Physics quiz question: ammeter reading after switch is closed

Physics 205B Quiz 5, spring semester 2018
Cuesta College, San Luis Obispo, CA

An ideal 6.0 V emf source is connected to two light bulbs, a resistor, and an ideal ammeter, and an open switch. When the switch is closed, the ammeter reading will:
(A) decrease.
(B) remain constant.
(C) increase.
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (A)

When the switch is open, the 1.0 Ω light bulb will be dark as no current will pass through it.  The current in this circuit will start at the 6.0 V emf, pass through the ammeter, go through the 7.5 Ω resistor, and then through the 2.0 Ω resistor, and back to the 6.0 V emf.  

The equivalent resistance R_eq of this circuit is 7.5 Ω + 2.0 Ω = 9.5 Ω.  

The current I through this circuit is (6.0 V)/(9.5 Ω) = 0.63 A, which is the ammeter reading.

When the switch is closed, then the 1.0 Ω light bulb is in parallel with the 7.5 Ω resistor.

The equivalent resistance R_eq of this circuit is 2.0 Ω + ((1/1.0 Ω) + (1/7.5 Ω))^–1 = 2.88 Ω.

The current I_emf through the emf is (6.0 V)/(2.88 Ω) = 2.08 A.

Now let's figure out how much current goes through the ammeter when the switch is closed, as the 2.08 A that passes through the 6.0 V emf will split with some either going through the switch path or going through the ammeter path, as given by Kirchhoff's junction rule:

I_emf = I_switch + I_ammeter.  

Let's apply Kirchhoff's loop rule for the clockwise emf-ammeter-7.5 Ω-2.0 Ω round trip path:

voltage rises = voltage drops,

6.0 V = ∆V_{7.5 Ω} + ∆V_{2.0 Ω},

and then apply Ohm's law to the right-hand side terms:

6.0 V = I_ammeter·(7.5 Ω) + I_emf·(2.0 Ω),

and since we already know I_emf= 2.08 A, then:

6.0 Ω = I_ammeter·(7.5 Ω) + (2.08 A)·(2.0 Ω),

0.25 A = I_ammeter,

which means the ammeter reading will decrease from its previous reading of 0.63 A when the switch was still open.

Sections 30882, 30883
Exam code: quiz05z0m6
(A) : 22 students
(B) : 8 students
(C) : 4 students
(D) : 0 students

Success level: 65%
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Physics quiz question: power dissipated by resistor in parallel circuit

Physics 205B Quiz 5, spring semester 2018
Cuesta College, San Luis Obispo, CA

An ideal 6.0 V emf source is connected to a resistor and two light bulbs. The electrical power used by the 4.0 Ω resistor is:
(A) 1.5 W.
(B) 9.0 W.
(C) 24 W.
(D) 96 W.

Correct answer (highlight to unhide): (B)

The basic equation for the power dissipated by the 4.0 Ω resistor is:

P_resistor = I_resistor·ΔV_resistor,

where the current through the resistor I_resistor is not equal to the current passing through the 6.0 V emf source (I_circuit = ΔV_resistor/R_eq), due to the junction rule. However, we do not need to find I_resistor, as we can appeal to Ohm's law:

I_resistor = ΔV_resistor/R_resistor,

such that we can substitute this into the basic power equation, and result in a "specialized" form of the power equation for this resistor:

P_resistor = I_resistor·ΔV_resistor,

P_resistor = (ΔV_resistor/R_resistor)·ΔV_resistor,

P_resistor = (ΔV_resistor)²/R_resistor.

To find the ΔV_resistor voltage used by the resistor, we apply the loop rule in the clockwise direction, starting from lower right-hand corner, through the 6.0 V emf source, then through the 4.0 Ω resistor before returning to the starting point in the lower right-hand corner (the loop rule can be applied to any round-trip loop in a circuit, even if there are other parts of this circuit):

"voltage supplied = voltage used,"

∆V_rises = ∆V_drops,

(6.0 V) = ΔV_resistor.

Then we can evaluate the "specialized" form for the power used by the resistor:

P_resistor = (ΔV_resistor)²/R_resistor,

P_resistor = (6.0 V)²/(4.0 Ω) = 9.0 W.

(This "specialized" equation for power should only be used if the voltage used by the circuit element is known already either from the loop rule (as was done here) or from Ohm's law.)

(Response (A) is the numerical value for the current flowing through the resistor; response (C) is ΔV_resistor·R_resistor; response (D) is ΔV_resistor)²·(R_resistor)².)

Sections 30882, 30883
Exam code: quiz05z0m6
(A) : 5 students
(B) : 19 students
(C) : 9 students
(D) : 1 student

Success level: 56%
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Physics quiz question: equivalent resistance of serio-parallel circuit

Physics 205B Quiz 5, spring semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Comprehensive Problem 18.112

An ideal 12 V emf source is connected to three resistors, as shown at right. The equivalent resistance of this circuit 
is:
(A) 3.4 Ω.
(B) 4.7 Ω.
(C) 9.3 Ω.
(D) 14.0 Ω.

Correct answer: (C)

The 2.0 Ω resistor and the 4.0 Ω resistor are directly connected in parallel to each other, such that their equivalent resistance is:

R_2,4 = [ (R₂)^–1 + (R₄)^–1 ]^–1,

R_2,4 = [ (2.0 Ω)^–1 + (4.0 Ω)^–1 ]^–1,

R_2,4 = 1.333... Ω.

Then the 8.0 Ω resistor is in series with the combined R_2,4 equivalent resistor, so the final equivalent resistance of the circuit is:

R_eq = R₈ + R_2,4,

R_eq = 8.0 Ω + 1.333... Ω = 9.3 Ω, to the significant tenths decimal place.

(Response (A) is where the 2.0 Ω resistor and the 4.0 Ω resistor are first combined in series, and their resulting equivalent resistance R_2,4 combined in parallel with the 8.0 Ω resistor; response (B) is the average of all three resistance values; response (D) is the sum of all three resistance values.)

Section 30882
Exam code: quiz04eQu7
(A) : 1 student
(B) : 4 students
(C) : 25 students
(D) : 2 students

Success level: 80%
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Physics quiz question: finding amount of unknown charge

Physics 205B Quiz 3, spring semester 2020
Cuesta College, San Luis Obispo, CA

A –4.5 nC point charge and an unknown positive point charge are held a distance of 3.0 cm apart. The magnitude of the electric force of the –4.5 nC charge on the unknown positive charge is 7.6×10^–5 N. The amount of the unknown positive charge is:
(A) 1.7×10^–9 C.
(B) 5.6×10^–8 C
(C) 1.7×10⁴ C.
(D) 4.5×10⁴ C.

Correct answer (highlight to unhide): (A)

The magnitude of the force on the unknown positive point charge is given by:

|F_{1 on 2}| = k·|q₁|·|q₂|/(r²),

where q₁ is the –4.5 nC point charge, and q₂ is the unknown positive charge. Then the absolute value of the unknown charge can then be solved for:

|q₂| = |F_{1 on 2}|·(r²)/(k·|q₁|),

|q₁| = |7.6×10^–5 N|·(0.030 m)²/((8.99×10⁹ N·m²/C²)·|–4.4×10^–9 C|),

|q₁| = 1.6907675195...×10^–9 C,

or to two significant figures, the amount of the unknown (positive) charge is (+)1.7×10^–9 C.

(Response (B) is |F_{1 on 2}|·r/(k·|q₁|); response (C) is |F_{1 on 2}|/|q₁| (which is the electric field of the unknown positive charge at the location of the –4.5 nC charge); and response (D) is k·|q₁|/r² (which is the electric field of the –4.5 nC charge at the location of the unknown positive charge).)

Sections 30882, 30883
Exam code: quiz03Cv1d
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(B) : 1 student
(C) : 4 students
(D) : 1 student

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Physics quiz question: object inside converging lens focal point

Physics 205B Quiz 2, spring semester 2020
Cuesta College, San Luis Obispo, CA

An object 1.0 cm in height is placed 16 cm in front of a converging lens with a focal length of +25 cm. The resulting image is:
(A) upright, diminished.
(B) upright, enlarged.
(C) inverted, diminished.
(D) inverted, enlarged.
(E) (No image would be produced.)

Correct answer (highlight to unhide): (B)

Solving for the location of the image d_i:

(1/d_o) + (1/d_i) = (1/f),

(1/d_i) = (1/f) – (1/d_o),

(1/d_i) = (1/(+25 cm)) – (1/(+16 cm)) = –0.0225 cm^–1,

d_i = 1/(–0.0225 cm^–1) = –44.444... cm,

where the negative sign by convention makes this a virtual image located to the right side of the lens (the side of the lens opposite the original object). The linear magnification m is given by:

m = h_i/h_o = –d_i/d_o = –(–44.444... cm)/(16 cm) = +2.777...,

which means that this virtual image is upright, due to the positive sign, and is enlarged, being about 2.8 times the size of the original object.

Sections 30882, 30883
Exam code: quiz02B3rD
(A) : 1 student
(B) : 21 students
(C) : 3 students
(D) : 10 students
(E) : 0 students

Success level: 60%
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Physics quiz question: comparing blue, yellow laser frequencies

Physics 205B Quiz 1, spring semester 2020
Cuesta College, San Luis Obispo, CA

"laser3.jpg"
©CrystaLaser
https://www.crystalaser.com/new/yellowlaser.html

Blue laser light has a wavelength of 445 nm light in air, while yellow laser light has a wavelength of 594 nm in air. The __________ laser light has a higher frequency in air.
(A) blue.
(B) yellow.
(C) (There is a tie.)
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (A)

Wave speed v depends on the properties of the medium. Frequency f depends on the properties of the source. These two parameters can be varied independently of each other.

The wavelength λ is the parameter dependent on both of the independent parameters:

λ = v/f.

Since the two laser light colors have the same speed v (as they travel through the same medium), then because of their different frequencies, they will have have different λ wavelengths (as this parameter depends on both the wave source and the properties of the medium):

λ₁ = v₁/f₁,

λ₂ = v₂/f₂,

where λ₁ < λ₂, as blue light (here, λ₁) has a much shorter wavelength than yellow light (here, λ₂). Since v₁ = v₂ (as both forms of laser light travel through the air), then f₁ > f₂, and thus blue laser light has a higher frequency than yellow laser light.

Sections 30882, 30883
Exam code: quiz01PxP7
(A) : 22 students
(B) : 8 students
(C) : 5 students
(D) : 0 students

Success level: 63%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.67

Physics quiz question: polarized light transmitted through two polarizers

Physics 205B Quiz 1, spring semester 2020
Cuesta College, San Luis Obispo, CA

Vertically polarized light is incident on a set of two ideal polarizers.


If the total fraction of the incident light intensity transmitted through both polarizers is 0.15, then the angle θ of the second polarizer is:
(A) 23°.
(B) 44°.
(C) 67°.
(D) 81°.

Correct answer (highlight to unhide): (C)

The fraction of the vertically polarized light that passes through polarizer 1 is cos²θ, where the angle θ = 0° is measured between the vertical polarization of the light entering polarizer 1, and the transmission axis of polarizer 1 (which is also vertical). The light after passing through polarizer 1, but before passing through polarizer 2 is still vertical (having a polarization that matches the transmission axis of polarizer 1).


The fraction of this vertically polarized light that passes through polarizer 2 is again cos²θ, where the angle θ measured between the vertical polarization of the light entering polarizer 2 and the transmission axis of polarizer 2 (rotated clockwise from the vertical) is unknown.


Since the total fraction of the light that passes through both polarizers is given, then this is the result of multiplying together the fractions that passed through each polarizer:

total fraction = (fraction 1)×(fraction 2),

0.15 = (1)×(cos²θ),

√(0.15) = cosθ

cos^–1(√(0.15)) = θ = 67.213502000402852°,

or to two significant figures, the clockwise angle θ that the transmission axis of polarizer 2 is rotated from vertical is 67°.

(Response (A) is cos^–1(√(0.85)); response (B) is (1/0.15)²; and response (D) is cos^–1(0.15).)

Student responses
Sections 30882, 30883
Exam code: quiz01PxP7
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(B) : 1 student
(C) : 30 students
(D) : 2 students

Success level: 86%
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Physics quiz archive: electromagnetic waves, polarization, reflection/refraction

Physics 205B Quiz 1, spring semester 2020
Cuesta College, San Luis Obispo, CA
Sections 30882, 30883, version 1
Exam code: quiz01PxP7



Sections 30882, 30883 results

0- 6 :	* [low = 6]
7-12 :	**
13-18 :	*********
19-24 :	*********** [mean = 21.9 +/- 5.9]
25-30 :	************ [high = 30]

Physics quiz question: concrete beam contraction

Physics 205A Quiz 7, fall semester 2019
Cuesta College, San Luis Obispo, CA

A concrete beam (linear expansion coefficient 9.8×10^–6 K^–1) is 12.2 m long at room temperature (293 K). In order to contract by 0.0010 m, its temperature should be decreased by:
(A) 1.2×10^–5 K.
(B) 8.2×10^–5 K.
(C) 0.38 K.
(D) 8.4 K.

Correct answer: (D)

The relation between the change in length ∆L due to a temperature change ∆T is given by:

α·∆T = ∆L/L,

such that:

∆T = ∆L/(α·L) = (–0.0010 m)/((9.8×10^–6 K^–1)·(12.2 m)) = –8.36400133824 K,

or to two significant figures, the decrease in temperature is 8.4 K or 8.4° C.

(Response (A) is L·α; response (B) is ∆L/L; response (C) is ∆L/(α·T₀).)

Sections 70854, 70855
Exam code: quiz07VlnC
(A) : 3 students
(B) : 8 students
(C) : 3 students
(D) : 37 students

Success level: 73%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.49

Physics quiz question: concrete vs. glued laminated wood beam contraction

Physics 205A Quiz 7, fall semester 2019
Cuesta College, San Luis Obispo, CA

"Finger joints in glue-lam beams, Polkky mill, Kuusamo2"
Eli Sagor
https://flic.kr/p/9xRT62

A concrete beam (linear expansion coefficient 9.8×10^–6 K^–1) and glued laminated wood beam (linear expansion coefficient 3.6×10^–6 K^–1[*]) have the same length at room temperature (293 K). If they both decrease in temperature by the same amount, the length of the concrete beam will be __________ the length of the wood beam. (A) shorter than.
(B) equal to.
(C) longer than.
(D) (Not enough information is given.)

[*] awc.org/pdf/codes-standards/publications/archives/lrfd/AWC-LRFD1996-Glulam-0203.pdf.

Correct answer: (A)

The relation between the change in length ∆L due to a temperature change ∆T is given by:

α_concrete·∆T = ∆L_concrete/L,

and similarly for the glued laminated beam:

α_wood·∆T = ∆L_wood/L,

where the original length L and the temperature change ∆T is the same for both the concrete and glued laminated beam. Solving for this common factor for both beams:

L·∆T = ∆L_concrete/α_concrete,

L·∆T = ∆L_wood/α_wood,

we can then set this common factor for both beams equal to each other:

L·∆T = L·∆T,

∆L_concrete/α_concrete = ∆L_wood/α_wood.

From inspection, since α_concrete > α_wood, then for the equality to hold, ∆L_concrete > ∆L_wood, and so the concrete beam would contract more, and as a result be shorter than the glued laminated beam.

Sections 70854, 70855
Exam code: quiz07VlnC
(A) : 18 students
(B) : 7 students
(C) : 26 students
(D) : 0 students

Success level: 35%
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Physics quiz question: comparing specific heat capacity values

Physics 205A Quiz 7, fall semester 2019
Cuesta College, San Luis Obispo, CA

A 0.25 kg glass sample (400 K initial temperature) and a 0.75 kg graphite sample (300 K initial temperature) are brought in contact with each other to reach thermal equilibrium with a final temperature of 330 K. Ignore heat exchanged with the environment. The specific heat capacity values of these two substances are not known. The __________ sample has the larger specific heat capacity value.
(A) glass.
(B) graphite.
(C) (There is a tie.)
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (A)

The transfer/balance energy conservation equation for this system is given by:

Q_ext = ∆E_glass + ∆E_graphite,

and since there is no heat exchanged with the environment, Q_ext = 0, such that:

0 = m_glass·c_glass·ΔT_glass + m_graphite·c_graphite·ΔT_graphite,

and:

–m_glass·c_glass·ΔT_glass = m_graphite·c_graphite·ΔT_graphite,

–(0.25 kg)·c_glass·(–70 K) = (0.75 kg)·c_graphite·(+30 K),

(17.6 kg·K)·c_glass = (22.5 kg·K)·c_graphite.

From inspection, for the equality to hold, c_glass > c_graphite, and thus glass must have a greater specific heat capacity value than graphite.

Sections 70854, 70855
Exam code: quiz07VlnC
(A) : 25 students
(B) : 17 students
(C) : 3 students
(D) : 6 students

Success level: 49%
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Physics quiz question: doubling sleeping pad thickness

Physics 205A Quiz 7, fall semester 2019
Cuesta College, San Luis Obispo, CA

"Klymit-Static-V-Luxe-side"
Rebecca J. Crawford
hikingmastery.com/gear/sleeping-pads/klymit-static-v2-sleeping-pad.html

A Klymit® Static V sleeping pad[*] (area of 1.07 m² and thickness of 0.0635 m) has a thermal resistance of 0.21 K/watts. If the pad were made twice as thick, its __________ would increase.
(A) thermal resistance.
(B) thermal conductivity.
(C) (Both of the above choices.)
(D) (Neither of the above choices.)

[*] klymit.com/static-v-camping-sleeping-pad.html.

Correct answer (highlight to unhide): (A)

The thermal resistance R of an object can be related to the thermal conductivity κ of its material:

R = d/(κ·A),

where d is the thickness of the object that heat must conduct through, and A is the cross-sectional area. As the thermal conductivity κ is an intensive property of the material, it would not change because of doubling the thickness of the pad, in contrast to the thermal resistance, which would increase as a result.

Sections 70854, 70855
Exam code: quiz07VlnC
(A) : 36 students
(B) : 6 students
(C) : 9 students
(D) : 0 students

Success level: 71%
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Physics quiz question: changing emissivity of a brick

Physics 205A Quiz 7, fall semester 2019
Cuesta College, San Luis Obispo, CA

"Red bricks"
Ramesh NG
flic.kr/p/8TmkRf

A red brick has an emissivity of 0.93[*], and has a surface temperature of 270 K. Ignore conduction and convection heat transfers to/from the environment. If the brick were coated with aluminum paint such that its emissivity was lowered to 0.45, the rate of heat per time the brick _________ would decrease.
(A) radiates to the environment.
(B) absorbs from the environment.
(C) (Both of the above choices.)
(D) (Neither of the above choices.)

[*] thermoworks.com/emissivity-table.

Correct answer (highlight to unhide): (C)

The net power (rate of heat per time) radiated is given by:

Power = –e·σ·A·((T_obj)⁴ + (T_env)⁴) = –e·σ·A·(T_obj)⁴ – e·σ·A·(T_env)⁴,

where the first (negative) term corresponds to the rate of heat being removed (radiated) from the object, while the second (positive) term corresponds to the rate of heat being put into (absorbed) by the object (in order to be consistent with the ±Q convention for removing heat from (–) or putting heat into (+) a thermodynamic system).

As it is a factor common to both the radiation and absorption terms, lowering the emissivity value from 0.93 to 0.45 would then decrease both the rate of heat per time radiated to the environment and the rate of heat per time absorbed from the environment.

Sections 70854, 70855
Exam code: quiz07VlnC
(A) : 10 students
(B) : 5 students
(C) : 35 students
(D) : 1 student

Success level: 69%
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Physics quiz question: Young's modulus of steel

Physics 205A Quiz 6, fall semester 2019
Cuesta College, San Luis Obispo, CA

A sample of "#4" steel reinforcing bar has a length of 4.9 m and a cross-sectional area of 1.3×10^–4 m². A force of 1.1×10⁵ N is applied to stretch this steel bar by 0.020 m. The Young's modulus of steel is:
(A) 3.5×10⁶ N/m².
(B) 8.5×10⁸ N/m².
(C) 4.2×10¹⁰ N/m².
(D) 2.1×10¹¹ N/m².

[*] webcivil.com/usrcrebar.aspx.

Correct answer (highlight to unhide): (D)

Hooke's law is given by:

(F/A) = Y·(∆L/L),

such that the Young's modulus would be:

(F/A)·(L/∆L) = Y = ((1.1×10⁵ N)/(1.3×10^–4 m²))·((4.9 m)/(0.020 m)),

Y = 207,307,692,307.692 N/m²,

or to two significant figures, the Young's modulus for steel is: 2.1×10¹¹ N/m².

(Response (A) is (F/A)·(∆L/L); response (B) is F/A; response (C) is F/(A·∆L).)

Sections 70854, 70855
Exam code: quiz06co6O
(A) : 5 students
(B) : 2 students
(C) : 2 students
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Physics quiz question: amount of mass attached to a spring

Physics 205A Quiz 6, fall semester 2019
Cuesta College, San Luis Obispo, CA

A mass is attached to a horizontal spring (with a spring constant of 
40 N/m), and has a 0.80 s period of oscillation. Neglect friction and drag. The mass attached to this spring is:
(A) 0.13 kg.
(B) 0.16 kg.
(C) 0.65 kg.
(D) 5.1 kg.

Correct answer (highlight to unhide): (C)

The period T of a mass m attached to a spring with spring strength constant k is given by:

T = 2·π·√(m/k),

such that the mass m will be:

T/(2·π) = √(m/k),

(T/(2·π))² = m/k,

k·(T/(2·π))² = m = 0.6484555753 kg,

or two significant figures, the mass attached to the spring is 0.65 kg.

(Response (A) is T/(2·π); response (B) is (2·π)/k; and response (D) is k·T/(2·π).)

Sections 70854, 70855
Exam code: quiz06co6O
(A) : 1 students
(B) : 4 students
(C) : 45 students
(D) : 2 students

Success level: 87%
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Physics quiz question: finding acceleration due to gravity from pendulum

Physics 205A Quiz 6, fall semester 2019
Cuesta College, San Luis Obispo, CA

The California Academy of Sciences in San Francisco has a pendulum with a period of 6.406 s, consisting of a 106.6 kg ball attached to a 10.20 m long wire hanging from the ceiling. Assume that the ball can be considered a simple point mass. Neglect friction and drag. The magnitude of the acceleration due to gravity g at that location is:
(A) 9.432 m/s².
(B) 9.620 m/s².
(C) 9.808 m/s².
(D) 9.813 m/s².

[*] kathleensf.files.wordpress.com/2008/07/the-foucault-pendulum-at-the-california-academy-of-sciences-december-14-2010.pdf.

Correct answer (highlight to unhide): (D)

The period of a pendulum is given by:

T = 2·π·√(L/g),

Since the period T = 6.406 s and string length L = 10.20 m are known, the acceleration due to gravity g can then be solved for:

T/(2·π) = √(L/g),

(T/(2·π))² = L/g,

g = L·((2·π)/T)²,

g = (10.20 m)·((2·π)/(6.406 s))² = 9.8126439271 m/s²,

or to four significant figures, the acceleration due to gravity is 9.813 m/s².

(Response (A) is (1/L)·((2·π)/T)²; response (B) is 10·((2·π)/T)²); response (C) is 10·((2·π)/T).)

Sections 70854, 70855
Exam code: quiz06co6O
(A) : 1 student
(B) : 5 students
(C) : 9 students
(D) : 37 students

Success level: 71%
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Physics quiz question: linear mass density of viola string

Physics 205A Quiz 6, fall semester 2019
Cuesta College, San Luis Obispo, CA

Transverse waves travel at a speed of 357 m/s along a viola's A-string, stretched to a tension of 52.9 N[*][**]. The linear mass density of this string is:
(A) 4.15×10^–4 kg/m.
(B) 2.20×10^–2 kg/m.
(C) 0.148 kg/m.
(D) 0.385 kg/m.

[*] theviolaworkshop.com/page16.html.
[**] gamutmusic.com/viola-tensions.

Correct answer (highlight to unhide): (A)

The speed v of transverse waves along the viola string depends on the tension F and the linear mass density (mass per unit length) (m/L):

v = √(F/(m/L)).

Solving for the linear mass density results in:

v² = F/(m/L),

(m/L) = F/(v²),

(m/L) = (52.9 N)/(357 m/s)² = 0.000415067988 kg/m,

or to three significant figures, 4.15×10^–4 kg/m.

(Response (B) is (F/v)²; response (C) is F/v; response (D) is √(F/v).)

Sections 70854, 70855
Exam code: quiz06co6O
(A) : 35 students
(B) : 7 students
(C) : 8 students
(D) : 2 students

Success level: 67%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.78

Physics quiz question: energy changes of barrel rolled up a ramp

Physics 205A Quiz 5, fall semester 2019
Cuesta College, San Luis Obispo, CA

A Physics 205A student uses a ramp and a trailer to deliver an empty barrel. Ignore friction and drag; the barrel rolls without slipping. As the barrel is rolled up the ramp with constant speed, its __________ increases.
(A) gravitational potential energy.
(B) translational kinetic energy.
(C) rotational kinetic energy.
(D) (Two of the above choices.)
(E) (All of the above choices.)
(F) (None of the above choices.)

Correct answer (highlight to unhide): (A)

The energy transfer-balance equation is given by:

W_nc = ∆KE_tr + ∆KE_rot ∆PE_grav + ∆PE_elas,

where ∆PE_elas = 0, as there is no spring involved in this process.

Just looking at the three remaining terms on the right-hand side of the energy transfer-balance equation, for the change in translational kinetic energy:

∆KE_tr = (1/2)·m·(v_f² – v₀²),

and since the speed is constant, v₀ and v_f have the same magnitude, then KE_tr is constant (∆KE_tr = 0).

Similarly for the change in rotational kinetic energy:

∆KE_rot = (1/2)·I·(ω_f² – ω₀²),

and since the angular speed is constant, ω₀ and ω_f have the same magnitude, then KE_rot is constant (∆KE_rot = 0).

Then for the change in gravitational potential energy:

∆PE_grav = m·g·(y_f – y₀),

since y_f is greater than y₀, then PE_grav increases (∆PE_grav is positive).

(In order for the equality to hold for the energy transfer-balance equation, the student must then be doing positive work on the barrel in order to increase its gravitational potential energy.)

Sections 70854, 70855
Exam code: quiz05Gu1L
(A) : 41 students
(B) : 1 student
(C) : 0 students
(D) : 7 students
(E) : 3 students
(F) : 0 students

Success level: 79%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.53

Physics quiz question: comparing torques on supported square

Physics 205A Quiz 5, fall semester 2019
Cuesta College, San Luis Obispo, CA

A uniform 0.40 m × 0.40 m square with 
a mass of 0.50 kg is pivoted at one corner. It is supported by a 45° diagonal force at the opposite corner such that the bottom edge is parallel to the ground. Calculate all torques with respect to the corner pivot. The torque exerted by the __________ has a greater magnitude. (A) weight force.
(B) support force.
(C) (There is a tie.)
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (C)

Since the square is in static equilibrium ("is supported") and does not rotate, then the net torque on it is equal to zero (Στ = 0), such that the counterclockwise torque of the weight force on the square must equal the clockwise torque of the support force on the square:

(ccw) τ_w = (cw) τ_Fsupport.

Sections 70854, 70855
Exam code: quiz05Gu1L
(A) : 23 students
(B) : 21 students
(C) : 7 students
(D) : 0 students

Success level: 13%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.24

Physics quiz question: weight torque on supported square

Physics 205A Quiz 5, fall semester 2019
Cuesta College, San Luis Obispo, CA

A uniform 0.40 m × 0.40 m square with 
a mass of 0.50 kg is pivoted at one corner. It is supported by a 45° diagonal force at the opposite corner such that the bottom edge is parallel to the ground. Calculate all torques with respect to the corner pivot. The magnitude of the torque exerted by the weight force is:
(A) 0.98 N·m.
(B) 1.4 N·m.
(C) 2.0 N·m.
(D) 2.8 N·m.

Correct answer (highlight to unhide): (A)

The weight force w of Earth acting on the beam acts at the center of gravity, directly downwards with a magnitude m·g. The perpendicular lever arm ℓ for the weight force on the beam must extend from the pivot point to perpendicularly intercept the weight force line of action (which lies along the weight force vector), such that this will be a horizontal line that is half the side of the square:

ℓ = 0.20 m.

The magnitude of the (counterclockwise) torque exerted by the weight force on the beam is then:

τ = w·ℓ,

τ = (m·g)·(0.20 m),

τ = ((0.50 kg)·(9.80 m/s²))·(0.20 m),

τ = 0.98 N·m.

(Response (B) is m·g·(0.20 m)/sin(45°); response (C) is m·g·(0.40 m); and response (D) is m·g·(0.40 m)/sin(45°).)

Sections 70854, 70855
Exam code: quiz05Gu1L
(A) : 26 students
(B) : 16 students
(C) : 6 students
(D) : 4 students

Success level: 50%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.65