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Physics 1-2
Mr. Chumbley
Chapters 5 and 6
• Our definition of physics at the start of the year was:
• Physics is the study of matter, energy, and the relationship between them
• So far we’ve studied
• How objects move (kinematics)
• Why objects move (forces)
• Now we want to look at why forces cause objects to move in the
way that they do
• We haven’t yet looked at energy, but have focused primarily on
describing characteristics of matter
• Energy is classically defined as the capacity to do work
• What does this mean? What is work?
Chapter 5, Section 1
• In everyday usage, work is generally referred to as anything
that requires some sort of physical or mental effort
• In physics, work is a specific, measurable quantity
• Work is done on an object when a force causes a an object to
move through a displacement
• Work is the product of the component of a force along the
direction of displacement and the magnitude of the
displacement
𝑊 = 𝐹𝑑
• Taking the SI units of force and displace net, the unit of work
could be N·m
• The N·m is condensed and called the joule (J)
How much work is done on a vacuum cleaner pulled 3.0 m by a
force of 50.0 N?
Given:
F = 50.0 N
d = 3.0 m
𝑊 = 𝐹𝑑
𝑊 = 50.0 N 3.0 m
𝑊 = 150 J
Unknown:
W=?
• Sometimes, forces are not applied in the same direction as the
motion of the object
• When this is the case, the work done is only the force done in
the direction of the motion
𝑊 = 𝐹𝑑 cos 𝜃
How much work is done on a vacuum cleaner pulled 3.0 m by a
force of 50.0 N at an angle of 30.0˚ above the horizontal?
Given:
F = 50.0 N
d = 3.0 m
θ = 30.0˚
Unknown:
W=?
𝑊 = 𝐹𝑑 cos 𝜃
𝑊 = 50.0 N 3.0 m cos(30.0˚)
𝑊 = 130 J
• Practice A (p. 156)
•#1-4
1. 1.50 × 107 J
2. 7.0 × 102 J
3. 1.8 × 103 J
4. 1.1 m
Chapter 5, Section 2
• Energy can be defined as the ability to do work
• This means that when work is done on an object, the amount of
energy the object has increases
• Also, if an object does work on another object, or on the
surrounding environment, then its energy decreases
• Objects in motion have a specific type of energy
• Kinetic energy is the energy of an object that is due to the object’s
motion
• If we apply Newton’s 2nd Law to the mathematical definition of work,
the change in kinetic energy due to work can be found
• In a general sense, the kinetic energy of an object is equal to half the
product of the object’s mass and its speed squared
1
𝐾𝐸 = 𝑚𝑣 2
2
A 7.00 kg bowling ball moves at 3.00 m/s. How fast must a 2.45g
ping-pong ball move in order to have the same kinetic energy as the
bowling ball?
𝐾𝐸bb =
1
𝑚𝑣2
2
Given:
mbb = 7.00 kg
v = 3.00 m/s 𝐾𝐸𝑏𝑏 = 1 7.00 kg 3.00 m/s
2
mpp = 2.45 g
Unknown:
KEbb = ?
vpp = ?
𝐾𝐸bb = 31.5 J
1
𝐾𝐸bb = (𝑚pp)(𝑣pp)2
2
2
𝑣pp =
2𝐾𝐸bb
𝑚pp
𝑣pp =
2(31.5 J)
(0.00245 𝑘𝑔)
𝑣pp = 1.60 × 102 m/s
• Practice B (p. 160)
•#1-5
1. 170 m/s
2. 38.8 m/s
3. The bullet with the greater mass has more kinetic energy by a
ratio of 2 to 1.
4. 2.4 J and 9.6 J; The bullet with the greater speed has more
kinetic energy by a ratio of 4 to 1.
5. 1.6 × 103 kg
• When work is done on object to change its motion, there is a
specific relationship between the work done and the change in
kinetic energy
• The work-kinetic energy theorem states that the net work done
by all the forces acting on an object is equal to the change in
the object’s kinetic energy
𝑊𝑛𝑒𝑡 = ∆𝐾𝐸
𝑊𝑛𝑒𝑡
1
1
2
= 𝑚𝑣𝑓 − 𝑚𝑣𝑖2
2
2
How much work is done by a catcher if he stops a 145 g baseball
moving at 40 m/s when it hits his glove?
What is the average force exerted by the catcher on the ball if
the glove moves back 7.0 cm as a result of the impact?
On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial speed of 2.2
m/s. How far does the sled move if the coefficient of friction between the sled and
the ice is 0.10?
𝑊𝑛𝑒𝑡 = 𝐹𝑛𝑒𝑡 𝑑
𝐹𝑛𝑒𝑡 = 𝐹𝑓 = 𝜇𝑘 𝐹𝑁 = −𝜇𝑘 𝑚𝑔
Given:
𝑣𝑖2
𝑑=−
m = 10.0 kg
2𝜇𝑘 𝑔
vi = 2.2 m/s
1
1
2
2
𝑊
=
∆𝐾𝐸
=
𝑚𝑣
−
𝑚𝑣𝑖2
𝑛𝑒𝑡
vf = 0 m/s
𝑓
m
2
2
2.2
s
μk = 0.10
𝑑=−
m
2 0.10 −9.81 2
s
1
1
Unknown:
𝐹𝑛𝑒𝑡 𝑑 = 𝑚𝑣𝑓2 − 𝑚𝑣𝑖2
2
2
d= ?
𝑑 = 2.5 m
1
𝜇𝑘 𝑚𝑔𝑑 = − 𝑚𝑣𝑖2
2
• Practice C (p. 162)
•#1-4
• Even when objects are not moving, they possess some capability
to gain kinetic energy
• Potential energy is the energy associated with an object due to
the position, shape, or condition of the object
• Two major types of potential energy in mechanics
• Gravitational Potential Energy
• Elastic Potential Energy
• When looking at falling objects, it is easy to identify the kinetic
energy that is being gained, but that energy needs to come
from somewhere
• Gravitational potential energy is the potential energy stored in
the gravitational fields of interacting bodies
• Gravitational potential energy (PEg) is equal to the product of
the mass of an object, the acceleration due to gravity, and the
height of the object
𝑃𝐸𝑔 = 𝑚𝑔ℎ
A 5.8 kg box is placed on a 2.5 m high shelf. What is the potential energy of the
box as it sits on the shelf?
Given:
m = 5.8 kg
g = 9.81m/s2
h = 2.5 m
Unknown:
PEg= ?
𝑃𝐸𝑔 = 𝑚𝑔ℎ
𝑃𝐸𝑔 = 5.8 kg 9.81m/s2 (2.5 m)
𝑃𝐸𝑔 = 140 J
• When looking at falling objects, it is easy to identify the kinetic
energy that is being gained, but that energy needs to come
from somewhere
• Gravitational potential energy is the potential energy stored in
the gravitational fields of interacting bodies
• Gravitational potential energy (PEg) is equal to the product of
the mass of an object, the acceleration due to gravity, and the
height of the object
𝑃𝐸𝑔 = 𝑚𝑔ℎ
• Similar to kinetic energy, there exists a relationship between
work and potential energy
• The amount of work done on an object by gravity or in
opposition to gravity is equal to the change in potential energy
𝑊 = 𝐹𝑑
𝐹 = 𝑚𝑔,
𝑑 = ∆ℎ
𝑊 = 𝑚𝑔∆ℎ = ∆𝑃𝐸𝑔
Chapter 5, Section 3
• When looking at changes in energy in a system, it is necessary
to look at all of the different types of energy
• However, to simplify the total energy, certain types can be
isolated
• Total mechanical energy (ΣME) is the sum of the kinetic and
potential energies
Σ𝑀𝐸 = 𝐾𝐸 + Σ𝑃𝐸
• One of the most fundamental scientific concepts is the
conservation of energy
• Looking at simply mechanical energy, conservation of energy
still generally applies
Σ𝑀𝐸𝑖 = Σ𝑀𝐸𝑓
• This situation only applies if work is not done on an object,
primarily from friction
Starting from rest, a 25.0 kg child zooms down a frictionless slide from an initial
height of 3.00 m. What is her speed at the bottom of the slide?
Given:
m = 25.0 kg
vi = 0 m/s
hi = 3.00 m
hf = 0 m
Unknown:
vf = ?
• Practice E (p. 171)
•#1-3
Chapter 5, Section 4
• So far, we have only looked at the amount of energy gained by
performing work on an object
• The rate at which work is done is also a valuable quantity
• Power is a quantity that measures the rate at which work is done or
energy is transformed
𝑊
𝑃=
∆𝑡
𝑃 = 𝐹𝑣
• The SI unit of power is the watt (W), which is defines as one
joule per second
• Another common unit of power that is horsepower, which is
equal to 746 watts
A 193 kg curtain needs to be raised 7.5 m, at a constant speed, in as close to
5.0 s as possible. The power ratings for three separate motors are listed as 1.0
kW, 3.5 kW, and 5.5 kW. Which motor is best for the job?
Given:
m = 193 kg
Δt = 5.0 s
Δx = 7.5 m
Unknown:
P= ?
𝑊
𝑃=
∆𝑡
𝑊 𝐹∆𝑥 𝑚𝑔∆𝑥
𝑃=
=
=
∆𝑡
∆𝑡
∆𝑡
(193 kg)(9.81 m/s2)(7.5 m)
𝑃=
(5.00 s)
𝑃 = 2839 𝑊 = 2.8 kW
• Practice F (p. 174)
•#1-5