W)
W
P
t
Fd
P
t





P – power (Watts)
W – work (Joules)
t – time (seconds)
F – Force ( Newtons)
d – distance (m)
Ex 4 One watt is one joule of work
performed in one second. This is very
small: calculate the power when lifting a
5N plate of apples 0.25m in 1 second.




P = W/t
P = (5N)(0.25m)/1sec
1.25 Watt
Since it is so small a unit, power is often
described in kilowatts.
How is this related to light
bulbs?

In terms of electromagnetism, one watt is the
rate at which work is done when one ampere
(A) of current flows through an electrical
potential difference of one volt (V).

Ex 5 Who has a higher power rating?

Stephanie pushes a 20 kg box with a net force of
60N. She moves it 30 m across the floor in 5
seconds.

Dustyn pushes a 20 kg box with a net force of
72N. He moves it 25 m across the floor in 4
seconds.

Who has a higher power rating?

Stephanie pushes a 20 kg box with a net force of
60N. She moves it 30 m across the floor in 5
seconds. 360 W

Dustyn pushes a 20 kg box with a net force of
72N. He moves it 25 m across the floor in 4
seconds. 450 W
What is horse power?





HP is a term used in the US for power.
1 HP = 746 watts
US unit for work = foot-pound
US unit for power = foot-pound/s
1 hp = 550 foot-pound/s
Ex 6: A winch lifts a maximum load of 600 kg at
a constant velocity for a distance of 4 meters
during a time period of 2 seconds. What power
in watts did this require? What would be the
horsepower? What power in kilowatts did this
require?
P = W/t = Fd/t
P = (600kg)(9.8m/s2)(4m) /2sec
P = 11760 Watts
P = (11760 Watts) )(1 hp / 746 Watts) = 15.76 hp
P = (11760 Watts) (1000 W / 1 kW) = 11.76 kW
energy

How many examples can you think of
What is energy?

The capacity to do work

Define mechanical energy
Mechanical Energy can be
classified as Potential Energy or
Kinetic Energy
How would you define
Potential Energy?
Potential Energy: stored energy
due to an object’s position or
condition in a field of force.
What factors do you think
effect EP?

What would be the units?
Gravitational Potential Energy
dependent on mass, height, &
gravitational acceleration

Ep = mgh
m
is mass (kg)
 g is gravity (9.8m/s2)
 h is the height (m) of the object
 Ep is Joules (J)
one J= kg*m2/s2
 SI
Unit for Energy is the Joule, also
referred to as the Newton-Meter
How would you define kinetic
energy?


Energy of motion
Types? Besides just moving: thermal, sound
Kinetic energy – energy of
motion. This is dependent upon
mass of moving object and the
square of it’s velocity.
Ek = .5mv2
Kinetic energy – energy of
motion. This is dependent upon
mass of moving object and the
square of it’s velocity.
Ek = .5mv2
m is mass (kg)
v is velocity (m/s)
Ek is Joules (J) one J= kg*m2/s2
Example 7: A panda lifts a barbell with a mass
of 85 kg above his head. The barbell is 243
cm above the ground. How much
gravitational potential energy does the
barbell have with respect to the ground?
(85 kg)(9.8 m/s2)(2.43 m) =
2020 J
Ex 8 An object with a mass of 2.0 kg is
moving horizontally with a velocity of
0.5 m/sec. Calculate the kinetic energy
possessed by the object.
(.5)(2.0kg)(0.5 m/s)2 = 0.25 J
Example 9: A 1400-kg car has a net forward force
of 4500 N applied to it. The car starts from rest
and travels down a horizontal highway. What are
its kinetic energy and speed after it has traveled
124 m?





What would you solve for first?
Acceleration…then it is a dvvat problem!
a= 4500 N/1400 kg = 3.21 m/s2
vf = √vi + 2ad = 28.2 m/s
KE = 557,000 J
Example 10:
A kitchen pot has a
weight of 180 N. (about
40 pounds!!!)
a. How much work is
required to lift the pot
to a height of 35 cm
above the ground with
nothing but the hair on
your face??
b. What was the
increase in GPE of the
kitchen pot at this
height?
63 J









If the pot fell from a height of 35 cm,
what would be its final velocity?
How would you determine this?
vf2 = vi2 + 2ad
vf = √vi + 2ad
vf = 2.62 m/s
What would be its kinetic energy?
We have the velocity, now determine
mass
18.4 kg
Solving for KE= 63J!!!
Work Energy Theorem
W = ΔE
Work is the transfer of energy by
means of forces
Work equals the change in energy in
a system
Ex 11: How would you solve
this?
A man pushes a 12.00 kg cart horizontally
across the floor starting at 0.625 m/s and
accelerating to 3.70 m/s. How much work is
done on the cart?
Find the change in KE
W = ΔKE
ΔKE = .5 m (vf2 – vi2)
or determine them separately and find
difference KEf – Kei
Ans: 79.8 J


How do you determine force when calculating
work done dragging an item at an angle?
You must determine FHOR (FT cosθ)

Pendulum(swing)
PE
PE
KE
The Law of Conservation of
Mechanical Energy
ME = EP + EK
•Mechanical Energy is equal to the sum of
potential and kinetic energy
•Energy cannot be created nor destroyed, but is
only changed from one form to another
•The amount of energy in the universe remains
constant.
EX 12 A 60-kg skier is at the top of a ski
slope. At this highest point the skier is 10
m vertically above the chalet.
What is the skier’s gravitational potential energy at
the peak?
(60 kg)(9.8 m/s2)(10 m) = 5880 J
What is the skier’s gravitational potential energy at
the chalet?
0J
What is the skier’s gravitational potential energy at
a point directly between the peak and the
chalet?
(60 kg)(9.8m/s2)(5 m) = 2940 J
A 60-kg skier is at the top of a ski slope. At
this highest point the skier is 10 m vertically
above the chalet.
What is the skier’s kinetic energy at this very
point? (directly between the peak and the chalet)
Solving using kinematics:
vf = √vi + 2ad = 9.90 m/s
KE = (.5)(60 kg)(9.90 m/s)2
2940 J
A 60-kg skier is at the top of a ski slope. At
this highest point the skier is 10 m
vertically above the chalet.

You can solve this problem very easily using
your knowledge of energy:
Alpha = peak
PE = 5880 J
Now you can use KE
to solve for velocities
at any point!
PE = KE = 2940 J
PE = 0 J
gamma = Chalet
What is the KE at
the base of the hill?
Example 13
A rocket with a mass of 7.8
kg is launched vertically with
a kinetic energy of 1420 J.
How high does it rise?
KEi = PEf
PEf = mghf
PEf/mg = hf
hf =1420 J /[(7.8 kg)(9.8m/s2)]
hf = 18.6 m
The Law of Conservation
of Mechanical Energy
(or total energy)
This can be expanded…
EM = EP + EK
“EXTREME”
Law of Conservation of
Mechanical Energy
The Law of Conservation of
Mechanical Energy
This can be expanded…
EM = EP + EK or
EMi = EMf or
EPi + EKi = EPf + EKf or
mghi + .5mv2i = mghf + .5mv2f
Ex 14
A rocket is launched vertically
with a velocity of 29 m/s. How
high does it rise?
PEi + KEi = PEf + KEf
mghi + .5mvi2 = mghf + .5mvf2
.5vi2 = ghf
.5 vi2 / g = hf
.5(29 m/s)2 / 9.80 m/s2 = hf
42.9 m = hf
Example 15: A dog is riding a sled. She
and the sled weigh 844 N. They move
down a frictionless hill through a vertical
distance of 12.8 m. Use the conservation
of mechanical energy to find the speed of
the dog and sled at the bottom of the hill,
assuming they are pushed off with an
speed of 5.29 m/s.
Dog on sled
PEi + KEi = PEf + KEf
mghi + .5mvi2 = mghf + .5mvf2
ghi + .5vi2 = .5vf2
(ghi + .5vi2)/.5 = vf2
[(9.80)(12.8 m) + [.5(5.29 m/s)2]/.5 = vf2
16.7 m/s = vf