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

W = F d cos 
If the cart (above) is pulled at constant speed with an applied force of 20N at an angle of 600 over
a distance of 15m, determine the work done by I) the applied force ii) gravity iii) friction iv) the net
work (note: the friction force is the same as the horizontal component of the applied force)
-
+
FA = 20N FF = 10 N  = 600 d = 15 m
i) WA = FA  d cos 
= (20N) (15m) cos 600
= +150 Nm
ii) W g = Fg  d cos  = mg  d cos (-900) = 0 Nm
iii) W F = FF  d cos 
= (10N) (15m) cos 1800
No work done when force is perpendicular to d
= - 150 Nm
iv) W NET = FNET  d cos  = WA + W F = +150 Nm + (-150 Nm) = 0 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
So..
or..
a = ( vf2 - vi2 ) / 2d
KEf - KEi
= KE
2 d
What is the net work done on a 1500 kg truck that decreases its speed from 40 m/s to
25 m/s? What’s the breaking force if the speed change occurs in a distance of 200 m
m = 1500 kg vi = 40 m/s vf = 25 m/s d = 200m WNET = ?
FNET = ?
WNET = KE = 1/2 m (vf2 - vi2) = 1/2 (1500kg) ( (25m/s)2 - (40m/s)2) = - 7.3 x 105 Nm
FNET = WNET / d
= (-7.3 x 105 Nm) / 200m
= - 3700 N
Note: negative means
opposite the car’s motion
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
(if g is -ve then +ve work is done lifting)
Because doing work always changes energy from one
form to another then….
-
WT
H
GPE = - mg H
= - mg (df - di)
A 409 kg pile driver falls from 20m to 5m. How much GPE does it lose??
m = 409 kg di = 20 m df = 5 m g = - 9.8 N/kg GPE = ?
GPE = - mg H
= - mg (df - di) = - (409kg)(-9.8 N/kg) (5m- 20m)
GPE = - 6.0 x 104 Nm = - 6.0 x 104 J
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
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?
12
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 = (20N cos600) = 10N
10
fo r c e ( N )
8
6
AREA (rectangle) = h x b = 10N x 15m = 150 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
fo r c e ( N )
20
AREA (triangle) =( h x b) / 2 = (25N x 0.1m) / 2 = 1.25 Nm
15
2
10
Note: This is the same as Fav d
5
0
0
2
4
distance (cm)
6
8
10
Spring Potential Energy
A physical expression for spring potential energy (SPE) can be derived using the workenergy theorem
Consider a spring that is compressed a certain distance, x
30
Hooke’s Law says that the applied force on a spring
force increases proportionally with displacement Fapplied = k x
or…..
Conversly, the restoring force provided by the spring
will be …. Fspring = - k x
25
fo r c e ( N )
20
15
2
10
5
0
0
2
4
distance (m)
6
8
10
Where k is the spring constant for the spring
From earlier, you saw that the area under the F-d
graph represents the work done, so….
Area = Work = 1/2 h b =
1/2 Fapplied x
But… Fapplied = k x
Work = 1/2 (k x) x = 1/2 k x2
Because doing work always changes energy from one
form to another then….
SPE = 1/2 k x2 = 1/2 k (df2 - di2 )
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)
KE + PE - WNC = 0
Written as an expression…
A 10g dart is put in a child’s dart gun and the spring (k = 200 N/m)
is compressed 15cm. Determine the exit speed of the dart if there
isn’t any friction in the barrel as the dart leaves.
k = 100 N/m mdart = 0.010 kg
xi = 0 m xf = 0.15 m  x = 0.15 m vi = 0 m/s vf = ?
KE = + 1.125 J
Energy Change
(magnitude)
KE + PE - WNC = 0
1/2mdart(vf2 - vi2) + 1/2 k ( x)2 = 0
mdart(vf2 ) + k ( x)2 = 0
(vf2 ) = k ( x)2 / mdart
If there was friction, the KE
would be reduced by this amount
- WNC
vf= (k ( x)2 / mdart
vf= ((100N/m)(0.15m)2 / 0.010 kg
Note…WNC would be Ffric (xf - xi)
= 15 m/s
SPE = - 1.125 J
Conservation of Energy
A 200g toy car travels down a 400 incline starting from rest. The length of the ramp is
2m, and a constant friction force of 0.5N acts against the car’s motion. Determine
the car’s speed at the bottom using a) dynamics and b) conservation of energy
+
FN
mc = 0.2 kg  d = 2m FF = - 0.5 N g = +9.8 m/s2
FF
a) FNET = Fg sin  + FF = mc g sin  + FF

Fg

 = 400
vi = 0 m/s
vf = ?
= mc a
= 3.8 m/s2
a = (mc g sin  + FF ) / mc = (0.2kg)(9.8m/s2) sin 400 - 0.5N) /
0.2kg
vf =  (vi2 + 2a  d ) =  (2a  d ) =  (2(3.8m/s2)(2m) ) = 3.9 m/s
b) KE + PE - WNC = 0
H =  d sin  = 2m sin 400 = 1.28m
1/2mC(vf2) + (- mgH) - FF  d = 0
vf =  ( (mgH + FF  d) / 1/2mc)
Energy Change
(magnitude)
1/2mC(vf2 - vi2) + (- mgH) - FF  d = 0
KE = + 1.5 J
- WNC = +1.0 J
vf =  ( (0.2)(9.8m/s2)(1.28m) + (-0.5N)(2m)
1/2 (0.2kg)
vf = 3.9 m/s
(check!!!)
GPE = - 2.5 J
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