| 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?
A ski lift carries skiers along a \(695 \, \text{m}\) slope inclined at \(34^\circ\). To lift a single rider, it is necessary to move \(72 \, \text{kg}\) of mass to the top of the lift. Under maximum load conditions, five riders per minute arrive at the top. If \(65\%\) of the energy supplied by the motor goes to overcoming friction, what average power must the motor supply?
At time \( t = 0 \), an object is released from rest at position \( x = +x_{\text{max}} \) and undergoes simple harmonic motion along the \( x \)-axis about the equilibrium position of \( x = 0 \). The period of oscillation of the object is \( T \). Which of the following expressions is equal to the object’s position at time \( t = \dfrac{T}{8} \)?
Block \( 1 \) of mass \( m_1 \) and Block \( 2 \) of mass \( m_2 = 2 m_1 \) are each attached to identical horizontal springs. Each block is displaced from equilibrium by an unknown amount and the blocks are released from rest simultaneously, undergoing simple harmonic motion. A student claims that Block \( 1 \) will make its first return to its equilibrium position before Block \( 2 \) first returns to its equilibrium position. Is this claim correct? Why or why not?
A proton (mp = 1.67 x10-27 kg) is being accelerated along a straight line at 3.6 ×1015 m/s2 in a machine. The proton has an initial speed of 2.4 x107 m/s and travels 3.5 cm.
Students attach a thin strip of metal to a table so that the strip is horizontal in relation to the ground. A section of the strip hangs off the edge of the table. A mass is secured to the end of the hanging section of the strip and is then displaced so that the mass-strip system oscillates, as shown in the figure. Students make various measurements of the net force \( F \) exerted on the mass as a result of the force due to gravity and the normal force from the strip, the vertical position \( y \) of the mass above and below its equilibrium position \( y \), and the period of oscillation \( T’ \) when the mass is displaced by different amplitudes \( A \). Which of the following explanations is correct about the evidence required to conclude that the mass undergoes simple harmonic motion?
From the figure above, determine which characteristic fits this collision best.
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?
An object at rest suddenly explodes into two fragments (\(m_1\) and \(m_2\)) by an explosion. Fragment \(m_1\) acquires \(3\) times the kinetic energy of the other. What is the ratio of \(m_1\) to \(m_2\)?
A \(81 \, \text{kg}\) student dives off a \(45 \, \text{m}\) tall bridge with an \(18 \, \text{m}\) long bungee cord tied to his feet and to the bridge. You can consider the bungee cord to be a flexible spring. What spring constant must the bungee cord have for the student’s lowest point to be \(2.0 \, \text{m}\) above the water?
\(\text{Negative}\)
\(160\,\text{J}\)
\(1.28\times10^{3}\,\text{N/m}\)
\(32\,\text{m/s^{2}}\)
\(1.27\,\text{Hz}\)
\(K(x)=160-640x^{2}\)
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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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