| 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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Three blocks of masses \(5 \, \text{kg}\), \(4 \, \text{kg}\), and \(3 \, \text{kg}\) are placed side by side in that order. A \(25 \, \text{N}\) force applied on the \(5 \, \text{kg}\) block accelerates all three blocks together to the right. Find the acceleration of the blocks and the normal force the \(4 \, \text{kg}\) block exerts on the \(3 \, \text{kg}\) block.
The distance from earth to sun is \(1 \, \text{AU}\). The distance from Saturn to sun is \(9 \, \text{AU}\). Find the period of Saturn’s orbit in years. You can assume that the orbits are circular.
If I weigh \( 741 \) \( \text{N} \) on Earth at a place where \( g = 9.80 \) \( \text{m/s}^2 \) and \( 5320 \) \( \text{N} \) on the surface of another planet, what is the acceleration due to gravity on that planet?
A person sitting in an enclosed train car, moving at constant velocity, throws a ball straight up into the air in her reference frame.
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?
A box is sliding down an incline at a constant speed of \( 2 \, \text{m/s} \). The angle of the incline is \( \theta \). The magnitude of the total of the opposing forces is \( 16 \, \text{N} \). Derive an equation for the force of gravity acting on the box.
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 child on Earth has a weight of \(500 \, \text{N}\). Determine the weight of the child if the Earth were to triple in both mass and radius (\(3M\) and \(3r\)).
A \(0.5 \, \text{mm}\) wire made of carbon and manganese can just barely support the weight of a \(70.0 \, \text{kg}\) person that is holding on vertically. Suppose this wire is used to lift a \(45.0 \, \text{kg}\) load. What maximum vertical acceleration can be achieved without breaking the wire?
An object of unknown mass is acted upon by multiple forces:
The coefficients of friction are \(\mu_s = 0.6\) and \(\mu_k = 0.2\). Starting from rest, the object travels \(10 \, \text{m}\) in \(4.5 \, \text{s}\). What is the mass of the unknown object?
\(4.4\,\text{m/s}^2\)
\(\mu_{\min}=0.45,\,\text{static}\)
\(0.24\,\text{m}\)
\(\text{less than}\)
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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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