| 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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A block of mass \( 0.5 \) \( \text{kg} \) is attached to a horizontal spring with a spring constant of \( 150 \) \( \text{N/m} \). The block is released from rest at position \( x = 0.05 \) \( \text{m} \), as shown, and undergoes simple harmonic motion, reaching a maximum position of \( x = 0.1 \) \( \text{m} \). The speed of the block when it passes through position \( x = 0.09 \) \( \text{m} \) is most nearly
The escape speed of an object of mass \( m \) from a planet of mass \( M \) and radius \( r \) depends on the gravitational constant and
One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance \(h\) above the floor. A block of mass \(M\) is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance \(x\). The block is released and strikes the floor a horizontal distance \(D\) from the edge of the table. Air resistance is negligible. Derive expressions for the following quantities only in terms of \(M, x, D, h,\) and any constants.
A roller coaster ride at an amusement park lifts a car of mass \( 700 \, \text{kg} \) to point \( A \) at a height of \( 90 \, \text{m} \) above the lowest point on the track, as shown above. The car starts from rest at \( A \), rolls with negligible friction down the incline and follows the track around a loop of radius \( 20 \, \text{m} \). Point \( B \), the highest point on the loop, is at a height of \( 50 \, \text{m} \) above the lowest point on the track.
A crate is pulled 2.5 m at constant velocity along a 25° incline. The coefficient of kinetic friction between the crate and the plane is 0.250. What is the efficiency of this procedure?
A block attached to a spring undergoes simple harmonic motion. The acceleration of the block has its maximum magnitude at the point where:
A simple pendulum consists of a sphere tied to the end of a string of negligible mass. The sphere is pulled back until the string is horizontal and then released from rest. Assume the gravitational potential energy is zero when the sphere is at its lowest point.
What angle will the string make with the horizontal when the kinetic energy and the potential energy of the sphere-Earth system are equal?
A uniform solid cylinder of mass \( M \) and radius \( R \) is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it. The string is then pulled with a constant force \( F \) , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle \( \theta \), what is the kinetic energy of the cylinder in terms of \( F \) and \( \theta \)?
What is the relationship between the period \( T \) and frequency \( f \) of an object in simple harmonic motion?
A snowboarder starts from rest and slides down a \(32^\circ\) incline that’s \(75 \, \text{m}\) long.
\(\text{Negative}\)
\(160\,\text{J}\)
\(1.28\times10^{3}\,\text{N/m}\)
\(32\,\text{m/s^{2}}\)
\(1.27\,\text{Hz}\)
\(K(x)=160-640x^{2}\)
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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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