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WORK AND ENERGY
AP Physics B
WORK
WORK
• Work is the transfer of energy to or from a system by the
application of forces exerted on the system by the
environment.
• Work is done on a system by forces from outside the system
(external forces).
• Internal forces—forces between objects within the system —
cause energy transformations within the system, but don’t
change the system’s total energy.
• In order for energy to be transferred as work, the system
must undergo a displacement. In other words, the system
must move during the time the force is applied.
WORK
• Work done by a constant force F in the direction of a
displacement ∆r is equal to the product of these two:
𝑾 = 𝑭∆𝒓
• The unit of work is N∙m. This unit is so important that it has
been given its own name, the joule.
• The joule is the unit of all forms of energy.
• Work is a scalar quantity.
EXAMPLE 1
WORK DONE IN PUSHING A CRATE
Sarah pushes a heavy crate 3.0 m along the floor at constant
speed. She pushes with a constant horizontal force of
magnitude 70 N. How much work does Sarah do on the crate?
𝑊 = 𝐹 ∆𝑟
= 70 3.0
= 210 J
FORCE AT AN ANGLE TO THE DISPLACEMENT
 A force does the greatest possible amount of work on an
object when the force points in the same direction as the
object’s displacement.
 Less work is done when the force acts at an angle to the
displacement.
 Work done by a constant force F at an angle to the
displacement is equal to:
𝑊 = 𝐹 ∆𝑟 cos 𝜃
 Quantities of F and ∆r are always positive, so the sign of W is
determined entirely by the angle between the force and the
displacement.
EXAMPLE 2
WORK DONE IN PULLING A SUITCASE
A strap inclined upward at a 45 ° angle pulls a suitcase through
the airport. The tension in the strap is 20 N. How much work
does the tension do if the suitcase is pulled 100 m at a
constant speed?
 We can use the work equation, with F = T, to find that the
tension does work:
𝑊 = 𝑇∆𝑟 cos 𝜃
= 20 100 cos 45
= 1400 J
KINETIC ENERGY
KINETIC ENERGY
 Consider a car being pulled by a tow rope. The rope pulls with
a constant force F while the car undergoes displacement ∆ r,
so the force does work W = F∆r on the car.
 Ignoring friction and drag, word done by F is transferred
entirely into the car’s energy of motion —its kinetic energy. The
change in the car’s kinetic energy is equal to the work done.
𝑊 = ∆𝐾 = 𝐾 − 𝐾𝑜
 Using kinematics, we can find another expression for the work
done.
1
𝐾 = 𝑚𝑣 2
2
 These two combined result in the work-energy theorem.
1
1
2
𝑊 = 𝐾 − 𝐾𝑜 = 𝑚𝑣 − 𝑚𝑣𝑜 2
2
2
EXAMPLE 3
SPEED OF A BOBSLED AFTER PUSHING
A two-man bobsled has a mass of 390 kg. Starting from rest,
the two racers push the sled for the first 50 m with a net force
of 270 N. Neglecting friction, what is the sled’s speed at the
end of the 50 m?
EXAMPLE 3
SPEED OF A BOBSLED AFTER PUSHING
 From the work -energy theorem, the change in the sled’s kinetic energy
is ∆K = K f – K o = W.
 The sled’s final kinetic energy is thus
𝐾 = 𝐾𝑜 + 𝑊
 Substituting the expressions for kinetic energy and work gives:
1
1
𝑚 𝑣 2 = 𝑚𝑣 𝑜 2 + 𝐹 ∆𝑟
2
2
 V o = 0, so the this equation simplifies to
1
𝑚𝑣 2 = 𝐹 ∆𝑟
2
 This can be solved for final velocity to get
𝑣=
=
2𝐹 ∆𝑟
𝑚
2(270)(50)
= 8.3 𝑚/𝑠
390
POTENTIAL ENERGY
POTENTIAL ENERGY
 When two or more objects in a system interact. It is
sometimes possible to store energy in the system in a way
that the energy can be easily recovered. This sort of stored
energy is called potential energy, since it has the potential to
be converted into other forms of energy.
 Forces due to gravity and springs are special in that they
allow for this storage of energy. Other interaction forces do
not.
 Interaction forces that can store useful energy are called
conservative forces.
 Forces that do not store energy are called nonconservative.
GRAVITATIONAL POTENTIAL ENERGY
 Think of lifting a book at constant speed. Constant speed
means the kinetic energy doesn’t change. Thus, the work done
on the book goes entirely into increasing the potential energy.
Representing potential energy with variable U:
∆𝑈𝑔 = 𝑊
 Because ∆𝑈𝑔 = 𝑈𝑔 − 𝑈𝑔,𝑜 the above equation can be written as
𝑈𝑔 = 𝑈𝑔,𝑜 + 𝑊
 The work done is 𝑊 = 𝐹 ∆𝑟, where ∆𝑟 is the vertical distance
that the book is lifted, or the height, h.
 𝐹 = 𝑚𝑔, so 𝑊 = 𝑚𝑔ℎ, and so
𝑈𝑔 = 𝑈𝑔,𝑜 + 𝑚𝑔ℎ
 The higher the object is lifted, the greater the gravitational
potential energy.
GRAVITATIONAL POTENTIAL ENERGY
 Because the change in potential energy is entirely dependent
on the change in height, we are free to choose a reference
level where we define 𝑈𝑔,𝑜 to be zero, so potential energy is
simplified as
∆𝑈𝑔 = 𝑚𝑔ℎ
 An important conclusion to take from this equation is that
gravitational potential depends only on the height of the
object and not the object’s horizontal position, or not on the
path the object took to get to that position.
EXAMPLE 4
RACING UP A SKYSCRAPER
In the Empire State Building Run -Up, competitors race up the
1576 steps of the Empire State Building, climbing a total
vertical distance of 320 m. How much gravitational potential
energy does a 70 kg racer gain during this race?
 We choose ℎ = 0 at the ground floor of the building, so, at the
top, the racer’s gravitational potential energy is
𝑈𝑔 = 𝑚𝑔ℎ
= 70 9.81 320
= 2.2 × 10 5 J
ELASTIC POTENTIAL ENERGY
 Energy can also be stored in a compressed or extended spring
as elastic potential energy.
 A force is required to compress the spring. This force does
work on the spring, transferring energy to the spring. So, how
much elastic potential energy is stored in the spring can be
found by calculating the amount of work needed to compress
the spring.
∆𝑈𝑠 = 𝑊
HOOKE’S LAW
 When no forces act on a spring to compress or extend it, it will
relax to its equilibrium length.
 If we compress the spring by a displacement 𝑥 with a force 𝐹 , by
Newton’s third law there is also an equal but opposite force that
tries to push the spring back to equilibrium.
 This force, called the spring force, 𝐹𝑠 , is propor tional to the
displacement, 𝑥 , of the end of the spring.
𝐹𝑠 = −𝑘 ∆𝑥
 The minus sign is because the spring force always points in the
opposite direction of the displacement.
 𝑘 is called the spring constant. Spring constants are measured in
N/m.
 The spring constant is a property that characterizes a spring.
Springs with a large spring constant are more difficult to
compress or extend than those with a small spring constant.
SPRING POTENTIAL ENERGY
 When dealing with potential energy, the reference level is the
point where potential energy is equal to zero. This is the point
where a spring is at equilibrium length, so any displacement
∆𝑥 is equal to the final position 𝑥 .
 The spring force is 𝐹𝑠 = −𝑘𝑥, so the applied force is 𝐹𝑎𝑝𝑝 = 𝑘𝑥.
 As the applied force pushes the end of the spring from its
equilibrium position to a final position 𝑥 , the applied force
increases from 0 to 𝑘𝑥.
 This is not a constant force —more compression requires a
larger applied force —so we can’t use 𝑊 = 𝐹 ∆𝑟 to find the work
done.
SPRING POTENTIAL ENERGY
 An average force can be used to calculate the work done on a
spring.
 Because the force varies in magnitude from 0 to 𝑘𝑥, the
1
average force used to compress the spring is 𝐹𝑎𝑣𝑔 = 𝑘𝑥 .
2
 Thus the work done by the applied force is
1
1
𝑊 = 𝐹𝑎𝑣𝑔 ∆𝑟 = 𝐹𝑎𝑣𝑔 𝑥 =
𝑘𝑥 𝑥 = 𝑘 𝑥 2
2
2
 This work is stored as potential energy in the spring, so
elastic potential energy for a spring is
1
𝑈𝑠 = 𝑘𝑥 2
2
EXAMPLE 5
PULLING BACK ON A BOW
An archer pulls back the string on her bow to a distance of 70
cm from its equilibrium position. To hold the string at this
position takes a force of 140 N. How much elastic potential
energy is stored in the bow?
 A bow is an elastic material, so we will model it as obeying
Hooke’s law, 𝐹𝑠 = −𝑘𝑥 , where 𝑥 is the distance the string is
pulled back.
 We can use the force required to hold the string, and the
distance it is pulled back, to find the bow’s spring constant 𝑘 .
 Then, we can use the equation for elastic potential energy.
EXAMPLE 5
PULLING BACK ON A BOW
 From Hooke’s law, the spring constant is
𝐹
𝑘=
𝑥
140
=
0.70
= 200 𝑁/𝑚
 Then the elastic potential energy of the flexed bow is
1
𝑈𝑠 = 𝑘 𝑥 2
2
= 1/2(200)(0.70) 2
= 49 J
THERMAL ENERGY
 Thermal energy is related to the microscopic motion of the
molecules of an object, and in turn is related to the kinetic
and elastic potential energy of said molecules.
 Increasing an object’s thermal energy also increases its
temperature.
 Imagine a box pulled across a floor at a constant speed. The
force required to pull the box does work on the system, which
therefore transfers energy into the system.
 Because the speed is constant, there is no change in kinetic
energy.
 Because the box’s height is constant, there is no change in
gravitational potential energy.
 The increased energy must be in the form of thermal energy.
THERMAL ENERGY
 Increase in thermal energy is a general feature of any system
where friction between sliding objects is present. Change in
thermal energy, ∆𝐸𝑡ℎ , is related to friction, 𝑓 , and
displacement, ∆𝑥 , as:
∆𝐸𝑡ℎ = 𝑓 ∆𝑥
 This equation does not only apply to object’s being acted on
by an external forces. For example, thermal energy can be
created by a moving object sliding to a halt on a rough
surface.
EXAMPLE 6
THERMAL ENERGY BY RUBBING
A 0.30 kg block of wood is rubbed back and forth against a
wood table 30 times in each direction. The block is moved 8.0
cm during each stroke and pressed against the table with a
force of 22 N. How much thermal energy is created in the
process? The coef ficient of friction between two wooden
surfaces is μ = 0.20.
 Use NSL in the y-direction to calculate normal force.
 Use 𝑓 = 𝜇𝑛 to calculate friction force.
 Use ∆𝐸𝑡ℎ = 𝑓 ∆𝑥 to calculate thermal energy created.
EXAMPLE 6
THERMAL ENERGY BY RUBBING
 The block is not accelerating in the y-direction, so Newton’s
second law gives
𝐹𝑦 = 𝑛 − 𝑤 − 𝐹 = 0
𝑛 = 𝑤 + 𝐹 = 𝑚𝑔 + 𝐹
= 0.30 9.81 + 22
= 25 N
 Friction force is then
𝑓 = 𝜇𝑛 = 0.20 25
= 5.0 N
 Total displacement of the block is
2 × 30 × 8.0 = 480 cm = 4.8 m
 So thermal energy created is
∆𝐸𝑡ℎ = 𝑓 ∆𝑥
= 5.0 4.8
= 24 J
LAW OF CONSERVATION
OF ENERGY
ENERGY TRANSFER AND CONSERVATION
 Recall that work is the transfer of energy to or from a system
by a force acting on the system. The total energy of a system
changes by the amount of work done on it.
 Like conservation of momentum, if we define a system that
has no external forces acting on it, then the total energy of
this isolated system is conserved, and change in energy is
zero.
 This applies to every form of energy (kinetic, chemical,
nuclear, etc.), but we often only concern ourselves with the
kinetic and potential energy of a system because these are
the energies associated with the motion and position of an
object. Together these energies are referred to as the
mechanical energy of the system.
CONSERVATION OF MECHANICAL ENERGY
 If the system is isolated and there’s no friction or friction is
negligible, the mechanical energy is conserved:
𝐾 + 𝑈𝑔 + 𝑈𝑠 = 𝐾𝑜 + 𝑈𝑔,𝑜 + 𝑈𝑠,𝑜
 If the system is isolated but there is friction within the
system, the total energy is conserved:
𝐾 + 𝑈𝑔 + 𝑈𝑠 + ∆𝐸𝑡ℎ = 𝐾𝑜 + 𝑈𝑔,𝑜 + 𝑈𝑠,𝑜
EXAMPLE 7
HITTING THE BELL
A the county fair, Katie tries her hand at the ring -the-bell
attraction. She swings the mallet hard enough to give the ball
an initial upward speed of 8.0 m/s. Will the ball ring the bell,
3.0 m from the bottom? Assume the track on which the ball
moves is frictionless.
 Identify the mechanical energies involved in this system, and
set the sum of the initial values of these energies equal to the
sum of the final values of these energies.
 Simplify this expression using values and/or the energies’
corresponding equations.
 Solve the expression for height, ℎ.
EXAMPLE 7
HITTING THE BELL
 Once the ball is in motion, only kinetic energy and gravitational
potential energy are involved, so conser vation of mechanical energy is
𝐾 + 𝑈𝑔 = 𝐾𝑜 + 𝑈𝑔,𝑜
 Substituting the equations for kinetic and gravitational potential
energy gives
1
1
𝑚 𝑣 2 + 𝑚𝑔ℎ = 𝑚 𝑣 𝑜 2 + 𝑚𝑔ℎ 𝑜
2
2
 The ball star ts at ℎ = 0, and at its highest point, 𝑣 = 0, so the above
expression simplifies to
1
𝑚𝑔ℎ = 𝑚 𝑣 𝑜 2
2
 Solving this for height gives
𝑣𝑜 2
ℎ=
2𝑔
(8.0) 2
=
2(9.81)
= 3.3 m
EXAMPLE 8
SPEED AT THE BOTTOM OF A SLIDE
Still at the county fair, Katie tries the water slide. The starting
point is 9.0 m above the ground. She pushes of f with an initial
speed of 2.0 m/s. If the slide is frictionless, how fast will Katie
be traveling at the bottom?
EXAMPLE 8
SPEED AT THE BOTTOM OF A SLIDE
 Conservation of mechanical energy gives
𝐾 + 𝑈𝑔 = 𝐾𝑜 + 𝑈𝑔,𝑜
Or
1
1
2
𝑚𝑣 + 𝑚𝑔ℎ = 𝑚𝑣𝑜 2 + 𝑚𝑔ℎ 𝑜
2
2
 Taking ℎ = 0,
1
1
2
𝑚𝑣 = 𝑚𝑣𝑜 2 + 𝑚𝑔ℎ 𝑜
2
2
 Which we can solve for 𝑣
𝑣 = 𝑣𝑜 2 + 2𝑔ℎ 𝑜
= (2.0) 2 +2(9.81)(9.0)
= 13 m/s
EXAMPLE 9
SPEED OF A SPRING-LAUNCHED BALL
A spring-loaded toy gun is used to launch a 10 g plastic ball.
The spring, which has a spring constant of 10 N/m, is
compressed by 10 cm as the ball is pushed into the barrel.
When the trigger is pulled, the spring is released and shoots the
ball back out. What is the ball’s speed as it leaves the barrel?
Assume that friction is negligible, and the spring is massless.
EXAMPLE 9
SPEED OF A SPRING-LAUNCHED BALL
 Energy conservation equation is 𝐾 + 𝑈𝑠 = 𝐾𝑜 + 𝑈𝑠 ,𝑜 . We can use
elastic potential of the spring to write this as
1
1
1
1
𝑚𝑣 2 + 𝑘𝑥 2 = 𝑚𝑣𝑜 2 + 𝑘 𝑥 𝑜 2
2
2
2
2
 We know that 𝑥 = 0 and 𝑣𝑜 = 0, so this simplifies to
1
1
𝑚𝑣 2 = 𝑘 𝑥 𝑜 2
2
2
 Which we can solve for 𝑣 to give
𝑣=
=
𝑘 𝑥𝑜 2
𝑚
(10)(−0.10) 2
0.010
= 3.2 m/s
POWER
ENERGY, WORK AND TIME
 In many situations, it is necessar y to know how quickly energy is
transformed or transferred. This implies a rate of transformation of
energy.
 The amount of energy ∆𝐸 transformed over a time inter val ∆𝑡 is called
the power 𝑃 and is defined as
∆𝐸
𝑃=
∆𝑡
 The unit of power is the watt, which is defined as 1 watt = 1 W = 1 J/s.
 Recall that ∆𝐸 = 𝑊 so a more conventional way of representing power
is
𝑊
𝑃=
∆𝑡
 One other useful was of representing the power formula is
𝑊
𝐹 ∆𝑥
∆𝑥
𝑃=
=
=𝐹
= 𝐹𝑣
∆𝑡
∆𝑡
∆𝑡
𝑃 = 𝐹𝑣
EXAMPLE 10
POWER TO PASS A TRUCK
Your 1500 kg car is behind a truck traveling at 60 mph (27
m/s). To pass it, you speed up to 75 mph (34 m/s) in 6.0 s.
What engine power is required to do this?
 Calculate the work required to speed up the car to the desired
final velocity. Recall that 𝑊 = ∆𝐸 .
 Calculate power using work and time.
EXAMPLE 10
POWER TO PASS A TRUCK
𝑊 = ∆𝐾 = 𝐾 − 𝐾𝑜
1
1
2
= 𝑚𝑣 − 𝑚𝑣𝑜 2
2
2
1
= 𝑚 𝑣 2 − 𝑣𝑜 2
2
1
=
1500 34 2 − 27 2
2
= 3.20 × 10 5 J
 To transform this amount of energy in 6 s, the power required
is
∆𝐾 3.20 × 10 5
𝑃=
=
= 53,000 W = 53 kW
∆𝑡
6.0
EXAMPLE 11
STOPPING A RUNAWAY TRUCK
A truck’s brakes can overheat and fail while descending
mountain highways, leading to an extremely dangerous runaway
truck. Some highways have runaway-truck ramps to safely bring
out-of-control trucks to a stop. These uphill ramps are covered
with a deep bed of gravel. The uphill slope and the large
coef ficient of rolling friction as the tires sink into the gravel
bring the truck to a safe halt.
EXAMPLE 11
STOPPING A RUNAWAY TRUCK
A 22,000 kg truck heading down a 3.5 ° slope at 20 m/s
suddenly has its brakes fail. Fortunately, there’s a runaway truck ramp 600 m ahead. The ramp slopes upward at an angle
of 10°, and the coef ficient of rolling friction between the
truck’s tires and the loose gravel is 𝜇 = 0.40. Ignore air
resistance and rolling friction as the truck rolls down the
highway.
a. Use conservation of energy to find how far along the
ramp the truck travels before stopping.
b. By how much does the thermal energy of the truck and
ramp increase as the truck stops?