Transcript What is the increase in the elastic potential energy

Physics in Context
Unit 7
Work and Energy 3
Key Objectives
 State the equation for calculating the
elastic potential energy of a spring, and
solve problems using this equation.
 Solve problems involving the inclined
plane as a simple machine.
 List other types of simple machines.
7.3 ENERGY
Elastic Potential Energy and Springs
• Consider the arrangement shown in the
diagram:
7.3 ENERGY
Elastic Potential Energy and Springs
• Here a spring is attached to a wall and to a
mass resting on a horizontal, fric tionless
table.
7.3 ENERGY
Elastic Potential Energy and Springs
• If we apply a force on the mass and
displace it to the right, we have done work.
7.3 ENERGY
Elastic Potential Energy and Springs
• This work has been converted into the
spring’s potential energy, a quantity we
call elastic potential energy.
7.3 ENERGY
Elastic Potential Energy and Springs
• We can calculate the work done in
stretching the spring as follows.
7.3 ENERGY
Elastic Potential Energy and Springs
• We know that springs obey Hooke’s law, a
graph of which is shown below.
7.3 ENERGY
Elastic Potential Energy and Springs
• Since the area under this graph equals the
work done, and we know that the area of a
triangle is equal to 1/2 (base height), we
use the equations
7.3 ENERGY
Elastic Potential Energy and Springs
• Then we have
7.3 ENERGY
Elastic Potential Energy and Springs
• and the potential energy of the spring
(PEs) is given by:
7.3 ENERGY
Elastic Potential Energy and Springs
• PROBLEM
• A spring whose constant is 2.0 newtons
per meter (k) is stretched 0.40 meter (x)
from its equilibrium position.
• What is the increase in the elastic
potential energy (∆PEs) of the spring?
7.3 ENERGY
Elastic Potential Energy and Springs
•
•
•
•
SOLUTION
∆PEs = ½ kx2
∆PEs = ½ (2.0 N/m)(0.40m)2
∆PEs = 0.16 J
Assessment Question 1
• A spring whose constant is 1.2 N/m (k) is
stretched 0.65 m (x) from its equilibrium
position.
• What is the increase in the elastic
potential energy (∆PEs) of the spring?
• ∆PEs = ½ kx2 = ½ (1.2 N/m)(0.65 m)2 =
A.0.25 J
B.1.6 J
C.4.7 J
D.20 J
7.3 ENERGY
Elastic Potential Energy and Springs
• SOLUTION
• What would happen if we released the
spring after stretching it?
• The force exerted on the mass by the
spring (the restoring force) would displace
the mass toward the wall (to the left).
7.3 ENERGY
Elastic Potential Energy and Springs
• SOLUTION
• The mass would then overshoot its
equilibrium position and compress the
spring.
• In turn, the spring would exert a, force on
the mass away from the wall (to the right).
7.3 ENERGY
Elastic Potential Energy and Springs
• SOLUTION
• Since friction is absent, this back-and-forth
motion, namely, SHM would continue
indefinitely.
7.3 ENERGY
Elastic Potential Energy and Springs
• SOLUTION
• As the spring moved back and forth, there
would be a continual exchange between
kinetic and potential energies, as shown in
the diagram.
7.3 ENERGY
Elastic Potential Energy and Springs
• SOLUTION
Assessment Question 2
If we released the spring after stretching it.
All of the following are true EXCEPT?
A. The force exerted on the mass by the spring (the
restoring force) would displace the mass toward the wall.
B. The mass would then overshoot its equilibrium position
and compress the spring, then the spring would exert a,
force on the mass away from the wall (to the right).
C. If friction is absent, this back-and-forth motion, namely,
Simple Harmonic Motion would continue indefinitely.
D. As the spring moved back and forth, the kinetic and
potential energies remain constant.
7.3 ENERGY
Elastic and Inelastic Collisions
• Imagine a ball bouncing repeatedly on a
sidewalk, as illustrated below.
7.3 ENERGY
Elastic and Inelastic Collisions
• After each successive bounce, the ball’s
height above the ground diminishes;
eventually the ball comes to rest on the
ground.
7.3 ENERGY
Elastic and Inelastic Collisions
• When the ball is on the ground, part of its
kinetic energy is lost and there is an incomplete
conversion to potential energy, a phenomenon
known as a partially inelastic collision.
7.3 ENERGY
Elastic and Inelastic Collisions
• If the ball stuck to the ground after its first
bounce, the collision would be termed totally
inelastic.
7.3 ENERGY
Elastic and Inelastic Collisions
• If, however, the ball rose repeatedly to the
same height, the collisions would be termed
elastic.
Assessment Question 3
If a ball is bouncing repeatedly on a sidewalk.
All of the following are true EXCEPT?
A. After each successive bounce, part of its kinetic energy is
lost and there is an incomplete conversion to potential
energy and the ball’s height above the ground diminishes
and the ball comes to rest on the ground which is known
as a partially inelastic collision.
B. If the ball stuck to the ground after its first bounce, the
collision would be termed totally inelastic collision.
C. If the ball rose repeatedly to the same height, the
collisions would be termed elastic collisions.
D. If the ball increased its height after each bounce this
would be termed a superelastic collision.
7.3 ENERGY
Elastic and Inelastic Collisions
• In an elastic collision both kinetic energy
and momentum are conserved, as outlined
below:
Assessment Question 4
• Two objects of the same mass moving at 10
m/s (vi) in opposite directions collide with 500
J (KEi) of force and stop on impact. What is
the mass of each object?
• KE1i + KE2i = KE1f + KE2f
• ½ mv1i2 + ½ mv2i2 = 500 J
• ½ m∙(10 m/s)2 + ½ m∙(10 m/s)2 = 500 J
• (100 m2/s2) m = 500 J m = 500 J / 100 m2/s2
A.5 kg
C. 2500 kg
B.100 kg
D. 50000 kg
7.3 ENERGY
Elastic and Inelastic Collisions
• In an elastic collision both kinetic energy
and momentum are conserved, as outlined
below:
7.3 ENERGY
Elastic and Inelastic Collisions
• In inelastic collisions, the kinetic energy
that is “lost” is converted into internal
energy (Q) (heat) of the objects by
frictional forces.
7.3 ENERGY
Elastic and Inelastic Collisions
• In a non ideal mechanical system, the total
energy is constant.
7.3 ENERGY
Elastic and Inelastic Collisions
• This form of energy will be discussed in
Chapter 8.
• Systems upon which frictional or other
external forces act are called non ideal
mechanical systems
Assessment Question 5
If a ball is bouncing repeatedly on a sidewalk.
All of the following are true EXCEPT?
A. In an elastic collision both kinetic energy and momentum
are conserved.
B. In inelastic collisions, the kinetic energy that is “lost” is
converted into internal energy (heat) (Q) of the objects by
frictional forces.
C. In a non ideal mechanical system, the final total energy is
greater than the initial total energy.
D. Systems upon which frictional or other external forces act
are called non ideal mechanical systems
7.3 ENERGY
Elastic and Inelastic Collisions
• The equation representing the laws of
conservation of energy on page 129 can
be expanded to include Q, and the total
energy (ET) is given by
Assessment Question 6
• An object with a mass of 5 kg (m) falls down
the side of a cliff from a distance of 10 m (h)
from the ground at 10 m/s (v). g = 9.8 m/s2
• If the total energy is 1000 J (ET), what is
the energy lost to heat (Q)?
• ET = PE + KE + Q
• Q = ET - (PE + KE) = ET – (mgh + ½ mv2)
Q = 1000 J -[(5 kg∙9.8 m/s2∙10 m) +½(5 kg∙(10
m/s))2]
A. 55 J
C. 260 J
B. 140 J
D. 530 J
7.4 SIMPLE MACHINES AND
WORK
• A simple machine is a device that allows
work to be done and offers an advantage
to the user.
• We need to use machines because the
force available to us is not always
adequate.
7.4 SIMPLE MACHINES AND
WORK
• For example, we cannot lift a 1000-newton
box using our muscles alone.
• However, a simple machine such as a
pulley or an inclined plane may allow us to
accomplish this task.
7.4 SIMPLE MACHINES AND
WORK
• Suppose we used a system of pulleys and
exerted a downward force of 250 newtons
to raise the 1000-newton box.
• We say that the pulley offers us a
mechanical advantage of 4.
7.4 SIMPLE MACHINES AND
WORK
• We calculate the mechanical advantage
from this relationship:
(Fout /Fin)
• In nature, however, one never gets
something for nothing and we pay for this
mechanical advantage by pulling the rope
of the pulley four times as far as the box is
lifted.
7.4 SIMPLE MACHINES AND
WORK
• Since energy must be conserved, we
could have deduced this fact from the
ideal relationship that work input must
equal work output:
7.4 SIMPLE MACHINES AND
WORK
• The relationship given above assumes
that no friction is present in the machine,
and the mechanical advantage is called
the ideal mechanical advantage (IMA).
7.4 SIMPLE MACHINES AND
WORK
• In reality, all machines have friction, so the
actual mechanical advantage (AMA) is
always less than the IMA.
• The IMA is based on the relative distances
involved in the operation of the machine
(din /dout) while the AMA is based on the
ratio Fout / Fin .
Assessment Question 7
All of the following are true EXCEPT?
A. A simple machine is a device that allows work to be done and offers
an advantage to the user.
B. If a person uses a system of pulleys and exerted a downward force
of 250 newtons to raise the 1000-newton box, the pulley system
offers a mechanical advantage of 4 and the person pulls the rope ¼
the distance that the box is lifted.
C. If no friction is present in the machine, the mechanical advantage is
called the Ideal Mechanical Advantage (IMA). In reality, all machines
have friction, so the Actual Mechanical Advantage (AMA) is always
less than the Ideal Mechanical Advantage IMA.
D. The Ideal Mechanical Advantage (IMA) is based on the relative
distances involved in the operation of the machine (din /dout) while the
Actual Mechanical Advantage (AMA) is based on the ratio Fout / Fin .
We calculate the mechanical advantage from the relationship: (Fout
/Fin)
7.4 SIMPLE MACHINES AND
WORK
• PROBLEM
• A 100-newton object is moved 2 meters up
an inclined plane whose end is lifted 0.5
meter from the floor.
• If a force of 50 newtons is needed to
accomplish this task, calculate the (a) IMA,
(b) AMA, and (c) efficiency of the inclined
plane.
7.4 SIMPLE MACHINES AND
WORK
• SOLUTION
• A diagram will help us to identify the
relevant quantities:
7.4 SIMPLE MACHINES AND
WORK
• SOLUTION
• The input force (Fin) is the force needed to
move the object along the plane (50 N).
7.4 SIMPLE MACHINES AND
WORK
• SOLUTION
• The output force (Fout) is the force that has
been lifted (100 N).
7.4 SIMPLE MACHINES AND
WORK
• SOLUTION
• The input distance (din) is the distance the
object is moved along the plane (2 m).
7.4 SIMPLE MACHINES AND
WORK
• SOLUTION
• The output distance (dout) is the distance
to which the object has been raised (0.5
m).
7.4 SIMPLE MACHINES AND
WORK
• SOLUTION
• The output distance (dout) is the distance
to which the object has been raised (0.5
m).
7.4 SIMPLE MACHINES AND
WORK
• PROBLEM
• A 100-newton object is moved 2 meters up
an inclined plane whose end is lifted 0.5
meter from the floor.
• If a force of 50 newtons is needed to
accomplish this task, calculate the (a) IMA,
7.4 SIMPLE MACHINES AND
WORK
• SOLUTION
• a) calculate the Ideal Mechanical
Advantage (IMA)
Assessment Question 8
• A 150 N (Fout) object is moved 15 m (din)
up an inclined plane whose end is lifted
2.5 m (dout) from the floor.
• If a force of 50 N (Fin) is needed to
accomplish this task, calculate the Ideal
Mechanical Advantage (IMA)
• IMA = din/dout = 15 m / 2.5 m =
A.0.5
C. 3
B.1
D. 6
7.4 SIMPLE MACHINES AND
WORK
• PROBLEM
• A 100-newton object is moved 2 meters up
an inclined plane whose end is lifted 0.5
meter from the floor.
• If a force of 50 newtons is needed to
accomplish this task, calculate the (b)
AMA
7.4 SIMPLE MACHINES AND
WORK
• SOLUTION
• b) calculate the Actual Mechanical
Advantage (AMA)
Assessment Question 9
• A 150 N (Fout) object is moved 15 m (din)
up an inclined plane whose end is lifted
2.5 m (dout) from the floor.
• If a force of 50 N (Fin) is needed to
accomplish this task, calculate the Actual
Mechanical Advantage (AMA)
AMA = Fout/Fin = 150 N / 50 N =
A. 0.5
C. 3
B. 1
D. 6
7.4 SIMPLE MACHINES AND
WORK
• PROBLEM
• A 100-newton object is moved 2 meters up
an inclined plane whose end is lifted 0.5
meter from the floor.
• If a force of 50 newtons is needed to
accomplish this task, calculate the (c)
efficiency of the inclined plane.
7.4 SIMPLE MACHINES AND
WORK
• SOLUTION
• calculate the (c) efficiency of the inclined
plane.
7.4 SIMPLE MACHINES AND
WORK
• The efficiency of the machine is the
AMA/IMA ratio; this value is always less
than 100%.
• Examples of simple machines include
inclined planes, pulleys, levers, wheels
and axles, and screwdrivers.
Conclusion
• An elastic collision is one in which
momentum and kinetic energy are
conserved.
• When gas molecules collide with the walls
of a container, these collisions are very
nearly elastic.
• Simple machines provide examples of how
work can be used to a person’s advantage
in performing such chores as lifting heavy
objects and exerting large forces that can
accomplish a task such as cutting through
steel.
Assessment Question 10
• A 150 N (Fout) object is moved 15 m (din)
up an inclined plane whose end is lifted
2.5 m (dout) from the floor.
• If a force of 50 N (Fin) is needed to
accomplish this task, calculate Efficiency:
Efficiency = AMA / IMA = (Fout/Fin) / (din/dout)
Efficiency = (150 N / 50 N) / (15 m / 2.5 m) =
A. 0.25
B. 0.5
C. 0.75
D. 1