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| 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 linear spring of negligible mass requires a force of \( 18.0 \, \text{N} \) to cause its length to increase by \( 1.0 \, \text{cm} \). A sphere of mass \( 75.0 \, \text{g} \) is then attached to one end of the spring. The distance between the center of the sphere \( M \) and the other end \( P \) of the un-stretched spring is \( 25.0 \, \text{cm} \). Then the sphere begins rotating at constant speed in a horizontal circle around the center \( P \). The distance \( P \) and \( M \) increases to \( 26.5 \, \text{cm} \).
A skier with a mass of \(58 \, \text{kg}\) glides up a snowy incline that forms an angle of \(28^\circ\) with the horizontal. The skier initially moves at a speed of \(7.2 \, \text{m/s}\). After traveling a distance of \(2.3 \, \text{m}\) up the slope, the skier’s speed reduces to \(3.8 \, \text{m/s}\).
A body starting from rest moves along a straight line under the action of a constant force. After traveling a distance \( d \) the speed of the body is \( v \). The speed of the body when it has travelled a distance \( \dfrac{d}{2} \) from its initial position is
A car of mass \( M \) moves around a circularly banked curve on a freeway off-ramp. The off-ramp has a radius of curvature \( R \) and is raised to an angle \( \theta \) from the horizontal. The road is slick, and friction is negligible.
In 2014, the European Space Agency placed a satellite in orbit around comet 67P/Churyumov-Gerasimenko and then landed a probe on the surface. The actual orbit was elliptical, but we can approximate it as a 50 km diameter circular orbit with a period of 11 days.
Two objects (49.0 and 24.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find the acceleration of the objects and the tension in the string.
The block is moving horizontally at a constant velocity. There are two applied forces on the object as shown in the image. In which direction is the friction force acting on the object?
According to Newton’s third law, each team in a tug of war pulls with equal force on the other team. What, then, determines which team will win?
A rocket launches upward by expelling exhaust gases downward. This is an illustration of Newton’s ____ Law.
A point \( P \) is subjected to three simultaneous forces of magnitudes \( F_A > F_B > F_C \). Point \( P \) is in equilibrium. Which of the following statements is not always true about the magnitudes of the forces?
\(\boxed{\mu \approx 0.51}\)
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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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