| 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 hockey puck glides on perfectly frictionless ice at constant velocity. Which statement is true?
The coefficient of static friction between hard rubber and normal street pavement is about \(0.85\). On how steep a hill (maximum angle) can you leave a car parked?
Determine the force needed to push a \( 150 \) \( \text{kg} \) body up a smooth \( 30^\circ \) incline with an acceleration of \( 6 \) \( \text{m/s}^2 \).
The speed of a \(40 \, \text{N}\) hockey puck, sliding across a level ice surface, decreases at the rate of \(0.61 \, \text{m/s}^2\). The coefficient of kinetic friction between the puck and ice is
A block of mass \( 4.0 \) \( \text{kg} \) rests on an inclined plane. The coefficient of static friction between the block and the plane \( \mu_s \) is \( 0.4 \). Which of the following gives the angle of inclination at which the block will start to slide?
A rescue helicopter lifts a 79 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted through a distance of 11 m.
Two students push a \(1750\, \mathrm{kg}\) car with a force of \(758\, \mathrm{N}\) along a perfectly level road at a constant velocity of \(4.00\, \mathrm{m/s}\). Find the force of friction.
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 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 \(3300 \, \text{m}\)-high mountain is located on the equator. How much faster does a climber on top of the mountain move than a surfer at a nearby beach? The Earth’s radius is \(6400 \, \text{km}\) and the Earth’s mass is \(5.97 \times 10^{24} \, \text{kg}\).
\(\boxed{\mu \approx 0.51}\)
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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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