| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[K_i = \frac{1}{2} m v_i^2\] | The box begins with kinetic energy \(K_i\). |
| 2 | \[W_s = K_f – K_i\] | The spring’s work \(W_s\) equals the change in kinetic energy \(\Delta K\). |
| 3 | \[K_f = 0\] | At maximum compression the box momentarily stops, so \(K_f = 0\). |
| 4 | \[W_s = -K_i\] | Since \(K_i > 0\) and \(K_f = 0\), the spring does negative work (it removes energy from the box). |
| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[K_i = \frac{1}{2}(20)(4.0)^2 = 160\,\text{J}\] | Compute the initial kinetic energy using \(m = 20\,\text{kg}\) and \(v_i = 4.0\,\text{m/s}\). |
| 2 | \[|W_s| = K_i = 160\,\text{J}\] | The magnitude of the spring’s work equals the lost kinetic energy. |
| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[W_s = -\frac{1}{2}k x_{\max}^2\] | Work done by a spring compressing from \(0\) to \(x_{\max}\). |
| 2 | \[-160 = -\frac{1}{2}k(0.50)^2\] | Insert \(|W_s| = 160\,\text{J}\) and \(x_{\max} = 0.50\,\text{m}\). |
| 3 | \[k = 1.28 \times 10^3\,\text{N/m}\] | Solve for the spring constant. |
| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[a_{\max} = \frac{k x_{\max}}{m}\] | For simple harmonic motion, acceleration magnitude is \(|a| = (k/m)|x|\); maximum occurs at amplitude. |
| 2 | \[a_{\max} = \frac{(1.28\times10^3)(0.50)}{20} = 32\,\text{m/s}^2\] | Substitute \(k\), \(x_{\max}\), and \(m\). |
| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\] | Frequency of a mass–spring system on a frictionless surface. |
| 2 | \[f = \frac{1}{2\pi}\sqrt{\frac{1.28\times10^3}{20}} \approx 1.27\,\text{Hz}\] | Insert \(k\) and \(m\) and simplify. |
| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[E = \frac{1}{2}kA^2 = 160\,\text{J}\] | Total mechanical energy \(E\) equals the initial kinetic energy; amplitude \(A = 0.50\,\text{m}\). |
| 2 | \[K(x) = E – \frac{1}{2}k x^2 = 160 – 640 x^2\] | Kinetic energy as a function of position for simple harmonic motion. |
| 3 | \[K(\pm0.50) = 0,\; K(0)=160\,\text{J}\] | Shows the endpoints and midpoint values used for sketching. |
| 4 | \[\text{Parabolic}\] | The graph is an inverted parabola opening downward, symmetric about \(x=0\), peaking at \(160\,\text{J}\) and touching the horizontal axis at \(x = \pm0.50\,\text{m}\). |
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The efficiency of a pulley system is 55%. The
pulleys are used to raise a mass of 90.0 kg to a height of
5.60 m. What force is exerted on the rope of the pulley
system if the rope is pulled for 22 m in order to raise
the mass to the required height?
Two blocks, \( m_2 > m_1 \), having the same kinetic energy, move from a frictionless surface onto a surface having friction coefficient \( \mu_k \). Which block will travel further before stopping.
What is the defining characteristic of the restoring force that causes an object to undergo Simple Harmonic Motion (SHM)?
An object of mass \( m = 3.0 \) \( \text{kg} \) is attached to one end of a string with negligible mass and length \( L = 0.80 \) \( \text{m} \). The object is released from rest at time \( t = 0 \), when the string is horizontal. At time \( t = t_1 \) the object is at the location shown in the figure, where the string is vertical. Which of the following is most nearly the magnitude of the tension in the string at time \( t = t_1 \)?
A ball of radius \( r \) rolls on the inside of a circular track of radius \( R \). If the ball starts from rest at the left vertical edge of the track, what will be its speed when it reaches the lowest point of the track, rolling without slipping? For a solid spherical ball, the moment of inertia is \(\frac{2}{5} m r^2\).
The graph represents the position \( x \) as a function of time \( t \) for an object undergoing simple harmonic motion. Which of the following equations could represent the position \( x \) as a function of time \( t \)?
A comet of mass \( m_c = 3.2 \times 10^{14} \) \( \text{kg} \) is orbiting a star with mass \( m_s = 1.8 \times 10^{30} \) \( \text{kg} \). The comet’s orbit is elliptical. At its closest point, the comet is a distance \( r_1 = 8.3 \times 10^{10} \) \( \text{m} \) from the star, and at its farthest point, the comet is a distance \( r_2 = 4.9 \times 10^{11} \) \( \text{m} \) from the star. What is the change in the kinetic energy of the comet as it moves along its orbit from distance \( r_2 \) to distance \( r_1 \) from the star?
A person holds a book at rest a few feet above a table. The person then lowers the book at a slow constant speed and places it on the table. Which of the following accurately describes the change in the total mechanical energy of the Earth–book system?
For an object oscillating in SHM, what is the relationship between its displacement, velocity, and acceleration graphs as a function of time?
\(\text{Negative}\)
\(160\,\text{J}\)
\(1.28\times10^{3}\,\text{N/m}\)
\(32\,\text{m/s^{2}}\)
\(1.27\,\text{Hz}\)
\(K(x)=160-640x^{2}\)
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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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