WorkandEnergy Theory
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Transcript of WorkandEnergy Theory
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General Physics 1
APHY 111
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Work and Energy
Contents
3.1 The work of a constant force
3.2 The work done by a spring
3.3 Kinetic energy and the work-energytheorem
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3.4 Potential energy3.5 Conservation of Mechanical energy
3.6 Mechanical energy with friction
3.7 General Problems
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3.1 The work of a constant force
• Formally, work is the product of a constant force F
through a parallel displacement s.
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FS W F =
• Work is a scalar quantity• The unit of work is a joule (J)
– 1 joule = 1 newton . 1 meter
– J = N · m
• Forces applied at angles must be resolved into
components.
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Only the component of the force parallel to the displacement contributes.
s F s F W )cos(|| φ ==
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Work is an Energy Transfer
• If the work is done on a system and it is positive,energy is transferred to the system
• If the work done on the system is negative,
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energy s rans erre rom e sys em
• If a system interacts with its environment, thisinteraction can be described as a transfer ofenergy across the system boundary – This will result in a change in the amount of energy
stored in the system
Work Done by Multiple Forces
• The total work is equal to the algebraic
sum of the work done by the individual
forces
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– Remember work is a scalar, so this is the
algebraic sum
= ∑net by individual forcesW W
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Example 1: Calculate the total work done on the above box. Take Fp=100 N,
m=20kg and µκ=0.2.
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3.2 Work Done Extending a Spring
• The force is NOT
constant (it varies
with position)
Law)s(Hooke' kx F =
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2
02
1kx FdxW
x
== ∫
2
1
2
2
2
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2
102
1
2
1
2
1
2
12121
kxkxW kxW kxW x x x x −=== →→→
3.3 Kinetic energy and the work-energytheorem
• Kinetic Energy is the energy of a particledue to its motion
Ekin = ½ mv 2
• Ekin is the kinetic energy (or simply K)
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• m is the mass of the particle• v is the speed of the particle
• A change in kinetic energy is one possibleresult of doing work to transfer energy intoa system
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Kinetic Energy, cont
• Relation between
work and K.E. for a
constant force:
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kini f
i f
E mvmvW
a
vvmaW
xmaW
∆=−=Σ
−=Σ
∆=Σ
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22
2
1
2
1
2).(
).(
.
energykineticinChange=ΣW
Work-Energy Theorem
• The Work-Kinetic Energy Theorem states
ΣW = Ekinf – Ekini = ∆Ekin• When work is done on a system and the only
change in the system is in its speed, the work
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one y e ne orce equa s e c ange nkinetic energy of the system. – The speed of the system increases if the work done
on it is positive
– The speed of the system decreases if the net work isnegative
– Also valid for changes in rotational speed
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Example 2: A force of 100N acts on a box that was initially stationary for a
distance of 10m. If the mass of the box is 10kg what will be it’s speed after
10m?
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Example 3: A block of mass 5.00 kg is set into motion up an inclined plane with an initial
velocity of 8.00 m/s. There is a constant force of friction acting on the block. The block comes to
rest after travelling 3.00m. For this motion calculate
(a) the change in the kinetic energy of the block
(b) The change in the potential energy
(c) The coefficient of friction.
(d) Repeat the calculation using the work-energy theorem.
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3.4 Potential Energy
• Potential energy is energy related to the
position of the different components of a
system.
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–
other
– Can be associated with only specific types of
forces
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Gravitational Potential Energy
• The system is the Earth and
the book
• Do work on the book by lifting
it slowly through a vertical
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displacement
• There is no change in kineticenergy since the book starts
and ends at rest
r mg W
r F W
∆=
∆=
)(
.
Gravitational Potential Energy, final
• The quantity mg∆r is identified as the change ingravitational potential energy Ug.
• If we choose Ug=0 at the surface of the earth then wecan define the gravitational potential as
Ug=mgh, where h=height above ground
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If h>0 the object is above the surface of the earthIf h
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Elastic Potential Energy, cont
• The work done inchanging a spring’s
length by x is called theelastic potential energy ofthe spring:
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s = x
• The elastic potentialenergy can be thought ofas the energy stored inthe deformed spring
• The stored potential
energy can be convertedinto kinetic energy
2
2
1kxU s =
0= sU
3.5 Conservation of Mechanical Energy• The sum of potential and kinetic energy is
called mechanical energy
E mech=K+U
K=kinetic energy U=potential energy
•
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conservative then the mechanical energyis conserved:
U
U U K K
U K U K
f ii f
ii f f
∆−=∆
−=−
+=+
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Example 4: A ball of mass m is dropped from a height h=10m above the ground.
(A) Neglecting air resistance, determine the speed of the ball when it hits the ground.
(B) Determine the speed of the ball when it is 1m above ground.
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Example 5: Use conservation of mechanical
energy to calculate the speed of an object afteris released from a spring that is compressed
by an amount x.
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3.7 Mechanical energy with friction.
• The work done by friction in a system is
always negative. Friction is removingmechanical energy (dissipating energy)from the system.
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• The mechanical energy is not conservedwhen friction is present.
• Instead we have that the change inmechanical energy equals the work done
by frictionfrictionmech W E =∆
Example – Block-spring System
• Example 5: – The mass is attached to
a spring, the spring iscompressed and thenthe mass is released
– A constant friction force
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ac s
– The block will be pushedby the spring and moveoff with some speed
– Block and surface is thesystem
– The force of friction isdecreasing the totalmechanical energy
What is the velocity of block after is released?
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Exercise 1: A parachutist of mass 60 kg jumps out of a helicopter at an altitude of
800 m. Her parachute opens immediately and she lands on the ground with a
speed of 5 m/s.
(a) How much energy has been lost due to air friction in this jump?
(b) What is the average frictional force exerted on the parachutist during her fall?
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Exercise 2: A block of mass m = 4 kg is posed on a horizontal table in front of a
spring of stiffness k = 200 N/m. The other end of the spring is connected to a
fixed vertical wall. The coefficient of kinetic friction between the mass and the
table is = 0,3. The spring is suppressed by x1 = 30 cm with the mass
remaining in contact with it, then released free to move.
Calculate:
(a) The elastic potential energy of the compressed spring.
(b) The kinetic friction on the mass.(c) The speed of the block when the spring reaches its natural length.
(d) The distance it moved further before it stopped.
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Exercise 3: A block of mass 5.00 kg is set into motion up an inclined plane with an initial velocity
of 8.00 m/s. There is a constant force of friction acting on the block. The block comes to rest after
travelling 3.00m. For this motion calculate
(a) the change in the kinetic energy of the block
(b) the change in the potential energy
(c) the force of friction and the coefficient of kinetic friction
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Exercise 4: A 10.0-kg block is released from point A as shown in the Figure. The track is
frictionless except for the portion between points B and C, which has a length of 6.00 m. The
block travels down the track, hits a spring of force constant 2250 N/m, and compresses the spring
0.300 m from its equilibrium position before coming to rest momentarily. Determine the
coefficient of kinetic friction between the block and the rough surface between B and C.
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Exercise 5: A bullet of mass m1 = 10 g moving with a speed u1 = 200 m/s hits on
a wall and passes through it. The bullet comes out of the wall with a speed
u2 = 150 m/s.
Find:
(a) The loss of energy ∆E of the bullet.
(b) If the wall is 8 cm thick, what is the resistive force F the wall exerts on the
bullet assuming that it remains constant during its motion through the wall?
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Exercise 6. A spring with a spring constant of 500 N/m is compressed by 10 cm.
(a) What is the potential energy stored to the spring?
A box of mass 3.5 kg is placed in front of the above compressed spring.
(b) Find the speed that this object will gain after it is released.
(c) If on the box is exerted a constant force of friction equal to 5 N, for how long this
force must act in order to stop the body?
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Exercise 7: A body of mass m = 1 kg is dropped from a height h = 40 cm onto a
spring of constant k = 2000 N/m. Suppose that after the body collides with the
spring it remains on it. Using energy considerations, or otherwise, evaluate the
maximum compression of the spring.
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