| 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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Determine the force needed to push a \( 150 \) \( \text{kg} \) body up a smooth \( 30^\circ \) incline with an acceleration of \( 6 \) \( \text{m/s}^2 \).
Two wires are tied to the \(500 \, \text{g}\) sphere as shown above. The sphere revolves in a horizontal circle at a constant speed of \(7.2 \, \text{m/s}\). What is the tension in the upper wire? What is the tension in the lower wire?
A box with a mass of \( M \) rests on a scale in an elevator that is moving downwards. The elevator slows with an acceleration of \( \dfrac{g}{4} \). Which of the following will give the reading of the scale?
Why do pilots sometimes black out while pulling out at the bottom of a dive?
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?
A ladder at rest is leaning against a wall at an angle. Which of the following forces must have the same magnitude as the frictional force exerted on the ladder by the floor?
In the diagram shown, a \(20 \, \text{N}\) force is applied to block \(B\) (\(7 \, \text{kg}\)). Block \(A\) has a mass of \(3 \, \text{kg}\). Assume frictionless conditions.
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 student is designing an experiment to find the spring constant \( k \) of a spring using only a set of known masses and a stopwatch. Which procedure would work?
A 84.4 kg climber is scaling the vertical wall. His safety rope is made of a material that behaves like a spring that has a spring constant of 1.34 x 103 N/m. He accidentally slips and falls 0.627 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?
\(4.4\,\text{m/s}^2\)
\(\mu_{\min}=0.45,\,\text{static}\)
\(0.24\,\text{m}\)
\(\text{less than}\)
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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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