BOSTES HSC Physics 2015 Question 27

Question: Maxwell’s theory of electromagnetism explained the nature of light and predicted the existence of other electromagnetic waves. Explain how Hertz performed experiments that validate Maxwell’s theory.

Marking Guidelines

Criteria	Marks
• Describes how Hertz performed experiments to test Maxwell’s theory and his predictions. • Describe how Hertz’s experiments validated Maxwell’s theory.	6

(Source: https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/guides/2015-hsc-mg-physics.pdf)

Comments: 

This question is about an understanding of Hertz’s experiments that validate Maxwell’s theory. Students are expected to describe how Hertz performed experiments to test Maxwell’s predictions. However, we could provide the following points: 

1. Oscillator: To produce electromagnetic waves, Hertz made an oscillator - an induction coil connected by two brass knobs - that generates sparks in an air gap between the two knobs under high voltages.

2. Detector: To detect electromagnetic waves, Hertz simply took a piece of copper wire, bent it into a circular shape, and created a short air gap between its two ends. 

3. A “surprising” phenomenon: When a spark is generated at the terminals of the induction coil, we could observe another spark in the air gap of the wire when the oscillator is reasonably close to the detector.

4. Theoretical explanation: According to Maxwell’s theory, electromagnetic waves are produced while the sparks are being generated. In addition, the electromagnetic waves propagate to the detector and set up oscillating electric and magnetic fields in the wire of the detector. (In other words, there is a resonance between the oscillator and the detector.)

5. Properties of electromagnetic waves: Hertz’s experiments show that electromagnetic waves have the following properties of light: 1. reflection; 2. refraction; 3. interference; 4. diffraction; 5. polarization.

6. Mathematical validation: Hertz was able to set up a stationary wave pattern by using a metal reflector, and determine the wavelength (λ) of the electromagnetic waves by measuring the distance between two nodes. Furthermore, by determining the oscillating frequency (f) of the electric current, Hertz calculated the speed of electromagnetic waves in air (v = fλ) which equals to the speed of light.

Essentially, Hertz developed the first primitive radio that can transmit electromagnetic waves. In a sense, Hertz’s experiments validate Maxwell’s equivalence of light and electromagnetic wave. Interestingly, Duhem, for example, proposed that Helmholtz’s electrodynamics could be another alternative to explain Hertz’s experiments (O’Rahilly, 1965). Thus, Hertz’s experiments do not validate Maxwell’s theory completely and conclusively. It is worth mentioning that Hertz intended to demonstrate that Maxwell’s theory is incorrect. In Hertz’s (1893) words, “I reflected that it would be quite as important to find out that electric force was propagated with an infinite velocity, and that Maxwell’s theory was false, as it would be, on the other hand, to prove that this theory was correct, provided only that the result arrived at should be definite and certain (p. 8).”

Feynman’s insights or goofs?:

Feynman has objections on the correctness of Maxwell’s theory. For instance, Feynman proposes that point charges interact only with other charges, but the interaction is related to both advanced and retarded waves. In addition, he explains that “[l]ight behaves like photons. It isn’t 100 percent like the Maxwell theory. So the electrodynamics theory has to be changed. We have already mentioned that it might be a waste of time to work so hard to straighten out the classical theory, because it could turn out that in quantum electrodynamics the difficulties will disappear or may be resolved in some other fashion. But the difficulties do not disappear in quantum electrodynamics (Feynman et al., 1964, section 28–5 Attempts to modify the Maxwell theory).” Essentially, light has both particle-like and wave-like characteristics. However, there are still difficulties in Maxwell’s theory even after modifications are made with quantum mechanics.

Feynman’s another objection is Maxwell’s ether. In an invited talk presented at the symposium The Past Decade in Particle Theory, Feynman (1970) elaborates that “in the case of ‘the ether never being found,’ it was ultimately realized that there wasn’t any ether at all, the ether was one of these scaffoldings to create a theory. It was later realized that the ether was an irrelevant complication and it may be that the partons are also nonexistent (p. 812).” In other words, Feynman believes that it is difficult to define the ether which cannot be detected conclusively. However, Einstein (1920) writes that, ‘[m]ore careful reflection teaches us however, that the special theory of relativity does not compel us to deny ether (p. 13).” In a sense, the concept of the ether does not fade away, but it is currently expressed by the term, fields (Wilczek, 1999).

Note:

1. In an article titled The Persistence of Ether, Wilczek (1999) writes that “[h]ow did I provoke Feynman? I asked him, ‘Doesn’t it bother you that gravity seems to ignore all we have learned about the complications of the vacuum?’ To which he immediately responded, ‘I once thought I had solved that one. I had a slogan: "The vacuum is empty. It weighs nothing because there’s nothing there."’ It was then he got wistful. I was deeply impressed to realize that Feynman had been wrestling with the problem of the cosmological term already in the 1940s, long before it became a widespread obsession and frustration. You have to admit that his slogan is catchy. So just maybe, despite everything I’ve said up to this point, eventually we really may have to do without ether (p. 13).”

2. Wilczek (1999) writes that “they are all surface manifestations of a single more basic entity, the electron field, an ether that pervades all space and time uniformly (p. 13)”.

References:

1. Einstein, A. (1920). Ether and the Theory of Relativity. In A. Einstein (1922). Sidelights on Relativity. London: Methuen. 

2. Feynman, R. P. (1970). Partons. In R. P., Feynman & L. M., Brown (Eds.), Selected Papers of Richard Feynman: With Commentary (pp. 773-813). Singapore: World Scientific.

3. Feynman, R. P., Leighton, R. B., & Sands, M. L. (1964). The Feynman Lectures on Physics, Vol II: Mainly electromagnetism and matter. Reading, MA: Addison-Wesley. 

4. Hertz, H. (1893). Electric Waves: Being Researches on the Propagation of Electric Action with Finite Velocity Through Space. London: Macmillan.

5. O’Rahilly, A. (1965). Electromagnetic Theory: A Critical Examination of Fundamentals, vol. 1. New York: Dover. 

6. Wilczek, F. (1999). The Persistence of Ether. Physics Today, 52(1), 11-13.

BOSTES HSC Physics 2015 Question 26

Question: 
We can use two mathematical models to calculate the work done when a 300 kg satellite is moved from Earth’s surface to an altitude of 200 km. Model A uses the mathematical equation, W = mg∆h, whereas Model B uses the mathematical equation, ΔE = −GMm/(R + h) – (–GMm/R).
(a) State the assumptions made on Earth’s gravitational field in models X and Y.
(b) Explain why models X and Y produce results that are approximately the same.

Marking Guidelines

Question 26 (a) Criteria	Marks
• Identifies correct assumptions.	2
• Identifies an assumption with correct information.	1

Possible answer: 
Assumption for Model X: Earth’s gravitational field is uniform.

Assumption for Model Y: Earth’s gravitational field varies with altitude.

Question 26 (b) Criteria	Mark
• Correct reason.	1

Possible answer: Variations in Earth’s gravitational field from the surface to an altitude of 200 km are sufficiently small.

(Source: https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/guides/2015-hsc-mg-physics.pdf)

Comments: 
In part (a), possible answer includes “the Earth’s gravitational field is uniform.” Alternatively, students might answer that the gravitational field is approximately constant on the Earth’s surface and up to an altitude of 200 km (as specified in the question). That is, we may assume the gravitational field strength to be constant (about 9.8 N/kg) if the satellite is not going beyond a certain height. Therefore, the work done in lifting the satellite can be mathematically expressed by the equation, W = mg∆h, because the gravitational force (mg) is assumed to be constant from Earth’s surface to the altitude (∆h) of 200 km. However, this assumption does not always hold for a low Earth orbit satellite which may have an altitude between 160 km and 2,000 km.

On the other hand, the sample answer states that “Model Y assumes the gravitational field changes with altitude.” To be precise, students could answer that the gravitational field follows the inverse square law (g = GM/r²). However, some students might include additional assumptions such as the density of the Earth is constant, the Earth is not rotating, and the Earth is spherical. Strictly speaking, the gravitational field is not even constant or the same at different locations on the Earth’s surface.

In part (b), the sample answer states that “variations in gravitational attraction from the Earth’s surface to an altitude of 200 km are sufficiently small.” However, 200 km is still quite a long distance. Thus, we should explain that the altitude of 200 km is negligible if it is compared to the radius of the Earth which is about 6,370 km. We can clarify this fact mathematically as shown below: 
Change in gravitational potential energy, ΔE = −GMm/(R + h) – (–GMm/R)
= (−GMm/R) (1 + h/R)^–1 – (–GMm/R) ≈ (−GMm/R) (1 – h/R – 1) 
= (GMmh/R²) = mgh
Note: By using (1 + x)^–1 = 1 – x for small values of x.

The approximation is appropriate because the altitude of 200 km is relatively short as compared to the radius of the Earth (i.e. h/R << 1). However, we can also explain that the error in work done increases as the altitude increases. The error is due to the neglected terms such as (−GMm/R) (h/R)².
Note: (1 + x)^–1 = 1 – x + x² – x³ + …

Feynman’s insights or goofs?:
Interestingly, Feynman explains that “[i]f we have a gravitational field that is uniform, if we are not going to heights comparable with the radius of the earth, then the force is a constant vertical force and the work done is simply the force times the vertical distance (Feynman et al., 1963, section 14-3 Conservative forces).” That is, the work done lifting an object is equal to the gravitational force times the vertical distance of the object raised. However, this simplification is possible only if the object is not moving to heights that are comparable to the radius of the earth. In other words, if the height increases, the error in calculating the work done increases.

More importantly, in Feynman’s words, “[t]he gravitational field of the earth is not precisely uniform, so a freely falling ball has a slightly different acceleration at different places — the direction changes and the magnitude changes (Feynman et al., 1964, section 42-5 Gravity and the principle of equivalence).” Simply put, the gravitational field of the Earth is not strictly constant and it can vary depending on the location of the Moon and the Sun. Essentially, the gravitational field of the Moon can affect the shape of the Earth and this further change the gravitational field of the Earth.

References:

1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.

2. Feynman, R. P., Leighton, R. B., & Sands, M. L. (1964). The Feynman Lectures on Physics, Vol II: Reading, MA: Addison-Wesley.

AP Physics 1 2015 Free Response Question 5

Question: 
A string is attached to an oscillator and its other end is attached to a block as shown below. The string passes over a pulley that is massless and frictionless. The distance between the oscillator and pulley is L and the mass of the block is M. The driving frequency of the oscillator is adjusted such that the string vibrates in its second harmonic. Identify locations on the string that have the greatest vertical speed.

Scoring Guidelines:

An indication that the string vibrates in its second harmonic or a wave is drawn such that λ2 = L.	1 point
Indicate points that are at the antinodes of any standing wave drawn on the string. (Full credit: the two points are located at one-fourth length of the string and three-fourths length of the string from the oscillator.)	1 point

(Source: http://apcentral.collegeboard.com/home)

Feynman's insights or goofs?:
The purpose of this question is to examine the properties of standing waves and apply the appropriate relationships among wavelength, frequency, and wave speed. Furthermore, students are expected to locate the nodes and antinodes of a string that is oscillating.

In this experiment, a string with one end is attached to an oscillator and its other end is attached to a block. Importantly, a node is not exactly located at the end that is attached to the oscillator because the oscillator vibrates up and down continuously. A definition of nodes is “[t]he points where there is no motion (Feynman et al., 1963, section 49-1 The reflection of waves).” If we observe the oscillations more carefully, the (virtual) node could be located slightly to the left of the oscillator. In a similar sense, an antinode does not occur exactly at the opening of the open pipe, but slightly beyond the opening where there is a lesser constraint for the air molecules to oscillate (See figure 1 below). The extra distance needed is sometimes called an end correction for the open-ended pipe. Thus, we should include the end correction for a more accurate calculation of the resonance frequency of the pipe. 

Fig 1

The scoring guidelines state that the two antinodes are located at one-fourth length of the string and three-fourths length of the string from the oscillator. Strictly speaking, the node does not occur exactly at the oscillator, and the antinodes can be located slightly to left of the one-fourth length of the string and three-fourths length of the string from the oscillator (See figure 2 below). Essentially, the points on the string that have the greatest average vertical speed can be located slightly to the left of the expected antinodes because of the end effect. Therefore, students could be penalized if the anti-nodes are not drawn exactly at the expected locations. Furthermore, some students might not indicate a node occurs at the oscillator which is not incorrect. 

Fig 2

In Feynman’s own words, “[s]uppose that the string is held at one end, for example by fastening it to an ‘infinitely solid’ wall. This can be expressed mathematically by saying that the displacement y of the string at the position x = 0 must be zero because the end does not move (Feynman et al., 1963, section 49–1 The reflection of waves).” However, there is a continuous vertical movement for the segment of the vibrating string attached to the oscillator. Thus, the node does not strictly occur at the location of the oscillator because the displacement of this segment of the string is not always zero. Furthermore, Feynman adds that “[t]he points where there is no motion satisfy the condition sin (ωx/c) = 0, which means that (ωx/c) = 0, π, 2π, …, nπ, … These points are called nodes (Feynman et al., 1963, section 49–1 The reflection of waves).” In short, the nodes should be defined as points that are always stationary.

Interestingly, Feynman mentions that “if we assume that the string is infinite and that whenever we have a wave going one way we have another one going the other way with the stated symmetry, the displacement at x = 0 will always be zero and it would make no difference if we clamped the string there (Feynman et al., 1963, section 49–1 The reflection of waves).” In a sense, Feynman has a slip of tongue when he simply says that “the string is infinite.” To be precise, the vibrating string is not infinitely long or massive, but that it is attached to an “infinitely rigid” wall. Thus, physics teachers could explain that Feynman is sometimes sloppy in his language usage.

More importantly, the fundamental frequency of the vibrating string is not simply dependent on the linear mass density and tension in the real world. Feynman explains that “[t]he idea that the natural frequencies are harmonically related is not generally true. It is not true for a system of more than one dimension, nor is it true for one-dimensional systems which are more complicated than a string with uniform density and tension. A simple example of the latter is a hanging chain in which the tension is higher at the top than at the bottom (Feynman et al., 1963, section 49–3 Modes in two dimensions).” Nevertheless, the stiffness of a real-life string can cause the wave velocity to be also dependent on the wavelength (Elmore & Heald, 1985).

Note: 

1. A node can be distinguished as a displacement node or a pressure node. For example, a “pressure node (corresponding to a displacement or velocity antinode) occurs at the open end of a tube, while a pressure antinode (corresponding to a displacement or velocity node) occurs at the closed end (Gregersen, 2011, p. 37).”

2. For another discussion of this question, you can visit the following website:
https://www.youtube.com/watch?v=z_KX8Xpxa-c

References:

1. Elmore, W. C., & Heald, M.A. (1985). Physics of Waves. New York: Dover.
2. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.
3. Gregersen, E. (2011). The Britannica Guide to Sound and Light. New York: The Rosen Publishing Group.

AP Physics 1 2015 Free Response Question 4

Question: 
Two identical spheres emerge from an electronic device at the same time (t = 0) and from the same height (h = H), as shown below. Sphere A has no initial velocity and moves downward in a straight line. Sphere B moves with an initial horizontal velocity (v₀) and covers a horizontal distance D when it hits the ground. The two spheres reach the ground at the same time (t_f) though sphere B moves parabolically and travels a longer distance before landing. (Assume there is no air resistance.)

Students have to explain why the two identical spheres reach the ground at the same time if they are dropped simultaneously.

Scoring Guidelines: 

Indicate the horizontal motions of the two spheres do not affect their vertical motions.	1 point
Indicate the two spheres move with the same vertical velocity.	1 point
Indicate the two spheres move with the same vertical acceleration.	1 point
Indicate the two spheres falling from the same height would take the same time.	1 point
There is/are no incorrect or irrelevant statement(s).	1 point

(Source: http://apcentral.collegeboard.com/home)

Comments:
The purpose of this question is to assess students’ understanding of two-dimensional motion, as well as equations behind one sphere that is dropped from rest and a second identical sphere that is projected with an initial horizontal speed from the same height. In this post, we discuss a part of the question which is about why a sphere that is falling vertically in a straight line can reach the ground at the same time as the other sphere that is falling in a parabolic path. Importantly, the explanation of this phenomenon could include theoretical and empirical knowledge.

1. Theoretical idealization: Based on the scoring guidelines, a point can be awarded if students indicate that the two spheres falling the same height, would take the same time. That is, the horizontal velocity of the sphere does not affect its vertical velocity. In other words, the horizontal motion and vertical motion are independent of each other. However, we have assumed that the falling of an object is independent of its mass. Strictly speaking, mass independence is not valid for an observer on the ground (Lehavi & Galili, 2009). By using Newton’s law of gravitation or Newton’s third law, the Earth can accelerate toward the falling object. Thus, the statement that “all objects fall to the ground with the same acceleration” is a theoretical idealization and it is only approximately correct.

2. Empirical knowledge: Based on the scoring guidelines, another point can be awarded if students indicate that the difference in horizontal motion does not affect the vertical motion of the spheres. However, we should not simply deduce the independence of horizontal and vertical motions to be a theory. Similarly, we should not assume that two objects of different mass will reach the ground at the same time by simply using a thought experiment. Importantly, the falling of two spheres to the ground can be verified by experiments. Note that the air resistance acting on the two spheres are in different directions while they are falling toward the ground. Furthermore, it is empirically verified that the frictional drag on an object is proportional to the velocity or proportional to the square of the velocity depending on how fast the object is moving.

Nevertheless, if Einstein were a student taking this examination, he might pose the following thought experiment: “I stand at the window of a railway carriage which is traveling uniformly, and drop a stone on the embankment, without throwing it. Then, disregarding the influence of the air resistance, I see the stone descend in a straight line. A pedestrian who observes the misdeed from the footpath notices that the stone falls to earth in a parabolic curve. I now ask: Do the ‘positions’ traversed by the stone lie ‘in reality’ on a straight line or on a parabola? (Einstein, 1961, p. 10).” In essence, the so-called straight line or parabolic path is dependent on the velocity of the observer. Simply phrased, we can explain the object’s motion in terms of the observer’s frame of reference. Perhaps, based on the scoring guidelines, Einstein or some students could be penalized in answering this question by using the concept of inertial reference frame? 

Feynman’s insights and goofs?:
We can understand the question from the perspectives of empirical knowledge and theoretical idealization.
1. Empirical knowledge: Feynman explains that “[a] falling body moves horizontally without any change in horizontal motion, while it moves vertically the same way as it would move if the horizontal motion were zero. In other words, motions in the x-, y-, and z-directions are independent if the forces are not connected. (Feynman et al., 1963, section 9–3 Components of velocity, acceleration, and force).” Essentially, the motions of an object can be resolved into perpendicular directions that are independent of each other. Furthermore, in the Fig. 7-3 of Volume I of The Feynman Lectures on Physics, there is an apparatus that demonstrate the independence of vertical and horizontal motions (Feynman et al., 1963). Thus, the independence of the motions of the object in the x-, y-, and z-directions can also be considered as an empirical fact.

2. Theoretical idealization: Feynman writes that “[w]hat happens if we shoot a bullet faster and faster? Do not forget that the earth’s surface is curved. If we shoot it fast enough, then when it falls 16 feet it may be at just the same height above the ground as it was before. How can that be? It still falls, but the earth curves away, so it falls ‘around’ the earth (Feynman et al., 1963, section 7–4 Newton’s law of gravitation).” In other words, we have idealized the ground to be flat if the object falls within a short distance. This approximation does not hold if the object moves at a relatively high horizontal speed. In short, there are theoretical idealization and approximations involved in the falling of the object. The earth is not perfectly flat.

Note:

1. If Galileo were a student taking this examination, he might explain that the two spheres reach the ground at the same time because we have assumed that the air resistance is negligible. For example, in the Fourth Day of Dialogues Concerning Two New Sciences, Galileo (1638) writes that “if we consider only the resistance which the air offers to the motions studied by us, we shall see that it disturbs them all and disturbs them in an infinite variety of ways corresponding to the infinite variety in the form, weight, and velocity of the projectiles (p. 252).”

2. For another discussion of this question, you can visit the following website:
https://www.youtube.com/watch?v=x7UHl3Umeeg

References: 

1. Einstein, A. (1961/1916). Relativity: The Special and The General Theory. New York: Random House.

2. Feynman, R. P., Leighton, R. B., & Sands, M. L. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.

3. Galileo, G. (1638/1954). Dialogues Concerning Two New Sciences (trans. H. Crew & A. de Salvio). New York: Dover.

4. Lehavi, Y., & Galili, I. (2009). The status of Galileo’s law of free-fall and its implications for physics education. American Journal of Physics, 77(5), 417-423.

AP Physics 1 2015 Free Response Question 3

Question: 
A block is initially stationary at position x = 0 and is in contact with a massless spring that is uncompressed. The block is then slowly compressed along a frictionless surface from position x = 0 to x = -D (See figure 1 below) such that ∆x = D. When the block is released, it reaches a rough part of the track at the right-hand side of x = 0 and it eventually slows down to rest at position x = 3D. Assume the coefficient of kinetic frictional force between the block and rough track is μ.

 

                                                                                 Fig. 1

In this question, students have to explain the correct and incorrect aspects of the following statement: “if the spring is compressed twice as long as before, the block has more energy when it leaves the spring, so it will slide farther along the track before stopping at position x = 6D.”

Scoring Guidelines:

Identify the correct aspect: the block has more energy when it leaves the spring.	1 point
Identify the incorrect aspect: the new final position of the block is not at x = 6D. (The spring’s elastic potential energy is proportional to the square of ∆x.)	1 point

(Source: http://apcentral.collegeboard.com/home)

Comments:

The main purpose of this question is to examine the relationship between the energy stored in a compressed spring-block system and the work done by the frictional (kinetic) force on the block after it leaves the spring. In other words, the question is about the transformation of the elastic potential energy of the block into its kinetic energy and thermal energy (microscopic internal energy). However, there are terminological and language issues in this question.

1. Terminological issues: It is surprising that the question and scoring guidelines adopt imprecise terminologies. For example, based on the scoring guideline for (b)(i), the student is correct because the block will have more energy when it leaves the spring. Note that the term “energy” could be changed to “kinetic energy.” To be more precise, it could be further changed to “translational kinetic energy” that is different from “rotational kinetic energy.” Next, based on the scoring guideline for (b)(ii), the student is incorrect because the spring’s energy does not scale linearly with its compression. However, the term “spring’s energy” could be changed to “elastic potential energy.” It is stated in AP 1: Algebra-Based Exam Description that, “Essential knowledge 4.C.1: The energy of a system includes its kinetic energy, potential energy, and microscopic internal energy. Examples should include gravitational potential energy, elastic potential energy, and kinetic energy.”

2. Language issues: The sentence “the block will have more energy when it leaves the spring” in the question is misleading. Firstly, it seems to suggest that the block will have more energy, and thus violating the conservation of energy. More importantly, one may expect the block to be in contact with the spring until the spring is no longer compressed. Thus, after the block leaves the spring, the block’s (kinetic) energy starts to decrease at a constant rate because of frictional force of the track. This contradicts the statement that “the block will have more energy when it leaves the spring.” Perhaps the scoring guidelines could recognize students who identify the incorrect reasoning pertaining to “the block will have more energy when it leaves the spring”? Or perhaps the sentence could be rephrased as “the block will have relatively more kinetic energy when there is an increase in the compression of the spring”?

Feynman’s insights or goofs?:
The question adopts the term “spring’s energy” instead of “elastic potential energy.” In Feynman’s words, “[e]lastic energy is the formula for a spring when it is stretched. How much energy is it? If we let go, the elastic energy, as the spring passes through the equilibrium point, is converted to kinetic energy and it goes back and forth between compressing or stretching the spring and kinetic energy of motion. There is also some gravitational energy going in and out, but we can do this experiment ‘sideways’ if we like (Feynman et al., 1963, section 4–4 Other forms of energy).” That is, the motion of a spring may involve elastic energy, kinetic energy, and gravitational energy. However, physics teachers can use the following terms that are more precise: elastic potential energy, translational kinetic energy, and gravitational potential energy.

On the other hand, this question assumes that the frictional force is constant. Conversely, Feynman explains that “the frictional drag on a ball or a bubble or anything that is moving slowly through a viscous liquid like honey, is proportional to the velocity, but for motion so fast that the fluid swirls around (honey does not but water and air do) then the drag becomes more nearly proportional to the square of the velocity (Feynman et al., 1963, section 12-2 Friction).” Furthermore, Feynman states that “the drag force on an airplane is approximately a constant times the square of the velocity, or F ≈ cv² (Feynman et al., 1963, section 12-2 Friction).” However, the drag force can also be approximated by F ≈ av + bv². Importantly, Feynman clarifies that “as we study this law of the drag on an airplane more and more closely, we find out that it is ‘falser’ and ‘falser’ (Feynman et al., 1963, section 12-2 Friction).”

Interestingly, Feynman does not mention that the frictional force is independent of contact area which can be found in many physics textbooks. He explains that when there is good contact between two solids, they can hold very tight together and thus have a larger frictional force. Most important, Feynman adds that “to a fairly good approximation, the frictional force is proportional to this normal force, and has a more or less constant coefficient; that is, F = μN, where μ is called the coefficient of friction. Although this coefficient is not exactly constant, the formula is a good empirical rule for judging approximately the amount of force that will be needed in certain practical or engineering circumstances. If the normal force or the speed of motion gets too big, the law fails because of the excessive heat generated. It is important to realize that each of these empirical laws has its limitations, beyond which it does not really work (Feynman et al., 1963, section 12-2 Friction).”

Note:

1. Feynman also explains that “[i]n experiments of the type described above, the friction is nearly independent of the velocity. Many people believe that the friction to be overcome to get something started (static friction) exceeds the force required to keep it sliding (sliding friction), but with dry metals, it is very hard to show any difference (Feynman et al., 1963, section 12-2 Friction).” 

2. Jonathan Thomas-Palmer, for example, strongly disagrees with the use of the symbol K for the kinetic energy of the block because k stands for spring constant in the same question. (Source: https://www.youtube.com/watch?v=6dlijzUQHw4)

3. For another discussion of this question, you can visit the following websites:

https://www.youtube.com/watch?v=qQM-IGnxi6g
https://www.youtube.com/watch?v=KH60uz5LNWk

Reference: 

Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.