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# Part (a): Solving Using Conservation of Energy
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] U_i + K_i = U_f + K_f [/katex] | 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 | [katex] U_i = m_g \cdot g \cdot h [/katex] | Initial potential energy is due to [katex] m_g [/katex] being at a height [katex] h [/katex] from the ground. [katex] m_y [/katex] is on the ground, so its potential energy is zero. |
| 3 | [katex] U_i = 38 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \cdot 2.5 \, \text{m} [/katex] | Calculate the numerical value of [katex] U_i [/katex]. [katex] g [/katex] (acceleration due to gravity) is 9.8 m/s² and [katex] h [/katex] is 2.5 m. |
| 4 | [katex] U_i = 931 \, \text{J} [/katex] | Multiplying the values provides the initial potential energy. |
| 5 | [katex] K_i = 0 [/katex] | Initially, the system is at rest so the initial kinetic energy is zero. |
| 6 | [katex] K_f = \frac{1}{2}(m_y + m_g + m_p) v^2 [/katex] | Final kinetic energy calculated as the sum of the kinetic energies of [katex] m_g [/katex], [katex] m_y [/katex] and the pulley. [katex] m_p [/katex] is the mass of the pulley. The pulley contributes translational kinetic energy due to its mass. |
| 7 | [katex] U_f = 0 [/katex] | At the final state, [katex] m_g [/katex] is on the ground, so its potential energy is zero. |
| 8 | [katex] 931 \, \text{J} = \frac{1}{2} (32 \, \text{kg} + 38 \, \text{kg} + 3.1 \, \text{kg}) v^2 [/katex] | Use [katex] U_i + K_i = U_f + K_f [/katex] and solve for [katex] v [/katex]. |
| 9 | [katex] v = \sqrt{\frac{2 \times 931 \, \text{J}}{73.1 \, \text{kg}}} [/katex] | Simplify the equation to solve for [katex] v [/katex]. |
| 10 | [katex] v = \sqrt{25.48 \, \text{m}^2/\text{s}^2} [/katex] | Calculate the value under the square root. |
| 11 | [katex] v = 5.05 \, \text{m/s} [/katex] | The final velocity of [katex] m_g [/katex] just before it strikes the ground. |
# Part (b): Solving Using Forces, Torque, and Kinematics
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] F = m \cdot a [/katex] | Newton’s second law, the net force applied on the system causes acceleration. |
| 2 | [katex] F_{my} – F_{mg} = (m_y + m_g + m_p) \cdot a [/katex] | Forces due to weights of [katex] m_y [/katex] and [katex] m_g [/katex], including the pulley’s inertia, where [katex] F_{my} = m_y \cdot g [/katex] and [katex] F_{mg} = m_g \cdot g [/katex]. |
| 3 | [katex] T = \frac{m_g g – m_y g}{1 + \frac{m_p}{m_y + m_g}} [/katex] | Balance the forces taking into account the pulley rotation effect, where [katex] T [/katex] is the net torque and [katex] m_p [/katex] is moment of inertia derived as [katex] \frac{1}{2} m_p R^2 [/katex] assuming a uniform cylinder. |
| 4 | [katex] a = \frac{m_g g – m_y g}{m_y + m_g + \frac{1}{2} m_p} [/katex] | Solving the earlier equation for [katex] a [/katex], the acceleration of the system, where [katex] \frac{1}{2} m_p [/katex] comes from dividing the pulley’s moment of inertia by [katex] R^2 [/katex]. |
| 5 | [katex] v_f^2 = 2 \cdot a \cdot 2.5 \, \text{m} [/katex] | Use kinematic equation for final velocity, incorporating the calculated acceleration and distance fallen, which is 2.5 m. |
| 6 | [katex] v_f = \sqrt{2 \cdot a \cdot 2.5 \, \text{m}} [/katex] | Solve for final velocity using the value for [katex] a [/katex]. |
| 7 | Approximate numerical calculation similar to Part (a) | The velocity is essentially the same result as obtained from energy conservation, demonstrating the physics consistency. |
Just ask: "Help me solve this problem."
Two uniform solid balls, one of radius R and mass M, the other of radius 2R and mass 8M, roll down a high incline. They start together from rest at the top of the incline. Which one will reach the bottom of the incline first?
A rotating, rigid body makes 10 complete revolutions in 10 seconds. What is its average angular velocity?
The system in the Figure is in equilibrium. A concrete block of mass 225 kg hangs from the end of a uniform strut whose mass is 45.0 kg.
A seesaw is balanced on a fulcrum, with a boy of mass [katex] M_1 [/katex] sitting on one end and a girl of mass [katex] M_2 [/katex] sitting on the other end. The seesaw is a uniform plank of length [katex]L[/katex] and mass [katex] M[/katex]. The fulcrum is located at the midpoint of the plank. Does [katex] M_1 = M_2 [/katex]. Justify your working.
A 150-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope.
What constant force must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500 rev/s in 2.00 s?
Note: [katex] I_\text{disk} = \frac{1}{2}mr^2 [/katex]
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [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 |
| Variable | SI 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] |
| Variable | Derived 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]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/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.
Start with the given measurement: [katex]\text{5 km}[/katex]
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]
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]
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] |
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