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# Part (a): Solving Using Conservation of Energy

Step Derivation/Formula Reasoning
1 U_i + K_i = U_f + K_f The total mechanical energy (kinetic + potential) at the initial state equals the total mechanical energy at the final state due to conservation of energy.
2 U_i = m_g \cdot g \cdot h Initial potential energy is due to m_g being at a height h from the ground. m_y is on the ground, so its potential energy is zero.
3 U_i = 38 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \cdot 2.5 \, \text{m} Calculate the numerical value of U_i . g (acceleration due to gravity) is 9.8 m/s² and h is 2.5 m.
4 U_i = 931 \, \text{J} Multiplying the values provides the initial potential energy.
5 K_i = 0 Initially, the system is at rest so the initial kinetic energy is zero.
6 K_f = \frac{1}{2}(m_y + m_g + m_p) v^2 Final kinetic energy calculated as the sum of the kinetic energies of m_g , m_y and the pulley. m_p is the mass of the pulley. The pulley contributes translational kinetic energy due to its mass.
7 U_f = 0 At the final state, m_g is on the ground, so its potential energy is zero.
8 931 \, \text{J} = \frac{1}{2} (32 \, \text{kg} + 38 \, \text{kg} + 3.1 \, \text{kg}) v^2 Use U_i + K_i = U_f + K_f and solve for v .
9 v = \sqrt{\frac{2 \times 931 \, \text{J}}{73.1 \, \text{kg}}} Simplify the equation to solve for v .
10 v = \sqrt{25.48 \, \text{m}^2/\text{s}^2} Calculate the value under the square root.
11 v = 5.05 \, \text{m/s} The final velocity of m_g just before it strikes the ground.

# Part (b): Solving Using Forces, Torque, and Kinematics

Step Derivation/Formula Reasoning
1 F = m \cdot a Newton’s second law, the net force applied on the system causes acceleration.
2 F_{my} – F_{mg} = (m_y + m_g + m_p) \cdot a Forces due to weights of m_y and m_g , including the pulley’s inertia, where F_{my} = m_y \cdot g and F_{mg} = m_g \cdot g .
3 T = \frac{m_g g – m_y g}{1 + \frac{m_p}{m_y + m_g}} Balance the forces taking into account the pulley rotation effect, where T is the net torque and m_p is moment of inertia derived as \frac{1}{2} m_p R^2 assuming a uniform cylinder.
4 a = \frac{m_g g – m_y g}{m_y + m_g + \frac{1}{2} m_p} Solving the earlier equation for a , the acceleration of the system, where \frac{1}{2} m_p comes from dividing the pulley’s moment of inertia by R^2 .
5 v_f^2 = 2 \cdot a \cdot 2.5 \, \text{m} Use kinematic equation for final velocity, incorporating the calculated acceleration and distance fallen, which is 2.5 m.
6 v_f = \sqrt{2 \cdot a \cdot 2.5 \, \text{m}} Solve for final velocity using the value for a .
7 Approximate numerical calculation similar to Part (a) The velocity is essentially the same result as obtained from energy conservation, demonstrating the physics consistency.

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1. v = 5.05 \, \text{m/s}
2. v = 5.05 \, \text{m/s} . The velocity is essentially the same result as obtained from energy conservation, demonstrating the physics consistency.

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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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