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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]F_{\text{friction}} = \mu F_{\text{normal}}[/katex] | The force needed to set the train in motion, is the force same as overcoming static friction. ([katex]F_{\text{friction}}[/katex]) is the product of the coefficient of static friction ([katex]\mu[/katex]) and the normal force ([katex]F_{\text{normal}}[/katex]). |
| 2 | [katex]\mu = 0.741[/katex] [katex]F_{\text{friction}} = 1.13 \times 10^{5} \, \text{N}[/katex] |
The coefficient of static friction and the force to overcome static friction are given. |
| 3 | [katex]F_{\text{normal}} = \frac{F_{\text{friction}}}{\mu}[/katex] | Rearrange the friction formula to solve for the normal force. |
| 4 | [katex]F_{\text{normal}} = \frac{1.13 \times 10^{5} \, \text{N}}{0.741}[/katex] | Substitute the known values into the rearranged equation. |
| 5 | [katex]F_{\text{normal}} = 1.52456 \times 10^{5} \, \text{N}[/katex] | Calculate the normal force. |
| 6 | [katex]F_{\text{normal}} = mg[/katex] | Note that the normal force acts opposite to the weight of the train. thus can also set the normal force equal to the weight of the train. |
| 7 | [katex]m = \frac{F_{\text{normal}}}{g}[/katex] | Rearrange the equation to solve for the mass (m). |
| 8 | [katex]m = \frac{1.524 \times 10^{5} \, \text{N}}{9.81 \, \text{m/s}^2}[/katex] | Substitute the calculated normal force from step 5 to find the mass. |
| 9 | [katex]m \approx 1.55 \times 10^4 \, \text{kg}[/katex] | The approximate mass of the train calculated from the given forces and coefficients. |
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A \( 25.0 \) \( \text{kg} \) block is initially at rest on a horizontal surface. A horizontal force of \( 75.0 \) \( \text{N} \) is required to set the block in motion, after which a horizontal force of \( 60.0 \) \( \text{N} \) is required to keep the block moving with constant speed.
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?
A person stands on a scale in an elevator. His apparent weight will be the greatest when the elevator
Two objects, \( A \) and \( B \), move toward one another. Object \( A \) has twice the mass and half the speed of object \( B \). Which of the following describes the forces the objects exert on each other when they collide and provides the best explanation?
A child slides down a slide with a \( 34^\circ \) incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.
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?
A snowboarder starts from rest and slides down a \(32^\circ\) incline that’s \(75 \, \text{m}\) long.
The Earth’s radius is \(6.37 \times 10^{6} \, \text{m}\). What is the radius of a planet that has the same mass as Earth but on which the free-fall acceleration is \(5.50 \, \text{m/s}^2\)?
A \(1 \, \text{kg}\) mass and an unknown mass \(M\) hang on opposite sides of a pulley suspended from the ceiling. When the masses are released, \(M\) accelerates downward at \(5 \, \text{m/s}^2\). Find the value of \(M\).
The Moon does not crash into the Earth because:
[katex]m \approx 1.55 \times 10^4 \, \text{kg}[/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] |
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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