work - work (w) is defined as a force moved over a distance - only the component of the force in the...
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WorkWork
- Work (W) is defined as a force moved over a distance
- Only the component of the force in the direction of motion does work
Units: N m
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
4
5530
370
b) FH / F = 3 / 5
FH = (3 / 5) F = (3 / 5) (20N) = 12 N
W = FH d = (12N)(15m)
W = 180 Nm
Transformation of EnergyTransformation 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
v
v
v
Fg
Fg Fg
Fg
No work
No energy change
Work done slowing down planet
Energy changes from kinetic to GPE
Work done increasing the planet’s speed
Energy changes from GPE to kinetic
No work
No energy change
EnergyEnergy
- Energy (E) is defined as the capacity to do work Units: Joule (J)
-Energy is the conceptual system for explaining how the universe works and accounting for changes in matter
1 Calorie (C) = 1 kcal = 4186 J
-There are many types of energy which are divided up into mechanical and non-mechanical forms
Form of Non-Form of Non-Mechanical Mechanical
EnergyEnergyAssociated with…Associated with…
Chemical
Thermal
Nuclear
Electromagnetic
bonds between atoms
vibration of atoms
bonds between protons and neutrons in nucleus
Vibration of electric charges
Form of Form of Mechanical Mechanical
EnergyEnergyAssociated with…Associated with…
Kinetic an object that is moving
Gravitational Potential
an object’s position in a gravitational field
Elastic Potential
stretched or compressed elastic materials
Spring Potential
stretched or compressed springs
Kinetic EnergyKinetic 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)
FNET
vi
FNET
vfChange in motion
WNET = FNET d = m a d But vf2 = vi
2 + 2 a d So.. a = ( vf2 - vi
2 ) / 2d
WNET = m ( vf2 - vi
2 ) d =
2 d
or.. KEf - KEi = KE
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
WNET = KE = 1/2 m (vf2 - vi
2)
m = 10 kg vi = 4 m/s vf = 15 m/s d = 5 m WNET = ? FNET = ?
= 1/2 (10kg) ( (15m/s)2 - (4m/s)2) = 1045 Nm
FNET = WNET / d = (1045 Nm) / 5m = 209 N
1/2 m vf2 - 1/2 m vi
2
Gravitational Potential EnergyGravitational 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
H
F
WT
+
-
F = WT = mg
W = F d
W = mg H
and… d = H
Because doing work always changes energy from one form to another then….
GPE = mg H = mg (df - di)
A 50 kg pile driver falls from 5m to 1m. How much GPE does it lose??
GPE = mg H = mg (df - di)
m = 50 kg di = 5 m df = 1 m g = 10 N/kg GPE = ?
= (50kg)(10N/kg) (1m- 5m)
GPE = - 2000 Nm = - 2000 J Note: negative means GPE has decreased
Transformation of EnergyTransformation of Energy
A device that changes energy from one form to another is called a machine
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)
Car Engine -
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 EnergyConservation of Energy
Conservative forces keep energy within a system (I.e. gravity)
Energy cannot be created nor destroyed, only transferred from one form to another
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 PEi = PEf = 0 KEf = ? E
ne
rgy C
ha
ng
e
(ma
gn
itud
e)
KE = KEf - KEi - 240 000J
- WNC = + 240 000JKEi + PEi + WNC = KEf + PEf
KEi + WNC = KEf
KEi + F d = KEf
320 000J + (-8000N) (30m) = KEf
80 000 J = KEf
Conservation of EnergyConservation 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 WNC = 0J vB
= ? vC = ? vD = ?
En
erg
y Ch
an
ge
A to
B
(ma
gn
itud
e)
KE = + 20 000 J
GPE = - 20 000 J
KEi + PEi + WNC = KEf + PEf
PEA = KEf + PEf
m g dA = 1/2 m vf2 + m g df
g dA = 1/2 vf2 + g df
g dA - g df = 1/2 vf2
2g( dA - df) = 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 = 28 m/s At D: same height as at B so same speed
Force-Displacement GraphsForce-Displacement Graphs- How much work is done by a person pulling the cart 15m?
0
2
4
6
8
10
12
14
0 3 6 9 12 15
distance (m)
forc
e (
N)
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.
AREA (rectangle) = h x b = 12N x 15m = 180 Nm
- How much work is done to stretch a spring in a spring scale 10cm?
0
5
10
15
20
25
30
0 2 4 6 8 10
distance (cm)
forc
e (
N)
2
The work done is the AREA under the applied force vs. displacement graph
AREA (triangle) =( h x b) / 2 = (25N x 0.1m) / 2 = 1.25 Nm
Note: This is the same as Fav d
PowerPower
Power is the rate at which work is donePower (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
PowerPower
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:
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
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)
= 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