| Derivation / Formula | Reasoning |
|---|---|
| \[F_{g\parallel}= m g \sin\theta\] | Component of weight acting down the incline; \(m\) is mass, \(g\) gravitational acceleration, and \(\theta\) the incline angle. |
| \[F – F_{g\parallel}= m a\] | Apply Newton’s second law \(\sum F = m a\) along the slope, taking the upward direction as positive. |
| \[F = m\bigl(a + g \sin\theta\bigr)\] | Solve algebraically for the applied force \(F\). |
| \[F = 150\bigl(6 + 9.8 \times 0.5\bigr)\] | Substitute \(m = 150\,\text{kg}\), \(a = 6\,\text{m/s}^2\), \(g = 9.8\,\text{m/s}^2\), and \(\sin 30^{\circ}=0.5\). |
| \[\boxed{F = 1635\,\text{N}}\] | Numerical evaluation gives the required push force. |
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A \(2,000 \, \text{kg}\) car collides with a stationary \(1,000 \, \text{kg}\) car. Afterwards, they slide \(6 \, \text{m}\) before coming to a stop. The coefficient of friction between the tires and the road is \(0.7\). Find the initial velocity of the \(2,000 \, \text{kg}\) car before the collision?
Block \(m_2\) is stacked on top of block \(m_1\). Block \(m_1\) 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 \(m_2\)?
A student kicks a soccer ball. The ball exerts a force back on the student’s foot. Why doesn’t the student’s foot accelerate backward as much as the ball accelerates forward?
During lunch, Alex and Jordan argue about inertia. Alex says if she spins a basketball faster, it will have greater inertia. Jordan argues that inertia only depends on the ball’s mass, not its speed. Who is correct?
A force \(F\) is used to hold a block of mass \(m\) on an incline as shown in the diagram above. The plane makes an angle of \(\theta\) with the horizontal and \(F\) is perpendicular to the plane. The coefficient of friction between the plane and the block is \(\mu\). What is the minimum force \(F\) necessary to keep the block at rest?
An elevator carrying a person of mass \( m \) is moving upward and slowing down. How does the magnitude \( F \) of the force exerted on the person by the elevator floor compare with the magnitude \( mg \) of the gravitational force?
Three students are pulling on a bag of skittles. Each is pulling with a horizontal force. If student 1 pulls Eastward with \(170 \, \text{N}\), student 2 pulls Southward with \(100 \, \text{N}\) and student 3 pulls with \(200 \, \text{N}\) at an angle of \(20^\circ\) west of north, what is the net force caused by the three students on the bag of skittles?
The two blocks of masses \( M \) and \( 2M \) travel at the same speed \( v \) but in opposite directions. They collide and stick together. How much mechanical energy is lost to other forms of energy during the collision?
A spring with a spring constant of \( 50. \) \( \text{N/m} \) is hanging from a stand. A second spring with a spring constant of \( 100. \) \( \text{N/m} \) is hanging from the first spring. How far do they stretch if a \( 0.50 \) \( \text{kg} \) mass is hung from the bottom spring?
A block starts from rest at the top of a \(50^\circ\) incline. The coefficient of kinetic friction between the block and the incline is \(0.4\). If the block reaches a velocity of \(7 \, \text{m/s}\) at the bottom of the incline, what is the length of the incline?
\( 1635 \) \( \text{N} \) (or \( 1650 \) if using \( g = 10 \))
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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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