| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\vec{F}_g = m_c\,\vec{g}\] | Weight acts downward; here \(m_c = 900\,\text{kg}\). |
| 2 | \[\vec{N}= -\vec{F}_g\] | Normal force from bed acts upward, balancing weight (level road, no vertical acceleration). |
| 3 | \[\vec{f}_s \text{ (leftward)}\] | Static friction on crate points left, supplying the horizontal deceleration so that the crate does not slide. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x = \frac{v_i + v_x}{2}\,\Delta t\] | Constant–acceleration displacement relation. |
| 2 | \[55 = \frac{25 + v_x}{2}\,(3.0)\] | Substitute \(\Delta x = 55\,\text{m},\ v_i = 25\,\text{m/s},\ \Delta t = 3.0\,\text{s}.\) |
| 3 | \[v_x = 11.7\,\text{m/s}\] | Solve for final speed after braking distance. |
| 4 | \[v_x = v_i + a\,\Delta t\] | Linear velocity–time equation. |
| 5 | \[11.7 = 25 + a(3.0)\] | Insert known values. |
| 6 | \[a = -4.4\,\text{m/s}^2\] | Negative sign shows acceleration is opposite the motion (deceleration). |
| 7 | \[\boxed{\,|a| = 4.4\,\text{m/s}^2\,}\] | Requested magnitude. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\vec{f}_s = m_c\,|a|\] | Static friction provides the horizontal net force \(m_c a\). |
| 2 | \[f_{s,\max}= \mu_s N = \mu_s m_c g\] | Maximum static friction formula with \(N = m_c g\). |
| 3 | \[\mu_s m_c g \ge m_c |a|\] | No sliding requires static friction capability to exceed demand. |
| 4 | \[\mu_{\min}= \frac{|a|}{g}= \frac{4.4}{9.8}=0.45\] | Mass cancels; compute numerical value. |
| 5 | \[\boxed{\,\mu_{\min}=0.45\;\text{(static)}\,}\] | The friction is static because no relative motion occurs. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[a = \frac{v_x – v_i}{\Delta t}= \frac{25 – 0}{10}=2.5\,\text{m/s}^2\] | Uniform acceleration from rest to \(25\,\text{m/s}\) in \(10\,\text{s}\). |
| 2 | \[F_s = m_c a\] | Spring force supplies crate’s required horizontal force. |
| 3 | \[F_s = kx\] | Hooke’s law for spring with constant \(k = 9200\,\text{N/m}\). |
| 4 | \[x = \frac{m_c a}{k}= \frac{900(2.5)}{9200}=0.24\,\text{m}\] | Solve for extension. |
| 5 | \[\boxed{\,x = 0.24\,\text{m}\,}\] | Numerical answer for maximum extension during acceleration. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[a = 0\] | Constant speed implies zero acceleration. |
| 2 | \[F_s = m_c a = 0\Rightarrow x = 0\] | No net horizontal force required, so spring force and thus extension become zero. |
| 3 | \[\boxed{\,x_{\text{const}} < x_{\text{accel}}\,}\] | Extension is less (indeed zero) compared with part (d) when acceleration was present. |
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A spacecraft somewhere in between the Earth and the Moon experiences zero net force acting on it. This is because the Earth and the Moon pull the spacecraft in equal but opposite directions. Find the distance \(D\) away from Earth such that the spacecraft experiences zero net force. The distance between the Moon and Earth is \(\sim 3.844 \times 10^8 \, \text{m}\).
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For linear motion the term “inertia” refers to the same physical concept of
\(4.4\,\text{m/s}^2\)
\(\mu_{\min}=0.45,\,\text{static}\)
\(0.24\,\text{m}\)
\(\text{less than}\)
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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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