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Objective: Determine the acceleration of the blocks and the tension in the cord given the masses, coefficients of friction, and incline angle given:

• [katex] m_A = m_B = 5.0 , \text{kg} [/katex]
• [katex] \mu_A = 0.20 [/katex]
• [katex] \mu_B = 0.30 [/katex]
• [katex] \theta = 32^\circ [/katex]
• [katex] g = 9.8 , \text{m/s}^2 [/katex] (acceleration due to gravity)

Part a: Calculate the acceleration of the blocks

Step Formula Derivation Reasoning
1 [katex] F_{\text{gravity, parallel A}} = m_A g \sin(\theta) [/katex] Parallel component of gravitational force for A.
2 [katex] F_{\text{gravity, parallel B}} = m_B g \sin(\theta) [/katex] Parallel component of gravitational force for B.
3 [katex] F_{\text{friction A}} = \mu_A m_A g \cos(\theta) [/katex] Frictional force on A.
4 [katex] F_{\text{friction B}} = \mu_B m_B g \cos(\theta) [/katex] Frictional force on B.
5 [katex] F_{\text{net A}} = F_{\text{gravity, parallel A}} – F_{\text{friction A}} [/katex] Net force on A.
6 [katex] F_{\text{net B}} = F_{\text{gravity, parallel B}} – F_{\text{friction B}} [/katex] Net force on B.
7 [katex] F_{\text{net}} = F_{\text{net B}} – F_{\text{net A}} [/katex] Total net force on the system.
8 [katex] a = \frac{F_{\text{net}}}{m_A + m_B} [/katex] Acceleration of the system.

Plug in the given values:

Step Formula Derivation Reasoning
9 [katex] a = \frac{(m_B g \sin(\theta) – \mu_B m_B g \cos(\theta)) – (m_A g \sin(\theta) – \mu_A m_A g \cos(\theta))}{m_A + m_B} [/katex] Substitute the net forces from steps 5 and 6.
10 [katex] a = \frac{(5 \cdot 9.8 \cdot \sin(32^\circ) – 0.30 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ)) – (5 \cdot 9.8 \cdot \sin(32^\circ) – 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ))}{5 + 5} [/katex] Substitute given values.
11 [katex] a = \frac{(5 \cdot 9.8 \cdot (\sin(32^\circ) – 0.30 \cdot \cos(32^\circ))) – (5 \cdot 9.8 \cdot (\sin(32^\circ) – 0.20 \cdot \cos(32^\circ)))}{10} [/katex] Simplify the expression.
12 [katex] a = \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{10} [/katex] Combine like terms.
13 [katex] a = \frac{9.8 \cdot (0.10 \cdot \cos(32^\circ))}{2} [/katex] Simplify further.

Part b: Calculate the tension in the cord

Step Formula Derivation Reasoning
1 [katex] T = m_A \cdot a + F_{\text{friction A}} [/katex] Tension equals the force to accelerate block A plus frictional force on A.
2 [katex] F_{\text{friction A}} = \mu_A \cdot m_A \cdot g \cdot \cos(\theta) [/katex] Frictional force opposing the motion of block A.
3 [katex] T = m_A \cdot a + \mu_A \cdot m_A \cdot g \cdot \cos(\theta) [/katex] Substitute the frictional force into the tension formula.
4 [katex] T = m_A \cdot \left( \frac{F_{\text{net}}}{m_A + m_B} \right) + \mu_A \cdot m_A \cdot g \cdot \cos(\theta) [/katex] Substitute the expression for from the acceleration calculation.
5 [katex] T = 5 \cdot \left( \frac{(5 \cdot 9.8 \cdot \sin(32^\circ) – 0.30 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ)) – (5 \cdot 9.8 \cdot \sin(32^\circ) – 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ))}{10} \right) + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) [/katex] Insert given values for masses, coefficients of friction, gravitational acceleration, and angle.
6 [katex] T = 5 \cdot \left( \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{10} \right) + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) [/katex] Simplify the expression for the net force component.
7 [katex] T = \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{2} + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) [/katex] Further simplify the tension formula.
8 [katex] T = \frac{5 \cdot 9.8 \cdot 0.10 \cdot \cos(32^\circ)}{2} + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) [/katex] Combine like terms for the final tension calculation.
9 [katex] T \approx 17.66 , \text{N} [/katex] Calculate the numeric value for tension.

[katex] \boxed{T = 17.66 , \text{N}} [/katex]

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1. .42 m/s2
2. 17.66

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KinematicsForces
[katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex][katex]F = ma[/katex]
[katex]v = v_i + at[/katex][katex]F_g = \frac{G m_1m_2}{r^2}[/katex]
[katex]a = \frac{\Delta v}{\Delta t}[/katex][katex]f = \mu N[/katex]
[katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex]
Circular MotionEnergy
[katex]F_c = \frac{mv^2}{r}[/katex][katex]KE = \frac{1}{2} mv^2[/katex]
[katex]a_c = \frac{v^2}{r}[/katex][katex]PE = mgh[/katex]
[katex]KE_i + PE_i = KE_f + PE_f[/katex]
MomentumTorque and Rotations
[katex]p = m v[/katex][katex]\tau = r \cdot F \cdot \sin(\theta)[/katex]
[katex]J = \Delta p[/katex][katex]I = \sum mr^2[/katex]
[katex]p_i = p_f[/katex][katex]L = I \cdot \omega[/katex]
Simple Harmonic Motion
[katex]F = -k x[/katex]
[katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex]
[katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex]
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: [katex]\text{5 km}[/katex]

2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity [katex] v_i [/katex] is written as [katex] u [/katex]; sometimes [katex] \Delta x [/katex] is written as [katex] s [/katex].
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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