| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[F = mg\] | Calculate the weight force acting on the springs using \(m\) and \(g\). |
| 2 | \[F = (0.50)(9.8) = 4.9\, \text{N}\] | Substitute \(m = 0.50\,\text{kg}\) and \(g = 9.8\,\text{m/s}^2\) to obtain the common tension \(F\). |
| 3 | \[\Delta x_{2} = \frac{F}{k_{2}}\] | Apply Hooke’s Law to the bottom spring where \(k_{2} = 100\,\text{N/m}\). |
| 4 | \[\Delta x_{2} = \frac{4.9}{100} = 0.049\, \text{m}\] | Compute the stretch of the bottom spring. |
| 5 | \[\Delta x_{1} = \frac{F}{k_{1}}\] | Apply Hooke’s Law to the top spring where \(k_{1} = 50\,\text{N/m}\). |
| 6 | \[\Delta x_{1} = \frac{4.9}{50} = 0.098\, \text{m}\] | Compute the stretch of the top spring. |
| 7 | \[\Delta x_{\text{total}} = \Delta x_{1} + \Delta x_{2} = 0.098 + 0.049 = 0.147\, \text{m}\] | Add the individual stretches to find the overall extension of the two–spring system. |
| 8 | \[\boxed{\Delta x_{1} = 0.098\, \text{m},\; \Delta x_{2} = 0.049\, \text{m},\; \Delta x_{\text{total}} = 0.147\, \text{m}}\] | Present the final boxed numerical answers for each stretch. |
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Two identical satellites are placed in orbit of two different planets. Satellite \(A\) orbits Mars, and Satellite \(B\) orbits Jupiter. The orbital speeds of each satellite are the same. Which satellite has a greater orbital radius?
A \( 1 \) \( \text{kg} \) mass on a \( 37^{\circ} \) incline is connected to a \( 3.0 \) \( \text{kg} \) mass on a horizontal surface, as shown. The surfaces and the pulley are frictionless. If \( F = 12 \) \( \text{N} \):
A block of mass \( m \), acted on by a force \( F \) directed horizontally, slides up an inclined plane that makes an angle \( \theta \) with the horizontal. The coefficient of sliding friction between the block and the plane is \( \mu \).
In the diagram above, block \(A\) has a mass of \(3.2 \, \text{kg}\) and block \(B\) a mass of \(2.4 \, \text{kg}\). The pulley is frictionless and has no mass.
The figure shows a truck pulling three crates across a rough road. Which of the following best describes the directions of all the horizontal forces acting on crate 2?
A ball falls straight down through the air under the influence of gravity. There is a retarding force \(F\) on the ball with magnitude given by \(F=bv\), where \(v\) is the speed of the ball and \(b\) is a positive constant. The ball reaches a terminal velocity after a time \(t\). The magnitude of the acceleration at time \(t/2\) is
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 stone hangs by a fine thread from the ceiling, and a section of the same thread dangles from the bottom of the stone. If a person gives a sharp pull on the dangling thread, where is the thread likely to break: below the stone or above it? What if the person gives a slow and steady pull?
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 \( 26.5 \, \text{cm} \).
An elevator carrying a person of mass \( m \) is moving upward and slowing down. How does the magnitude \( F \) of the force exerted on the person by the elevator floor compare with the magnitude \( mg \) of the gravitational force?
\(\Delta x_{\text{top}} = 0.098\,\text{m}\)
\(\Delta x_{\text{bottom}} = 0.049\,\text{m}\)
\(\Delta x_{\text{total}} = 0.147\,\text{m}\)
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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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