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# Part (a) – To determine the acceleration of the system

Step Derivation/Formula Reasoning
1 T_1 – T_2 = I \alpha This equation represents the relationship between the tensions on either side of the pulley (T_1 and T_2), the moment of inertia of the pulley (I), and the angular acceleration (\alpha).
2 I = \frac{1}{2} m_p r^2 The moment of inertia (I) of a disk-shaped pulley, where m_p is the mass of the pulley and r is its radius.
3 \alpha = a/r The angular acceleration (\alpha) is related to the linear acceleration (a) of the falling mass by the radius of the pulley (r).
4 f_k = \mu_k T_s The frictional force f_k is calculated using the coefficient of kinetic friction (\mu_k) and the tangential component of the tension T_s, which can be approximated as T_s = (T_1 + T_2)/2.
5 T_1 – T_2 – f_k r = I \alpha Combine the effects of tension and frictional force (considering the friction opposes the motion thus negative), and include the effect of the pulley’s inertia.
6 (m_2 g – T_2) – (T_1 – m_1 g) = (m_1 + m_2) a The forces acting on the blocks (m_1 g – T_1 downwards on m_1 and m_2 g – T_2 downwards on m_2, assuming m_1 < m_2 so m_2 descends) produces the net force, which equals the total system mass times the system’s acceleration.
7 Solve for a using the equations Combine all equations replacing \alpha with a/r and solve for a to get the system’s acceleration, involving simplifications and algebraic manipulation.

# Part (b) – To determine the tension in the rope

Step Derivation/Formula Reasoning
1 m_1 g – T_1 = m_1 a For mass m_1, the net force is the difference between the weight and the tension, which equals its mass times acceleration.
2 T_1 = m_1 g – m_1 a Solving the above equation for T_1 gives the tension in the rope on the side of mass m_1.
3 T_1 calculation Plug the values of g, m_1, and a obtained from part (a) to find T_1.

# Part (c) – To determine the magnitude and direction of the frictional force exerted on the pulley

Step Derivation/Formula Reasoning
1 f_k = \mu_k \frac{T_1 + T_2}{2} Using the approximation that the mean tension provides the force of friction calculation against the direction of motion.
2 f_k calculation Calculate the value by substituting \mu_k, T_1, and T_2 values (where T_2 can be similarly calculated like T_1).
3 f_k direction determination The frictional force opposes the direction of relative motion between the rope and the pulley, which will be clockwise due to the motion of m_2 falling downward.

This sequence of tables will help in understanding the steps involved thoroughly to solve the given problem numerically. Please note the precise solutions would require the use of actual numerical values derived or given directly in the problem (e.g., gravitational acceleration g = 9.8 \, \text{m/s}^2 ).

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

General Metric Conversion Chart

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

1. Start with the given measurement: \text{5 km}

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

3. Perform the multiplication: \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}

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

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

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 v_i is written as u ; sometimes \Delta x is written as s .
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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