| 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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An object undergoing simple harmonic motion has a maximum displacement of \(6.2\) \(\text{m}\) at \(t = 0.0\) \(\text{s}\). If the angular frequency of oscillation is \(1.6\) \(\text{rad/s}\), what is the object’s displacement when \(t = 3.5\) \(\text{s}\)?
A block rests on a flat plane inclined at an angle of \(30^\circ\) with respect to the horizontal. What is the minimum coefficient of friction necessary to keep the block from sliding?
Why do pilots sometimes black out while pulling out at the bottom of a dive?
A \( 25.0 \) \( \text{kg} \) block is placed at the top of an inclined plane set at an angle of \( 35 \) degrees to the horizontal. The block slides down the \( 1.5 \) \( \text{m} \) slope at a constant rate. How much work did friction do on the block?
A car suddenly stops and a passenger lurches forward. This motion is best explained by Newton’s ____ Law.
A crate rests on a horizontal surface and a woman pulls on it with a 10-N force. No matter what the orientation of the force, the crate does not move. From least to greatest, rank the normal force on the crate.
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?
Why do you push down harder on the pedals of a bicycle when first starting out than when moving at constant speed? Why do you need to pedal at all when cycling at constant speed?
A small block moving with a constant speed \(v\) collides inelastically with a block \(M\) attached to one end of a spring \(k\). The other end of the spring is connected to a stationary wall. Ignore friction between the blocks and the surface.
A 1100 kg car accelerates from 32 m/s to 8.0 m/s in 4.0 sec. What amount of force was needed to slow it down?
\(4.4\,\text{m/s}^2\)
\(\mu_{\min}=0.45,\,\text{static}\)
\(0.24\,\text{m}\)
\(\text{less than}\)
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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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