| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \frac{1}{2}m v_{0}^{2} = m g \Delta x \sin(34^\circ) \] | This is the energy conservation for a frictionless slide where all the gravitational potential energy \(m g \Delta x \sin(34^\circ)\) is converted into kinetic energy \(\frac{1}{2}m v_{0}^{2}\) at the bottom. |
| 2 | \[ v_{0} = \sqrt{2g \Delta x \sin(34^\circ)} \] | Solve for the frictionless final speed \(v_{0}\) by isolating it in the energy equation. |
| 3 | \[ \frac{1}{2}m v_{x}^{2} = m g \Delta x \sin(34^\circ) – \mu \; m g \Delta x \cos(34^\circ) \] | For a slide with kinetic friction, the work done by friction \(\mu m g \Delta x \cos(34^\circ)\) is subtracted from the available gravitational potential energy. |
| 4 | \[ v_{x} = \frac{1}{2}v_{0} \] | It is given that the child’s speed at the bottom with friction is exactly half the frictionless speed. |
| 5 | \[ \frac{1}{2}m \left(\frac{1}{2}v_{0}\right)^2 = m g \Delta x \sin(34^\circ) – \mu \; m g \Delta x \cos(34^\circ) \] | Substitute \(v_{x} = \frac{1}{2}v_{0}\) into the energy equation with friction. |
| 6 | \[ \frac{1}{2} \left(\frac{1}{2}v_{0}\right)^2 = g \Delta x \sin(34^\circ) – \mu \; g \Delta x \cos(34^\circ) \] | Cancel the mass \(m\) from both sides since it appears throughout. |
| 7 | \[ \frac{1}{2} \left(\frac{1}{4}v_{0}^2\right) = \frac{1}{8}v_{0}^2 = g \Delta x \sin(34^\circ) – \mu \; g \Delta x \cos(34^\circ) \] | Simplify the left side by computing \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\) and then multiplying by \(\frac{1}{2}\). |
| 8 | \[ \text{Since} \; \frac{1}{2}v_{0}^2 = g \Delta x \sin(34^\circ), \; \text{we have} \; \frac{1}{8}v_{0}^2 = \frac{1}{4}g \Delta x \sin(34^\circ) \] | Replace \(\frac{1}{8}v_{0}^2\) using the frictionless energy equation for consistency. |
| 9 | \[ \frac{1}{4}g \Delta x \sin(34^\circ) = g \Delta x \sin(34^\circ) – \mu \; g \Delta x \cos(34^\circ) \] | Equate the expression obtained from energy with friction to the simplified form of frictionless energy. |
| 10 | \[ \frac{1}{4}\sin(34^\circ) = \sin(34^\circ) – \mu \cos(34^\circ) \] | Cancel \(g \Delta x\) from both sides since they are nonzero. |
| 11 | \[ \sin(34^\circ) – \frac{1}{4}\sin(34^\circ) = \mu \cos(34^\circ) \] | Simplify the right side by subtracting \(\frac{1}{4}\sin(34^\circ)\) from \(\sin(34^\circ)\). |
| 12 | \[ \frac{3}{4}\sin(34^\circ) = \mu \cos(34^\circ) \] | This gives the relationship that relates \(\mu\) to the sine and cosine of \(34^\circ\). |
| 13 | \[ \mu = \frac{\frac{3}{4}\sin(34^\circ)}{\cos(34^\circ)} = \frac{3}{4}\tan(34^\circ) \] | Solve for \(\mu\) by dividing both sides by \(\cos(34^\circ)\). |
| 14 | \[ \boxed{\mu \approx 0.51} \] | Substitute \(\tan(34^\circ) \approx 0.67\) to get a numerical value \(\mu \approx \frac{3}{4} \times 0.67 \approx 0.50-0.51\). This is the coefficient of kinetic friction. |
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A horizontal, uniform board of weight \( 125 \, \text{N} \) and length \( 4 \, \text{m} \) is supported by vertical chains at each end. A person weighing \( 500 \, \text{N} \) is hanging from the board. The tension in the right chain is \( 250 \, \text{N} \).
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 \(5.50 \, \text{m/s}^2\)?
Two masses, \( m_1 \) and \( m_2 \), are connected by a cord and arranged as shown in the diagram, with \( m_1 \) sliding along a frictionless surface and \( m_2 \) hanging from a light, frictionless pulley. What would be the mass of the falling mass, \( m_2 \), if both the sliding mass, \( m_1 \), and the tension, \( T \), in the cord were known?
What is the mass of a dog that weighs \(58 \, \text{N}\) on Earth?
A \(10 \, \text{kg}\) box is pushed to the right by an unknown force at an angle of \(25^\circ\) below the horizontal while a friction force of \(50 \, \text{N}\) acts on the box as well. The box accelerates from rest and travels a distance of \(4 \, \text{m}\) where it is moving at \(3 \, \text{m/s}\).
The steepest street in the world is Baldwin Street in Dunedin, New Zealand. It has an inclination angle of \( 38.0^\circ \) with respect to the horizontal. Suppose a wooden crate with a mass of \( 25.0 \) \( \text{kg} \) is placed on Baldwin Street. An additional force of \( 59 \) \( \text{N} \) must be applied to the crate perpendicular to the pavement in order to hold the crate in place. If the coefficient of static friction between the crate and the pavement is \( 0.599 \), what is the magnitude of the frictional force?
Late one morning, a mosquito collides with the windshield of a speeding truck. The force of the truck on the mosquito is ____ the force of the mosquito on the truck; the resulting acceleration of the mosquito is ____ the acceleration of the truck.
There are two cables that lift an elevator, each with a force of \(10{,}000 \, \text{N}\). The \(1{,}000 \, \text{kg}\) elevator is lifted from the first floor and accelerates over \(10 \, \text{m}\) until it reaches its top speed of \(6 \, \text{m/s}\). What is the mass of the people in the elevator?
Two ice skaters push off against each other. Skater A has twice the mass of Skater B. Which statement is correct?
A person stands on a scale in an elevator. His apparent weight will be the greatest when the elevator
\(\boxed{\mu \approx 0.51}\)
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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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