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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\vec{F}_{\text{net}} = m \vec{a}\] | Newton’s second law links the net force to the acceleration \( \vec{a} \). |
| 2 | \[\vec{a} = \frac{\vec{F}_{\text{net}}}{m}\] | Because a single non-zero force gives \( \vec{F}_{\text{net}} \neq 0 \) and mass \( m \) is finite, \( \vec{a} \neq 0 \). Hence zero acceleration is impossible. |
| 3 | \[v = v_i + a t\] | One–dimensional kinematic relation for constant acceleration. |
| 4 | \[0 = v_i + a t_{\text{stop}}\] | If \( v_i \) and \( a \) have opposite signs, a time \( t_{\text{stop}} = -\frac{v_i}{a} \) exists when the instantaneous velocity becomes zero even though \( a \neq 0 \) (e.g., a ball momentarily at rest at the peak of its flight). |
| 5 | \[\boxed{\vec{a} \neq 0,\ \ v = 0\ \text{possible}}\] | Conclusion: zero acceleration cannot occur under a lone non-zero force, but zero velocity can occur momentarily. |
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While flying horizontally in an airplane, you notice that a string dangling from the overhead luggage compartment hangs at rest at 15° away from the vertical toward the back of the plane. Using this observation, you can conclude that the airplane is:
A spring launches a \(4 \, \text{kg}\) block across a frictionless horizontal surface. The block then ascends a \(30^\circ\) incline with a kinetic friction coefficient of \(\mu_k = 0.25\), stopping after \(55 \, \text{m}\) on the incline. If the spring constant is \(800 \, \text{N/m}\), find the initial compression of the spring. Disregard friction while in contact with the spring.
A skateboarder, with an initial speed of \( 20.0 \, \text{m/s} \), rolls to the end of friction-free incline of length \( 25 \, \text{m} \). At what angle is the incline oriented above the horizontal?
An object of unknown mass is acted upon by multiple forces:
The coefficients of friction are \(\mu_s = 0.6\) and \(\mu_k = 0.2\). Starting from rest, the object travels \(10 \, \text{m}\) in \(4.5 \, \text{s}\). What is the mass of the unknown object?
A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.
Block \(m_1\) is stacked on top of block \(m_2\). Block \(m_2\) is connected by a light cord to block \(m_3\), which is pulled along a frictionless surface with a force \(F\) as shown in the diagram above. Block \(m_1\) is accelerated at the same rate as block \(m_2\) because of the frictional forces between the two blocks. If all three blocks have the same mass \(m\), what is the minimum coefficient of static friction between block \(m_1\) and block \(m_2\)?
Imagine a hypothetical planet that has two moons. Moon \(\#1\) is in a circular orbit of radius \(R\) and has a mass \(M\).
A constant force of 8.0 N is exerted on a 16 kg object initially at rest. How much speed will the object gain after 4 seconds?
Two balls have their centers \( 2.0 \) \( \text{m} \) apart. One ball has a mass of \( 8.0 \) \( \text{kg} \). The other has a mass of \( 6.0 \) \( \text{kg} \). What is the gravitational force between them?
A block of mass \( m \), acted on by a force \( F \) directed horizontally, slides up an inclined plane that makes an angle \( \theta \) with the horizontal. The coefficient of sliding friction between the block and the plane is \( \mu \).
Zero acceleration cannot occur with a single non-zero force, but zero velocity can occur momentarily.
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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] |
Metric Prefixes
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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