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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] U_{\text{spring}} = \frac{1}{2} k x^2 [/katex] | The potential energy stored in the compressed spring, where [katex] k [/katex] is the spring constant and [katex] x [/katex] is the compression distance. |
| 2 | [katex] W_{\text{friction}} = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d [/katex] | Work done against friction while moving up the incline: [katex] \mu_k [/katex] is the coefficient of kinetic friction, [katex] m [/katex] is mass, [katex] g [/katex] is acceleration due to gravity, [katex] \theta [/katex] is the angle of incline, and [katex] d [/katex] is the distance traveled up the incline. |
| 3 | [katex] PE = m \cdot g \cdot h [/katex] | The gravitational potential energy gained by the block at the end of the incline. Here, [katex] h [/katex] is the vertical height, calculated further in Step 5. |
| 4 | [katex] U_{\text{spring}} = W_{\text{friction}} + PE [/katex] | Conservation of mechanical energy: The total mechanical energy (spring potential energy) initially equals the sum of the work done against friction and the gravitational potential energy at the end. |
| 5 | [katex] h = d \cdot \sin(\theta) [/katex] | Using the distance [katex] d [/katex] traveled up the incline and the angle [katex] \theta [/katex] to calculate the height [katex] h [/katex]. |
| 6 | Substituting [katex] h [/katex], [katex] W_{\text{friction}} [/katex], and [katex] PE [/katex] in step 4: | [katex] \frac{1}{2} k x^2 = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d + m \cdot g \cdot d \cdot \sin(\theta) [/katex] |
| 7 | Simplifying and solving for [katex] x [/katex]: | [katex] x^2 = \frac{2(\mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d + m \cdot g \cdot d \cdot \sin(\theta))}{k} [/katex] [katex] x = \sqrt{\frac{2d(mg(\mu_k \cdot \cos(\theta) + \sin(\theta)))}{k}} [/katex] |
| 8 | [katex] x = \sqrt{\frac{2 \cdot 55 \cdot (4 \cdot 9.8 \cdot (0.25 \cdot \cos(30^\circ) + \sin(30^\circ)))}{800}} [/katex] | Substitute values for [katex] d = 55 \, \text{m} [/katex], [katex] m = 4 \, \text{kg} [/katex], [katex] g = 9.8 \, \text{m/s}^2 [/katex], [katex] \mu_k = 0.25 [/katex], [katex] \theta = 30^\circ [/katex], and [katex] k = 800 \, \text{N/m} [/katex] to find [katex] x [/katex]. |
| 9 | [katex] x = \boxed{1.965\, m} [/katex] | Final value. |
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A 0.4 kg object is attached to a horizontal spring undergoes SHM with the total energy of 0.2 J. The potential energy as a function of position presented by the graph.
A 20 g piece of clay moving at a speed of 50 m/s strikes a 500 g pendulum bob at rest. The length of a string is 0.8 m. After the collision the clay-bob system starts to oscillate as a simple pendulum.
A 0.50-kg mass is attached to a spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. What is the total energy of the system?
Two boxes are tied together by a string and are sitting at rest on a frictionless surface. Between the two boxes is a massless compressed spring. The string trying the two boxes is then cut and the spring expands, pushing the boxes apart. The box on the left has four times the mass of the box on the right.
Two blocks, [katex] m_2 > m_1 [/katex], having the same kinetic energy, move from a frictionless surface onto a surface having friction coefficient [katex] \mu_k [/katex]. Which block will travel further before stopping.
[katex] x = \boxed{1.965\, m} [/katex]
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| Kinematics | Forces |
|---|---|
| [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 Motion | Energy |
|---|---|
| [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] |
| Momentum | Torque 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] |
| 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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