Download Mechanics VI

Definitions 1
Energy
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Different kinds of energy
• Kinetic energy
Kinetic energy is motion energy. The faster it moves
the more kinetic energy it possesses.
• Potential energy
Potential energy is position energy. The higher the
object is above the ground the more potential energy
it possesses.
Definitions 2
Energy
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Different kinds of energy
•Mechanical energy
Sum of kinetic energy and potential energy
•Work
When we lift an object from the ground up to the
height of a table, energy is transferred from us to
the load (chemical energy to potential energy).
The amount of energy transferred is called work.
Work 1
Energy
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Work
•If a force F is used to move an object for a distance
s as shown, we define
F
F
s
Work = force  distance moved (W = F  s)
Unit : newton metre (N m) or joule (J)
1 J  1 N 1 m
•Work is a scalar. It has magnitude only.
Work 2
Energy
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Properties of work
• If the force F is not in the same direction of the object
moves, we consider the component of F,

F

F
s
W = F cos  s
• No displacement implies no work is done. Diagram
• Force  direction of motion ( = 90) implies no
word is done. Diagram
Kinetic Energy 1
Energy
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Kinetic Energy
• Consider a mass m is pulled by a constant force F
on a smooth surface by a distance s as shown. The
initial velocity and final velocity are u and v
respectively.
v
u
F
F
s
v2  u 2
1 2 1 2
Workdone  Fs  m(
)  mv  mu
2
2
2
Kinetic Energy 2
Energy
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•The work done calculated is equal to the amount of
energy transferred to the object, i.e. increase in K.E.
•If the object starts from rest (u = 0),
1 2
K.E. gained  mv
2
•Work done by net force = change in K.E.
•Hence, we define
1 2
K.E. of a moving body  mv
2
Potential Energy 1
Energy
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Potential Energy
• Consider we lift a mass m up
by a height of h from the
ground as shown. The minimum
force F required is equal to mg
so that the mass is moving
upwards with constant velocity
(no change in K.E.)
Workdone  Fs  mgh
F = mg
weight = mg
h
F = mg
weight = mg
Potential Energy 2
Energy
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• The work done calculated is equal to the amount
of energy transferred to the object, i.e. increase in
P.E. In this case, there is no increase in K.E.
(Why?)
• If the object starts on the ground,
P.E. gained  mgh
• Hence, we define
P.E. of a body  mgh
Conservation of energy 1
Energy
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Two different principles about energy
• Conservation of energy
Energy can never be created or destroyed. It
can only be changed from one form to another.
• Conservation of mechanical energy
When there is no collision and/or friction, the
sum of K.E. and P.E. conserves provided that
no work is done by external force.
Conservation of energy 2
Energy
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Different Examples
• Elastic collision Calculation
• An object is pulled on a rough surface Calculation
• An object is falling from a height Calculation
• Energy changes in a simple pendulum Calculation
Power 1
Energy
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Power
• Power is defined as the rate at which energy is
transferred with respect to time
energy transfer
E
Power 
 force  speed  P   Fv
time taken
t
Unit : joule per second (J s-1 ) or watt (W) 1 W  1 J s-1
• A car is moving on a rough road
Calculation
END of Energy
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Energy
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• A person is pushing a fixed wall as shown. No matter how
hard he pushes, no work is done.
fixed wall
pushing force
• However, he feels tired and his chemical energy has been
used up. All the chemical energy has been changed to
internal energy of his body. He feels very hot.
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Energy
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• A person is now holding a load and moving forward. Again
no work is done on the load. It is because the direction of
the force is perpendicular to the direction of motion.
holding force
direction of motion
weight
Conservation of energy 2
Energy
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• Two objects (mass 2 kg) are moving towards each other as
shown with speed 3 m s-1.
3ms
-1
mass 2 kg
3 m s -1
mass 2 kg
• After collision,
Case 1
3 m s -1
mass 2 kg
3 m s -1
mass 2 kg
Case 2
1 m s -1
1 m s -1
mass 2 kg
mass 2 kg
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Energy
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• In both cases, the conservation of momentum holds.
Total K.E. initially  1  2 kg  (3 m s 1 ) 2  2  18 J
2
Total K.E. finally (case 1)  1  2 kg  (3 m s 1 ) 2  2  18 J
2
Total K.E. finally (case 2)  1  2 kg  (1 m s 1 ) 2  2  2 J
2
• In case 1, there is no loss in total K.E. in the collision, it is
an elastic collision. Elastic collision is a collision such that
no loss in total K.E.
• In case 2, there is loss in total K.E. in the collision, it is not
an elastic collision.
Conservation of energy 2
Energy
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• An object (mass : 2 kg) is pulled by a force 10 N on a
horizontal rough surface with friction 4 N for 6 m. The
initial velocity is zero and the final velocity is 6 m s-1.
(Why?!)
1
6ms
10 N
4N
rough
surface
4N
6m
Apply v 2  u 2  2as,
v 2  02  2  3 m s  2  6 m  v  6 m s -1
10 N
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Energy
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• Work done by the pulling force = 10 N  6 m = 60 J
• This is the amount of chemical energy used by the man.
• Work done by the net force = (10 - 4) N  6 m = 36 J
• This is the amount of K.E. gained by the object.
1
K.E. gained   2 kg  (6 m s -1 ) 2
2
• Work done against friction = 4 N  6 m = 24 J
• This is the amount of internal energy gained by the
surface. It is part of energy input by the man.
• Conservation of M.E. does not hold since friction exists.
Conservation of energy 2
Energy
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• An object (mass 2 kg) are released from rest from a height
5 m. What is its speed just before it hits the ground?
Assume the air resistance can be neglected.
u  0 m s-1
5m
v
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Energy
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• There is no friction (air resistance) and external force, so
we can apply the conservation of M.E.
• By conservation of M.E.
K.E.initial  P.E.initial  K.E.final  P.E.final
1
0 J  2 kg  10 m s  5 m   2 kg  v 2  0 J
2
v  10 m s -1
-2
Conservation of energy 2
Energy
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• An object (mass 2 kg) is released from rest from a height
0.2 m higher than the lowest point in a simple pendulum,
what is its speed when it is at the lowest point? Assume the
air resistance can be neglected.
0.2 m
ground
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Energy
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• There is no friction (air resistance). Moreover,
although we have external force (tension), it does no
work on the object since it is always perpendicular to
the direction of motion of the object, so we can apply the
conservation of M.E.
• By conservation of M.E.
K.E.initial  P.E.initial  K.E.final  P.E.final
1
0 J  2 kg  10 m s  0.2 m   2 kg  v 2  0 J
2
v  2 m s -1
-2
Power 1
Energy
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• A car (mass 1000 kg) is now moving on a rough horizontal
road with a constant velocity 20 m s-1 . The friction acting
on the car by the road is 1000 N. What is the force
generated by the engine? What is the power of the engine?
20 m s-1
friction 1000 N
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Energy
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• By Newton’s second law, the force generated by the engine
must be exactly equal to the friction, that means 1000 N.
• In 1 s, the car has traveled for 20 m.
• In 1 s, the work done against friction
= 1000 N  20 m = 20000 J
20000 J
Power of the engine 
 20000 W
1s