Transcript Document

Work
- Work (W) is defined as a force moved over a distance
Units:
Nm
- Only the component of the force in the direction of motion does work

The cart (above) is pulled at constant speed with a force of 20N over a distance of 15m.
Determine the work done by the applied force if the handle is pulled a) in a straight line and b) at an
angle of 530 ,
a) F = 20 N
d = 15 m
W = FH  d
W = (20N)(15m)
W = 300 Nm
3
530
5
370
4
b) FH / F = 3 / 5
FH = (3 / 5) F = (3 / 5) (20N) = 12 N
W = FH  d = (12N)(15m)
W = 180 Nm
Transformation of Energy
When work is done, energy is transformed from one form into another
Consider a planet moving in an elliptical orbit around the sun
v
Work done slowing down planet
Energy changes from kinetic to GPE
v
Fg
Fg
Fg
v
No work
Fg
No energy change
v
Work done increasing the planet’s speed
Energy changes from GPE to kinetic
No work
No energy change
Energy
- Energy (E) is defined as the capacity to do work
Units:
Joule (J)
1 Calorie (C) = 1 kcal = 4186 J
-Energy is the conceptual system for explaining how the universe works and accounting for changes in
matter
-There are many types of energy which are divided up into mechanical and non-mechanical forms
Form of NonMechanical
Energy
Associated with…
Chemical
bonds between atoms
Thermal
Nuclear
Electromagnetic
vibration of atoms
bonds between protons
and neutrons in nucleus
Vibration of electric charges
Form of
Mechanical
Energy
Associated with…
Kinetic
an object that is moving
Gravitational
Potential
an object’s position in a
gravitational field
Elastic
Potential
Spring
Potential
stretched or compressed
elastic materials
stretched or compressed
springs
Kinetic Energy
A physical expression for kinetic energy can be derived using the work-energy theorem
Consider an object that has a net force (FNET) applied to it over a distance (d)
Change in motion
vf
FNET
vi
FNET
WNET = FNET d = m a d
WNET = m ( vf2 - vi2 ) d
=
But vf2 = vi2 + 2 a d
1/2 m vf2 - 1/2 m vi2
a = ( vf2 - vi2 ) / 2d
So..
or..
KEf - KEi
= KE
2 d
What is the net work done on a 10 kg cart that increases its speed from 4 m/s to 15
m/s? What’s the force needed if the speed change occurs in a distance of 5 m
m = 10 kg vi = 4 m/s vf = 15 m/s d = 5 m WNET = ?
FNET = ?
WNET = KE = 1/2 m (vf2 - vi2) = 1/2 (10kg) ( (15m/s)2 - (4m/s)2)
FNET = WNET / d
= (1045 Nm) / 5m
= 209 N
= 1045 Nm
Gravitational Potential Energy
A physical expression for gravitational potential energy (GPE) can be derived using
the work-energy theorem
Consider an object that is lifted a certain height at constant speed in a constant
gravitational field
W = F d
and… d = H
F = WT = mg
+
F
W =
mg H
Because doing work always changes energy from one
form to another then….
-
WT
H
GPE = mg H
=
mg (df - di)
A 50 kg pile driver falls from 5m to 1m. How much GPE does it lose??
m = 50 kg di = 5 m df = 1 m g = 10 N/kg GPE = ?
GPE =
mg H
=
mg (df - di)
GPE = - 2000 Nm = - 2000 J
=
(50kg)(10N/kg) (1m- 5m)
Note: negative means GPE has
decreased
Transformation of Energy
A device that changes energy from one form to another is called a machine
Car Engine -
A car engine changes chemical energy into kinetic (moving
car), gravitational potential energy (if car drives up a hill), and
thermal energy (engine gets hot - exhaust gasses)
Work is done by expanding gasses in a car engine cylinder
pushing on the piston which is free to move
Plants -
Plants are natural machines. Nuclear energy in the sun is
converted into radiant (EM) energy which is changed into
chemical energy in the plant
Work is done by molecular transport ( ionic pump) across
the plant (or animal) cell
Conservation of Energy
Energy cannot be created nor destroyed, only transferred from one form to another
Conservative forces keep energy within a system (I.e. gravity)
Non-conservative forces transfer energy out of a system (I.e. friction)
Written as an expression…
KEi + PEi + WNC = KEf + PEf
Consider a car with 320 000J of KE braking on the flat with a force
of 8000 N over a distance of 30m. What is the final energy of the
car?
KEi = 320 000J
 d = +30 m
F = -8000N
KEi + WNC = KEf
KEi + F  d = KEf
Energy Change
(magnitude)
KEi + PEi + WNC = KEf + PEf
PEi = PEf = 0
KEf = ?
- WNC = + 240 000J
320 000J + (-8000N) (30m) = KEf
80 000 J = KEf
KE = KEf - KEi - 240 000J
Conservation of Energy
What is the speed of the 50 kg jumper at B, C and
D? Assume that there is no friction
m = 50 kg g = 10 m/s2 KEA = 0J PEA = 50 000 J
dA = 100m dB = dD = 60m dc = 30m W NC = 0J vB
= ? vC = ? vD = ?
PEA = KEf + PEf
m g dA = 1/2 m vf2 + m g df
g dA = 1/2 vf2 + g df
g dA - g df
2g( dA - df)
1/2 vf2
=
=
Energy Change A to B
(magnitude)
KEi + PEi + WNC = KEf + PEf
KE = + 20 000 J
GPE = - 20 000 J
vf
At B: vB = 2g( dA - dB)= 2(10m/s2) (100m - 60m) = 28 m/s
At C: vC = 2g( dA - dC)=
2(10m/s2)
(100m - 30m) = 37 m/s
At D: same height as at B so
same speed = 28 m/s
Force-Displacement Graphs
- How much work is done by a person pulling the cart 15m?
14
The work done is the AREA under the applied force vs.
displacement graph where the applied force is the component in
the direction of motion.
12
force (N)
10
8
6
AREA (rectangle) = h x b = 12N x 15m = 180 Nm
4
2
0
0
3
6
9
12
15
distance (m)
- How much work is done to stretch a spring in a spring scale 10cm?
30
The work done is the AREA under the applied force vs.
displacement graph
25
force (N)
20
15
AREA (triangle) =( h x b) / 2 = (25N x 0.1m) / 2 = 1.25 Nm
2
10
Note: This is the same as Fav d
5
0
0
2
4
6
distance (cm)
8
10
Power
Power is the rate at which work is done
Power (P) = Work / Time = W / t
Units:
Nm / s or J/s or Watts (W)
P = Fav d / t = Fav vav
James Watt (1783) wanted to standardize the measure of
power using something that everyone was familiar with
….. the power output of a horse.
If a large draft horse can pull 150 lbs while walking at 2.5
mi/h determine how many Watts one “horsepower”
represents.
1 lb = 4.448 N
1 m/s = 2.237 mi/h
P = Fav vav = (150 lb) (4.448 N/lb) (2.5 mi/h) (1 m/s / 2.237 mi/h) = 746 W
Power
An engine is used to raise a 2000 lb load 200 m vertically up a mine shaft. If the load travels
upwards at a constant speed of 3 m/s calculate:
a)
The power rating of the engine in i) Watts and ii) Horsepower
Assume that the engine is 100% efficient (4.448 N = 1 lb)
Fav = 2000 lb
v = 3 m/s
d = 200 m
i) P = Fav vav = (2000 lb) (4.448 N/lb) (3 m/s) = 26 688 W
= 30 000 W
ii) P (hp) = P (W) (1hp / 746 W) = 26 688 W (1hp / 746 W) = 36 hp = 40 hp
b)
What is the power rating (hp) of the engine if it is only 70% efficient?
0.7 WIN = WOUT
0.7 WIN / t
= WOUT / t
0.7 PIN = POUT
0.7 PIN = 36 W
Therefore… PIN = 36 W / 0.7 = 51 hp = 50 hp