Transcript Chapter 5 PowerPoint - Derry Area School District

AP Physics
Chapter 5
Work and Energy
Chapter 5: Work and Energy
5.1
5.2
5.3
5.4
5.5
5.6
Work Done by a Constant Force
Work Done by a Variable Force
The Work-Energy Theorem: Kinetic Energy
Potential Energy
The Conservation of Energy
Power
Chapter 5: Learning Objectives
Work and the Work-energy Theorem
a)
Students will understand the definition of work, including when it is
positive, negative, or zero, so they can:
1)
2)
3)
Calculate the work done by a specified constant force on an object
that undergoes a specified displacement.
Relate the work done by a force to the area under a graph of force as
a function of position, and calculate this work in the case where the
force is a linear function of position.
Use the scalar product operation to calculate the work performed by
a specified constant force F on an object that undergoes a
displacement in a plane.
Chapter 5: Learning Objectives
Work and the Work-energy Theorem
a) Students will understand and be able to apply the workenergy theorem, so they can:
1)
2)
3)
Calculate the change in kinetic energy or speed that results from
performing a specified amount of work on an object.
Calculate the work performed by the net force, or by each of the
forces that make up the net force, on an object that undergoes a
specified change in speed or kinetic energy.
Apply the theorem to determine the change in an object’s kinetic
energy and speed that results from the application of specified
forces, or to determine the force that is required in order to bring an
object to rest in a specified distance.
Chapter 5: Learning Objectives
Forces and Potential Energy
a) Students will understand the concept of potential energy, so
they can:
1)
2)
Write an expression for the force exerted by an ideal spring and for
the potential energy of a stretched or compressed spring.
Calculate the potential energy of one or more objects in a uniform
gravitational field.
Chapter 5: Learning Objectives
Conservation of Energy
a) Students will understand the concept of mechanical energy
and of total energy, so they can:
1)
2)
Describe and identify situations in which mechanical energy is
converted to other forms of energy.
Analyze situations in which an object’s mechanical energy is changed
by friction or by a specified externally applied force.
b) Students will understand conservation of energy, so they can:
1)
2)
3)
Identify situations in which mechanical energy is or is not conserved.
Apply conservation of energy in analyzing motion of systems of
connected objects, such as an Atwood’s machine.
Apply conservation of energy in analyzing the motion of objects that
move under the influence of springs.
Chapter 5: Learning Objectives
Power
a) Students will understand the definition of power, so they can:
1)
Calculate the power required to maintain the motion of an object with
constant acceleration (e.g., to move an object along a level surface, to
raise an object at a constant rate, or to overcome friction for an object
that is moving at a constant speed.)
2)
Calculate the work performed by a force that supplies constant power,
or the average power supplied by a force that performs a specified
amount of work.
Homework for Chapter 5
• Read Chapter 5
• HW 5.A : p. 165 3, 6-11, 16.
• HW 5.B: p. 167: 19-26, 28, 31.
• HW 5.C: pp.167-168: 32, 34, 38-40, 45-52.
• HW 5.D: p.168: 60-64, 66, 67, 70.
• HW 5.E: p.170: 72,73,75-81.
5.1 Work Done by a Constant Force
Warmup: Work, Work, Work!
Physics Daily WarmUps # 76
The quantity work is defined as the product of the force applied to an object multiplied
by the distance through with the force is applied. This means that if no displacement of
the object occurs, no work is done on the object even though the force applied may be
quite large.
Reference to the work we do is a large part of our daily conversation. We all have
opinions as to which jobs require more of less work. However, this common usage of the
word work does not always match up with the physics definition. Use the column on the
left to arrange the list of jobs in order of hardest work to easiest in your opinion. Use the
column on the right to arrange them in order of most work to least work according to
physics.
Opinion (hardest to easiest)
Jobs
(hardest)
store clerk
furniture mover
accountant
package delivery driver
package delivery driver
store clerk
furniture mover
accountant
(easiest)
Physics (most to least)
5.1 Work Done by a Constant Force
Definition: Work done by a constant force is equal to the product of the
magnitudes of the displacement and the component of the force parallel to the
displacement.
• W = (F cosӨ)d, where
• F is the magnitude of the force vector
• d is the magnitude of the displacement vector
• Ө is the angle between the two vectors (Warning – this angle is not
necessarily measured from the horizontal)
• When the angle is zero, cos Ө = 1, and W = F·d
• When the angle is 180°, cos Ө = -1 and W = - F·d (yes, negative work!)
• example: the force of brakes to slow down a car.
• Although force and displacement are vectors, work is a scalar quantity.
• The SI unit of work is the N·m, which is called a joule (J).
5.1 Work Done by a Constant Force
d
d
5.1 Work Done by a Constant Force
If there is no
displacement, no
work is done: W = 0.
For a constant force in the
same direction as the
displacement, W = Fd.
For a constant force at an
angle to the displacement,
W = (FcosӨ) d
5.1 Work Done by a Constant Force
Example 1: A 500 kg elevator is pulled upward by a constant force of 5,500 N for a distance of
50.0 m.
Fup
d
a) Find the work done by the upward force.
b) Find the work done by the gravitational force.
c) Find the work done by the net force
(the net work done on the elevator).
w (weight)
Solution:
Given:
Fup = 5,500 N
w = mg = (500 kg)(9.80 m/s2) = 4,900 N
d = 50.0 m
Unknown: a) Wup b) Wgrav c) Wnet
a)
The displacement is upward and the upward force is (of course) upward, so the angle
between them is zero. Therefore,
Wup = (F cos Ө) d = (Fup cos 0°) d = (5,500 N)(1)(50.0m) = 2.75 x 105 J
b)
The angle between displacement and the gravitational force (weight vector) is 180°.
So, Wgrav = (w cos 180°) d = (4,900 N)(-1)(50.0 m) = -2.45 x 105 J
c)
The work done by the net force is equal to the net work done on the elevator.
Wnet = Wup + Wgrav = 2.75 x 105 J + (-2.45 x 105 J) = 3.0 x 104 J
Note: Wnet is also equal to Wnet = (Fnet cos Ө) d, where Fnet = F - w
5.1 Work Done by a Constant Force
• The work done by a force is equal to the area under the curve in a force vs. position
graph.
Example 2: A force moves an object in the direction of the force. The graph shows
the force versus the object’s position. Find the net work done when the object moves
a) from 0 to 2.0 m
F (N)
30
b) from 2.0 to 4.0 m
c) from 4.0 to 6.0 m
20
d) from 0 to 6.0 m
10
0
Solution:
0.0
1.0
2.0 3.0
4.0
5.0 6.0
s (m)
Work done is equal to the area under the curve.
a) The area under the curve from 0 to 2.0 m is a triangle. The area is
½ (base x height), so W0-2 = ½ (2.0 m - 0)(20 N) = 20 J
b) The area under the curve is W2-4 = (4.0 m – 2.0 m)(20 N) = 40 J
c) The area under the curve is W4-6 = ½ (6.0 m – 4.0 m)(20 N) = 20 J
d) The sum of the areas is W0-2 + W2-4 + W4-6 = 20 J + 40 J + 20 J = 80 J
5.1 Work Done by a Constant Force: Check for Understanding
5.1 Work Done by a Constant Force: Check for Understanding
5.1 Work Done by a Constant Force: Check for Understanding
5.1 Work Done by a Constant Force: Check for Understanding
5.1 Work Done by a Constant Force: Check for Understanding
5.1 Work Done by a Constant Force: Check for Understanding
Homework for
Section 5.1
• HW 5.A : p. 165 3, 6-11,16.
5.2 Work Done by a Variable Force
5.2 Work Done by a Variable Force
• Forces that do work are not always constant.
example: To move a sofa, you push harder and harder until you can
overcome static friction.
example: Stretching or compressing a spring.
• As the spring gets stretched (or compressed) farther and farther, the
restoring force of the spring (the force that opposes stretching) gets greater
and greater.
•The applied force is directly proportional to the change in length of the
spring from its unstretched length.
• The spring constant, k, describes the stiffness of the spring.
•The resulting equation is known as Hooke’s Law: Fs = -kx
• The minus sign indicates the spring force is opposite to the direction of
displacement.
5.2 Work Done by a Variable Force
Spring Force
a) An applied force F
stretches the spring,
and the spring exerts
an equal and opposite
force Fs on the hand.
b) The magnitude of the
force depends on the
change in the spring’s
length. This is often
referenced to the
position of a mass on
the end of the spring.
5.2 Work Done by a Variable Force
• The work done by an external force in stretching or compressing a spring
(overcoming the spring force) is
W = ½ kx2
where x is the stretch or compression distance and k is the spring constant.
F
F = kx
work = area under the curve
area of the triangle = ½ (base x height)
work = ½ (x)(kx) = ½ kx2
x
5.2 Work Done by a Variable Force
The reference position xo is chosen for convenience.
a) xo may be chosen to be at the end of the
spring in its unloaded position.
b) xo may be at the equilibrium position
when a mass is suspended on a spring.
This is convenient when the mass
oscillates up and down on the spring.
5.2 Work Done by a Variable Force
Example:
A spring of spring constant 20 N/m is to be compressed by 0.10 m.
a) What is the maximum force required?
b) What is the work required?
Solution:
Given:
Unknown:
k = 20 N/m
a) Fs(max)
x = -0.10 m
b) W
a) From Hooke’s law, the maximum force corresponds to the
maximum compression.
Fs(max) = -kx = -(20 N/m)(-0.10 m) = 2.0 N
b) W = ½ kx2 = ½ (20 N/m)(-0.10 m)2 = 0.10 J
5.2 Work Done by a Variable Force: Check for Understanding
4
3
2
1
5.2 Work Done by a Variable Force
Homework for
Section 5.2
• HW 5.B: p. 167: 19-26, 28, 31.
The Work Energy Theorem
5.3 Kinetic Energy
5.4 Potential Energy
5.3 The Work-Energy Theorem: Kinetic Energy
v = velocity
5.3 The Work-Energy Theorem: Kinetic Energy
5.3 The Work-Energy Theorem: Kinetic Energy
Example:
An object hits a wall and bounces back with half of its original speed.
What is the ratio of the final kinetic energy to the initial kinetic energy?
Solution:
Given:
Unknown:
v = vo/2
K/Ko
Ko = 1/2 mvo2
and
So, K/ Ko = 1/4
K = ½ mv2 = ½ m (vo/2)2 = ¼ (1/2 mvo2)
5.3 The Work-Energy Theorem: Kinetic Energy
• The work-energy theorem offers a convenient way to solve kinematics
and dynamics problems.
Example:
The kinetic friction force between a 60.0 kg object and a horizontal
surface is 50.0 N. If the initial speed of the object is 25.0 m/s, what
distance will it slide before coming to a stop?
Given: m = 60.0 kg
Unknown: d
vo = 25.0 m/s
v=0
fk = 50.0 N
The kinetic friction force fk is the only unbalanced force and the angle
between the friction force and the displacement is 180°.
Wnet = (FcosӨ) d
= (fk cos 180°) d
= (50.0 N)(-1) d
and, Wnet = K- Ko, = ½ mv2 – ½ mvo2
= ½(60.0 kg)(0)2 – ½ (60.0 kg)(25.0 m/s)2
Solving, d = 3.75 x 102 m
5.3 The Work-Energy Theorem: Kinetic Energy: Check for Understanding
5.3 The Work-Energy Theorem: Kinetic Energy: Check for Understanding
5.3 The Work-Energy Theorem: Kinetic Energy: Check for Understanding
5.3 The Work-Energy Theorem: Kinetic Energy: Check for Understanding
5.4 The Work-Energy Theorem: Potential Energy
5.4 The Work-Energy Theorem: Potential Energy
5.4 The Work-Energy Theorem: Potential Energy
The work done in lifting an
object is equal to the change in
gravitational potential energy.
W = F∆ h = m g (h – ho)
5.4 The Work-Energy Theorem: Potential Energy
Kinetic and Potential Energy
Throwing a ball in the air
converts its kinetic energy to
potential energy, and back
again.
5.4 The Work-Energy Theorem: Potential Energy
Reference Point and Change in Potential Energy
a) The reference point (zero height) you choose may give rise to negative
potential energy. This is call a potential energy well.
b) The well may be avoided by choosing a new reference point.
5.4 The Work-Energy Theorem: Potential Energy
5.4 The Work-Energy Theorem: Potential Energy
5.4 The Work-Energy Theorem: Potential Energy
Example 5.6: A 10.0 kg object is moved from the second floor of a house 3.00
m above the ground to the first floor 0.30 m above the ground. What is the
change in gravitational potential energy?
5.4 The Work-Energy Theorem: Potential Energy
Example 5.7: A spring with a spring constant of 15 N/m is initially compressed
by 3.0 cm. How much work is required to compress the spring an additional 4.0
cm?
5.4 The Work-Energy Theorem: Potential Energy: Check for Understanding
5.4 The Work-Energy Theorem: Potential Energy: Check for Understanding
5.4 The Work-Energy Theorem: Potential Energy
5.4 The Work-Energy Theorem: Potential Energy: Check for Understanding
5.4 The Work-Energy Theorem: Potential Energy: Check for Understanding
5.4 The Work-Energy Theorem: Potential Energy: Check for Understanding
Or, use opp = hyp (sin 30) = 0.5 m
Homework for
Sections 5.3 & 5.4
• HW 5.C: pp.167-168: 32, 34, 38-40, 45-52.
5.5 The Conservation of Energy
Warmup: Road Hazard
5.5 The Conservation of Energy
5.5 The Conservation of Energy
Remember, mechanical energy is kinetic or potential.
If there is a nonconservative force such as friction doing work on the
system, the total mechanical energy of the system is not conserved, but
total overall energy is conserved.
5.5 The Conservation of Energy
Example 5.8: A 70 kg skier starts from rest on the top of a 25 m high slope.
What is the speed of the skier on reaching the bottom of the slope? (Neglect
friction)
5.5 The Conservation of Energy
• If there is a nonconservative force doing work in a system, the total
mechanical energy of the system is not conserved.
• However, the total energy (not mechanical!) of the system is still conserved.
• Some of the total energy is used to overcome the work done by the
nonconservative force.
• The difference in mechanical energy is equal to the work done by the
nonconservative force, that is
Wnc = E – Eo = E
5.5 The Conservation of Energy
Example 5.10: In Example 5.8, if the work done by the kinetic friction force is
-6.0 x 103 J (the work done by kinetic friction force is negative because the
angle between the friction force and the displacement is 180˚). What is the
speed of the skier at the bottom of the slope?
5.5 The Conservation of Energy
Example 5.9: A 1500 kg car moving at 25 m/s hits an initially uncompressed
horizontal spring with spring constant 2.0 x 106 N/m. What is the maximum
compression of the spring? (Neglect the mass of the spring.)
5.5 The Conservation of Energy: Check for Understanding
5.5 The Conservation of Energy: Check for Understanding
5.5 The Conservation of Energy: Check for Understanding
5.5 The Conservation of Energy: Check for Understanding
5.5 The Conservation of Energy: Check for Understanding
5.5 The Conservation of Energy: Check for Understanding
5.5 The Conservation of Energy: Check for Understanding
5.5 The Conservation of Energy: Check for Understanding
5.5 The Conservation of Energy
Explanation of the Math for Pulley Problem:
Uo = K
Just block A: K = ½ m vA2
Both blocks together: K = ½ 3m vAB2
Since the change in energy is the same for
both cases, set them equal to each other:
½ 3m vAB2 = ½ m vA2
vAB2 = vA2/ 3
vAB = vA / √3
Homework for
Section 5.5
• HW 5.D: p.168: 60-64, 66, 67, 70.
5.6 Power
Warmup: Rock-It
5.6 Work
5.6 Power
A common British unit of power is the horsepower (hp) and 1 hp = 746 W.
5.6 Power
5.6 Power
5.6 Power
Example 5.11: A 1500 kg car accelerates from 0 to 25 m/s in 7.0 s. What is
the average power delivered to the car by the engine? Ignore all frictional and
other losses.
5.6 Power: Check for Understanding
5.6 Power: Check for Understanding
5.6 Power: Check for Understanding
5.6 Power: Check for Understanding
Homework for
Section 5.6
• HW 5.E: p.170: 72-73, 75-81.
Chapter 5 Formulas
K = ½ mv2
kinetic energy
Ug = mgh
gravitational potential energy
W = F r cos 
work
Pave = W
t
power as defined by work over time
P = Fv cos 
power if the force is making an angle  with v
Fs = -kx
Hooke’s Law; relates spring force with spring constant and
change in length
Us = ½ kx2
elastic potential energy