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Review
Work = Force x distance
(Joules = N·m)
(J)
Example:
How much work is done by a crane that lifts 120 kg up 75 m?
F = mg
W = F·d = m·g·d = 120 x 9.8 x 75
= 88000 J
(note the distance here
is the height)
Power = rate of doing work
Power = rate of using energy
Power = Work / time
(Watts = J/s)
(W)
Example:
Find the power output of a 60 kg athlete who climbs the
Grouse Grind in 40 minutes. (elev. gain = 800m)
Power = Work = (mg)d = 60 x 9.8 x 800 = 196 W
time
t
2400
(in seconds)
Gravitational Potential Energy
The work done to lift (or lower) something is equal to its change
in gravitational potential energy
e.g.: Potential Energy (PE) (at height “h”) = work done to lift
the object to that height.
m
h
h=0 (reference point)
PE = F·d = m·g·h
PE = m·g·h
Example:
Find the potential energy of a 60 kg student sitting on a 80
cm high stool…
a) relative to the floor in the room.
b) relative to the floor in the room 1 floor below (3.0 m lower)
a) PE = mgh
= 60 x 9.8 x 0.8
= 470 Joules
b) PE = mgh
= 60 x 9.8 x 3.8
= 2200 Joules
Energy & Work
An object is accelerated from initial velocity, vo, to velocity, v, by
a force, F (net force), over a distance, d.
What is the work done?
F = ma
F·d = ma·d
F·d = m(a·d)
2ad = v2 – vo2
F·d = m(v2 - vo2)
2
W = ½mv2 - ½mvo2
Kinetic Energy (energy of motion)
KE = ½mv2
W = KE - KEo
W = ΔKE (Work – Energy Theorem)
Work – Energy Theorem
W = ΔKE
The work done on an object by the net force equals
the change in kinetic energy of that object.
Example:
A 1000 kg car going 20 m/s brakes and comes to a stop.
How much work is done? What happens to the energy?
KE = ½mvo2 = ½(1000)(20)2 = 200000 J
W = ΔKE = 0 – 200000 J = –200000 J (lost energy)
Energy is dissipated as heat.
Homework
 Practice
Problems: Pg. 221 (1, 2)
 Practice Problems: Pg. 224 (5, 7)
 Pg. 236 R.C. (5, 7) A.C. (1, 3)
 Pg. 236 Problems (2, 4, 5, 10)
Conservation of Mechanical
Energy
Total Mechanical Energy of an object moving in a gravitational
field:
E = PE + KE
E = mgh + ½mv2
Conservation Law:
If no other force (than gravity) does any work on an object in
a gravitational field, the total mechanical energy is
conserved.
Eo = Efinal
Example:
1. A soccer ball of mass 0.5 kg is kicked with an initial velocity
v
of 15 m/s. Find its…
a) velocity when it is 4.0 m high.
h
b) height when it is going 5.0 m/s.
15m/s
a)
Eo = Efinal
mgho + ½mvo2 = mgh + ½mv2
0 + ½(0.5)(15)2 = (0.5)(9.8)(4.0) + ½(0.5)v2
56.3 = 19.6 + 0.25v2
v = 12 m/s
b)
56.3 = (0.5)(9.8)h + ½(0.5)(5.0)2
h = 10 m
2. Tarzan, a 75 kg Ape-man, swings from a branch 3.0 m
above the ground with an initial speed of 5.0 m/s.
a) Find his velocity when he swings past ground level.
b) Find the maximum height he swings to.
c) What should his initial speed be if he is to just reach a
branch 4.0 m high?
b) & c)
3.0 m
a)
h=0
Note: in the physics jungle there is no air
resistance and the vines have no mass
and are unstretchable!
Note: The vine does NO WORK because it is always applying
its force perpendicular to the direction of motion
a)
mgho + ½mvo2 = mgh + ½mv2
gho + ½vo2 = 0 + ½v2
m cancels out; h = 0
v = √(vo2 + 2gho)
v = √((5.0)2 + 2(9.8)(3.0))
v = 9.2 m/s horizontal
b)
gho + ½vo2 = 0 + gh
h = vo2 + ho
2g
v=0
h = 4.3 m
c)
gho + ½vo2 = 0 + gh
vo = √(2gh - 2gho)
vo = 4.4 m/s
v=0
3. Tarzan’s less famous brother, George, swings from the
same 3.0 m-high branch with an initial speed of 3.5 m/s.
Unfortunately, he hits a tree 1.5 m above the ground. Find
his speed just before impact and the work done by the tree.
His mass is 90 kg.
3.0 m
1.5 m
mgho + ½mvo2 = mgh + ½mv2
m cancels out
gho + ½vo2 = gh + ½v2
v = √(vo2 + 2gho – 2gh)
v = √((3.5)2 + 2(9.8)(3.0 – 1.5))
v = 6.5 m/s
W
W
W
W
= ΔKE = KE - KEo
= ½mv2 – ½mvo2
= 0 – ½(90)(6.5)2
= -1900 J
Efficiency
Definition: Efficiency = useful output x 100%
input
Example:
1. A crane operates at 15 kW and lifts a 700 kg weight 50 m in
2 minutes. Find the efficiency.
Input:
Power = Energy / time
Ei = P x t = 15000W x 120s = 1.8 x 106 J
Ouput: Eo= PE
= mgh
= 700 x 9.8 x 50
= 3.43 x 105 J
Efficiency: Eff = 3.43 x 105 x 100%= 19%
1.8 x 106
2. A 6.5 hp forklift is 15% efficient when lifting a 220 kg box.
How long does it take to lift the box 1.5m?
(1 hp = 746 W)
Ouput: Eo = mgh
= 220 x 9.8 x 1.5
= 3234 J
Efficiency: Eff = Eo = 0.15
Ei
Input:
Ei = Eo = 3234 = 21560 J
0.15 0.15
Ei = P x t
21560 = (6.5 x 746) x t
t = 4.5s
3. An 800 kg, 100 hp car is 65% efficient at transferring
energy to the wheels. How long does it take to go from
0  100 km/h?
Efficiency: Eff = Eo x 100% = ½mv2 x 100%
Ei
Pxt
0.65 = ½ (800) (27.8)2
(100 x 746) t
0.65 = 309000
74600 x t
t = 6.4s
Homework