Work

1
Work
- Work (W) is defined as a force moved over a
distance
Units N m
- 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 ,
5
530
3
b) FH / F 3 / 5
a) F 20 N ?d 15 m
370
W FH ? d
FH (3 / 5) F
(3 / 5) (20N)
12 N
4
W (20N)(15m)
W FH ? d
(12N)(15m)
W 300 Nm
W 180 Nm
2
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 No energy change
No work No energy change
Fg
v
Work done increasing the planets speed Energy
changes from GPE to kinetic
3
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 Non-Mechanical Energy Associated with




Form of Mechanical Energy Associated with




Chemical
bonds between atoms
Kinetic
an object that is moving
Gravitational Potential
an objects position in a gravitational field
Thermal
vibration of atoms
bonds between protons and neutrons in nucleus
Elastic Potential
stretched or compressed elastic materials
Nuclear
Spring Potential
stretched or compressed springs
Electromagnetic
Vibration of electric charges
4
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
vi
vf
FNET
FNET
WNET FNET ?d
m a ?d
But vf2 vi2 2 a ?d
So.. a ( vf2 - vi2 ) / 2?d
1/2 m vf2 - 1/2 m vi2
WNET m ( vf2 - vi2 ) ?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? Whats
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)
1045 Nm
FNET WNET / ?d
(1045 Nm) / 5m
209 N
5
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
F WT mg
and ?d H

F
W mg H
Because doing work always changes energy from one
form to another then.
H
WT
-
?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)
(50kg)(10N/kg) (1m- 5m)
?GPE - 2000 Nm - 2000 J
Note negative means GPE has decreased
6
Transformation 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
7
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
PEi PEf 0 KEf ?
KEi PEi WNC KEf PEf
- WNC 240 000J
Energy Change (magnitude)
KEi WNC KEf
KEi F ? d KEf
320 000J (-8000N) (30m) KEf
80 000 J KEf
?KE KEf - KEi - 240 000J
8
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 WNC 0J
vB ? vC ? vD ?
KEi PEi WNC KEf PEf
Energy Change A to B (magnitude)
?KE 20 000 J
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
?GPE - 20 000 J
?2g( dA - df) vf
At B vB ?2g( dA - dB)
?2(10m/s2) (100m - 60m)
28 m/s
At D same height as at B so same speed
At C vC ?2g( dA - dC)
?2(10m/s2) (100m - 30m)
37 m/s
28 m/s
9
Force-Displacement Graphs
- How much work is done by a person pulling the
cart 15m?
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?
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
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
11
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

• 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