Transcript Tutorial_cons_o_energy

You are the technical advisor for the David Letterman Show. Your task is to design a
circus stunt in which Super Dave, who weighs 750 N, is shot out of a cannon that is 40o
above the horizontal. The “cannon” is actually a 1m diameter tube that uses a stiff spring
to launch Super Dave. The manual for the cannon states that the spring constant is 1800
N/m. The spring is compressed by a motor until its free end is level with the bottom of the
cannon tube, which is 1.5m above the ground. A small seat is attached to the free end of
the spring for Dave to sit on. When the spring is released, it extends 2.75m up the tube.
Neither the seat nor the chair touch the sides of the 3.5m long tube, so there is no friction.
After a drum roll, the spring is released and Super Dave will fly through the air. You have
an airbag 1m thick for Super Dave to land on. You know that the airbag will exert an
average retarding force of 3000 N in all directions. You need to determine if the airbag is
thick enough to stop Super Dave safely – that is, he is slowed to a stop by the time he
reaches ground level. Consider the seat and spring to have negligible mass. Ignore air
resistance.
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Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Question 9
Question 10
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Question 11
Question 12
Question 13
Question 14
Question 15
Question 16
Question 17
Question 18
Question 19
Paired Problem 1
1. Which of the following physics
principles are most suited to solve this
problem? i) Kinematical considerations
ii) Linear momentum conservation
iii) Mechanical energy conservation
iv) Work-Energy Theorem
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A : i only
B : ii only
C : iii and iv only
D : all of the above
Choice: A
Incorrect
This is a tedious way because it involves
vectors. Also, it only applies to the
“projectile motion” part of the problem.
Choice: B
Incorrect
There is only a collision at the airbag. We
need to analyze other parts of the
problem. Considering the energy of the
system at different points will be more
helpful.
Choice: C
Correct
Using these principles makes the problem easier to
solve. The work-energy theorem is useful in situations
where you need to relate a body’s speed at two different
points in its motion. The energy approach is useful
when a problem includes motion with varying forces
along a curved path. However, conservation of total
mechanical energy requires that only conservative
forces do work.
Choice: D
Incorrect
All of these principles are applicable. We
are asked to find the easiest way.
Since we are not dealing with time
explicitly, we should be applying the workenergy theorem and conservation of
mechanical energy.
2. Which form of energy does the spring-Daveearth system possess just before the spring is
released? Assume the reference height is the
floor.
i) Kinetic Energy (K)
ii) Spring Potential Energy
iii) Gravitational Potential Energy
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A : i only
B : ii only
C : iii only
D : ii and iii only
Choice: A
Incorrect
There is no kinetic energy before
the spring is released because
Super Dave has zero initial
velocity (vo=0).
Choice: B
Incorrect
This is one type of potential energy (PE)
associated with this system, but there is
another type of PE associated with this
system as well. Remember that Super
Dave is displaced vertically before the
spring is released.
Choice: C
Incorrect
This is one of the forms of potential
energy (PE) associated with this system,
but there is another type of PE associated
with this system as well. Remember that
the spring is initially compressed.
Choice: D
Correct
The system has gravitational and spring
potential energy, because Super Dave
begins above the floor (where PE=0) and
the spring is initially compressed.
3. Which of the following conditions are
required for the Law of Conservation of
Mechanical Energy to hold for a system?
final.
forces
-U is elastic potential energy.
-The subscripts i and f stand for initial and
-W is work done by non-conservative
-E is total mechanical energy.
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A: The work done by non-conservative forces must be zero.
B: Energy is not created or destroyed, but can change forms.
C: U K  0
D: Ef=Ei or (Uf+Kf=Ui+Ki)
Choice: A
Correct
The other choices are either statements
of conservation of total mechanical
energy (not the requirements) or are
incorrect.
Choice: B
Incorrect
This is how the Law can be
expressed in words, but there are
more requirements.
Choice: C
Incorrect
This is the mathematical
expression of the Law, but there
are more requirements.
Choice: D
Incorrect
This is the same as choice C, just
in a different form. There are
more requirements.
4. Since the gravitational and elastic forces are
conservative, and we ignore air resistance, we can apply
the Law of Conservation of Mechanical Energy to this
system. How do we (mathematically) express the initial
mechanical energy of the system before the spring is
released?
h=height
A: Ei=mgh
 
B: Ei  1kx2
2 
1  2
E

mgh

 kx
C: i
2 


D: Ei=0
x=compression of
the spring
k=spring constant
g=gravitational
acceleration (9.8m/s2)
m= Dave’s mass
Choice: A
Incorrect
Think about the spring, which
also adds to the initial
mechanical energy since it is
compressed.
Choice: B
Incorrect
What about the gravitational
potential energy? Recall that the
system begins above the floor.
Choice: C
Correct
Gravity and the compressed spring
contribute to total mechanical
energy. These forms of energy are
expressed correctly.
Choice: D
Incorrect
Kinetic energy is the only type of
mechanical energy that is initially
zero.
5. What happens to the energy that
was stored in the spring right after the
spring is released and Super Dave is
launched?
• A: It transforms to gravitational PE only.
• B: It transforms to K only.
• C: Some of it transforms to gravitational PE and
some to K.
• D: It is lost.
Choice: A
Incorrect
Dave gains kinetic energy as the
spring is released, because he gains
a non-zero velocity.
Recall :

1  2
K   mv
2 
where v is Super Dave’s velocity.
Choice: B
Incorrect
Since Super Dave is shot at an angle
 above the horizontal, he gains
height which increases his
gravitational PE (mgh).
Choice: C
Correct
The elastic PE transforms partly
to gravitational PE and partly to
kinetic energy, because Super
Dave gains height and velocity.
Choice: D
Incorrect
Since we are ignoring air resistance,
there is no non-conservative force
involved. Therefore, there is no
energy loss.
6. Dave’s energy right at the instance
of impact with the airbag consists of
which form of energy?:
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A: only K
B: only gravitational PE
C: both gravitational PE and K
D: none of the above
Choice: A
Incorrect
The top of the airbag is 1m above the
floor, so there should be some
gravitational potential energy at the point
where Super Dave first makes contact
with the airbag.
Choice: B
Incorrect
If Super Dave had no kinetic energy, we
wouldn’t have to worry about him getting
hurt. Super Dave is traveling with a nonzero velocity. Therefore, he has potential
energy.
Choice: C
Correct
Dave’s energy at this point consists of
both kinetic energy and gravitational PE,
because he is traveling with a non-zero
velocity and is still displaced vertically.
Choice: D
Incorrect
He has both kinetic energy and
gravitational PE. Remember that he is
traveling with a non-zero velocity and is
not quite at floor level at this point.
7. Can we still use the principle of
mechanical energy conservation
after Dave hits the air bag?
• A: yes
• B: no
• C: we don’t have enough information to decide
Choice: A
Incorrect
Since the airbag softens Dave’s landing,
there is a retarding force, which is a nonconservative force. Therefore, we can not
use the principle of conservation of
mechanical energy.
Choice: B
Correct
The retarding force (non-conservative
force), which comes from the airbag
resisting Dave’s motion, does work.
Thus, we can not use this principle.
Choice: C
Incorrect
We do have enough information.
The airbag causes a non-conservative
force to do work on Dave. This retarding
force disallows the use of the principle of
mechanical energy conservation.
8. After the impact with the airbag,
which physics principles should we
use to solve the problem?
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A: kinematics
B: work-energy theorem
C: Impulse-Momentum Theorem
D: all are possible ways
Choice: A
Incorrect
Kinematical considerations will
not be useful here.
Choice: B
Correct
Using the work-energy
theorem will be the best
method to solve the problem.
Choice: C
Incorrect
This method is not useful due
to insufficient information.
Choice: D
Incorrect
Kinematics and the impulse-momentum
theorem will not be helpful here. This
problem can be solved by the workenergy theorem.
9. Let’s put the concepts that we have considered in
questions 1-8 into mathematical form.
Which of the following equations correctly describes our
application of conservation of total mechanical energy from
the point just before the release of the spring to the point
just before impact?
A: (1/2)kx  (1/2)mv
2
B:
2
(1/2)kx  mghi  (1/2)mv
2
2
(1/2)kx  mghi  (1/2)mv  mgh f
2
mgh

(1/2)mv
 mgh f
D:
i
C:
2
2
Choice: A
Incorrect
There is gravitational PE at the points just
before release and just before impact,
because Dave is above floor level (which
we are defining as the zero of
gravitational potential energy), in both
cases.
Choice: B
Incorrect
What about the gravitational PE
just before impact when Dave is
still at least 1 m above the floor?
Choice: C
Correct
This expression contains all of the
forms of energy that are involved at
these two points.
Choice: D
Incorrect
In addition to the gravitational PE just
before release, remember that the
compressed spring stores elastic potential
energy which can change into other forms
of mechanical energy.

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
10. From the equation we found in
question 9, which one of the following
expressions is true for v2 after
simplification?
• A : v2  kx2  mghi  mgh f
2
(kx
 2mg(hi  hf ))
• B : v2 
m
• C : v  2g(hi  hf )
2
Choice: A
Incorrect
Check your algebra. Pay attention
to the mass and the factor of 2.
Choice: B
Correct
Correct algebra.
Choice: C
Incorrect
Spring PE is set to zero in this
expression, which is incorrect.
11. From the information given in the
original problem statement, what is the
value of the initial compression of the
spring x ?
A: 1.5 m
B: 3.5 m
C: 1800 N/m
D: 2.75 m
E: 1 m
Choice: A
Incorrect
This is the initial height of Dave in his
seat atop the compressed spring.
Choice: B
Incorrect
This is the length of the tube and
has nothing to do with the physics
of the problem.
Choice: C
Incorrect
This is the stiffness (spring
constant k) of the spring.
Choice: D
Correct
This is the initial compression of the spring.
(The amount the spring is compressed from
its equilibrium position.)
Choice: E
Incorrect
This is the height of the top
surface of the airbag from the
floor.
12. Also from the information given in the
original problem:
What does the 750 N value represent?
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A : Dave’s mass (m)
B : Dave’s weight (W=mg)
C : the retarding force of the air
D : the spring constant
Choice: A
Incorrect
The problem states that Super Dave
weights 750 N. Weight differs from
mass by a factor of g (the value of
gravitational acceleration near Earth).
Choice: B
Correct
This value is given as Super Dave’s
weight in the problem statement.
mg = W = weight of Dave
Choice: C
Incorrect
We can ignore the retarding force of air in
this problem. It is stated in the problem that
Super Dave weighs 750 N.
W = 750 N = mg
Choice: D
Incorrect
The spring constant (k) is in units of
N/m and is given elsewhere in the
original problem. Super Dave is said to
weigh 750 N.
13. Now that we have Identified the
relevant values:
Calculate the numerical value of v.
Recall that v is Super Dave’s velocity
just before impact with the airbag.
W
Dave’s mass is 76.53 kg. from m  g
Reasoning:
2
(kx
 2mg(hi  hf ))
2
v 
m
1800N/m2.75m
2
v2 

 276.53kg9.8m/s2 1.5m 1m
76.53kg
v2  187.7m2 /s2
v  187.7m2 /s2  13.7m/s
Now that we found Dave’s speed just before
impact, we can focus on the “airbag-Dave”
system.
Here we will assume that Super Dave hits the
air bag at normal incidence (straight down
onto the air bag). This makes the calculation
much simpler, but is only true as a limiting
case.
14. Here, we are only concerned with Dave’s motion
in
the vertical direction.
Finish the following statement correctly.
The retarding force Fretarding from the airbag on
Dave…
• A : is perpendicular to Dave’s direction of
motion.
• B : is in the same direction as Dave’s direction
of motion.
• C : is opposite to Dave’s direction of motion.
• D : has no direction.
Choice: A
Incorrect
There could be no energy loss if this
were the case and Dave would get hurt
regardless of the strength of the
airbag’s force.
Choice: B
Incorrect
This is nonsense. Landing on an
airbag would not provide a force
that would cause one to accelerate
downward.
Choice: C
Correct
The retarding force of the airbag
opposes the direction of motion. This
force does work on Dave. It causes him
to slow down and hopefully helps him
to land safely.
Choice: D
Incorrect
It does have a specific direction.
Force is a vector quantity.
15. What does the work-energy
principle say?
• A : A system’s change in PE is equal to the work done
on the system by the resultant force.
• B : the work done is equal to the energy dissipated in
the form of heat
• C : the work done by the resultant force acting on the
system is equal to the change in the system’s K
• D : there is no such principle
Choice: A
Incorrect
This statement is true if all of
the work is done by
conservative forces. Even so, it
is not the Work-Energy
Theorem.
Choice: B
Incorrect
This is not always true, and is
not relevant to this problem.
Choice: C
Correct
This is the correct explanation of the work-energy
principle.
In mathematical form:
K f  K i  Wtotal
1
1
mv f 2  mvi2  Wtotal
2
2

Choice: D
Incorrect
There is such a principle.
16. In general, total work can be expressed as:
Wtotal  Ftotal ycos()
Where y represents change vertical distance and  is the
angle between the direction of motion and the direction
of force, remember we are only considering the vertical
component of the retarding force.

What is Wtotal in the Dave-airbag system?
A: Wtotal  (Fretarding )y
B: Wtotal  Fretarding mgy
C Wtotal  mgy
Choice: A
Incorrect
After Dave hits the airbag, the total force (we only care
about the y-component) acting upon him is a
combination of gravitational force and the retarding force
from the airbag. These forces are in opposite directions,
making the total force equal to:
Ftotal  (Fretarding  mg)

Choice: B
Correct
This is the correct expression for total
work in this system, because gravitational
force and the retarding force from the
airbag act on Dave in opposite directions.
Also, cos()=-1 because =180º.
Choice: C
Incorrect
After Dave hits the airbag, the total force (we only care
about the y-component) acting upon him is a
combination of gravitational force and the retarding force
from the airbag. These forces are in opposite directions,
making the total force equal to:
Ftotal  (Fretarding  mg)

17. Assuming that Dave is going to come to a stop at
some point after he hits the airbag and applying the workenergy principle, which one of the following expressions is
correct for the Dave-airbag system? (vi is the Dave’s
velocity just before impact. It’s the velocity v that we found
earlier.)
1
1
2
• A : mv f  mvi2  0
2
2



1
2

mv
• B:
i  Fretarding mgy
2
• C : 1 mv f 2  Fretarding mgy
2
Choice: A
Incorrect
There must be some work done if Dave
is going to stop. The retarding force
from the airbag does work on him.
Choice: B
Correct
Since Kf = 0, this is the correct
expression.
Choice: C
Incorrect
The kinetic energy before impact with
the airbag should be included. Since
Dave stops moving, vf=0.
18. For Dave to land safely, what must be true
about y?
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A : y > 1m
B:y=0
C : y < 1m
D : y = 1m
Choice: A
Incorrect
This wouldn’t be safe because
he would hit the hard floor.
Choice: B
Incorrect
This implies that vi = 0
Choice: C
Correct
If y<1m, Super Dave would stop
before the airbag is fully compressed
and would not hit the floor.
Choice: D
Incorrect
This is risky.
Dave would hit the floor.
19. What is the numerical value of y?
Solution
Solution
Applying the principle of conservation of mechanical energy from the point just
before the release of the spring to the point just before impact, we found:
(kx2  2mg(hi  hf ))
2
vi 
m
2
vi  187.7m2 /s2
vi  187.7m2 /s2  13.7m/s
In analysis after the impact, we applied the work-energy theorem. Finding:

1 2
mvi  Fretarding mgy
2

Rearranging to solve for y
we get:
2

1 
mvi
y  


2 (Fretarding  mg) 

1 76.5kg(187.7m2 /s2 )  Continue
y  

2  3000N  750N 
y  3.2m
Since y is greater than 1m, Super Dave
will not land safely. We need a thicker
airbag, or one that has a stronger
retarding force.
Paired Problem
1. In the track shown below, section AB is a quadrant of a circle
of 1.0m radius. A block is released at point A and slides without
friction until it reaches point B. The horizontal part is not smooth.
If the block comes to rest 3.0m after point B, what is the
coefficient of kinetic friction k on the horizontal surface?