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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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The gravitational force that the Moon exerts on Earth is often cited as the source of the tides we witness. However, the gravitational force the Sun exerts on Earth is over \(100\) times greater than the force the Moon exerts on Earth.
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Two objects are attracted to each other by a gravitational force \( F \). If each mass is tripled, so that each becomes \( 3 \) times its original value, and the distance between the objects is cut in half to \( \dfrac{1}{2} \) of its original separation, what is the new gravitational force between the objects in terms of \( F \)?
In an experiment where a constant horizontal force pulls on a box across a rough floor starting from rest, what would happen to the acceleration of the box if its mass were doubled but the pulling force remained unchanged?
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A ball of mass \( m \) is suspended from two strings of unequal length as shown above. The magnitudes of the tensions \( T_1 \) and \( T_2 \) in the strings must satisfy which of the following relations?
A train consists of \(50\) cars, each of which has a mass of \(6.1 \times 10^{3} \, \text{kg}\). The train has an acceleration of \(8.0 \times 10^{-2} \, \text{m/s}^2\). Ignore friction and determine the tension in the coupling at the following places:
Which of the following must be true for an object at translational equilibrium?
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A block is given a brief push so that it slides up a ramp. After the block reaches its highest point, it slides back down, but the magnitude of its acceleration is less on the descent than on the ascent. Why?
When a horizontal force of \( 4.5 \, \text{N} \) acts on a block on a resistance-free surface, it produces an acceleration of \( 2.5 \, \text{m/s}^2 \). Suppose a second \( 4.0 \, \text{kg} \) block is dropped onto the first. What is the magnitude of the acceleration of the combination if the same force continues to act? Assume that the second block does not slide on the first block.
\(\boxed{\mu \approx 0.51}\)
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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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