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| 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 sphere starts from rest and rolls down an incline of height \( H = 1.0 \) \( \text{m} \) at an angle of \( 25^\circ \) with the horizontal, as shown above. The radius of the sphere \( R = 15 \) \( \text{cm} \), and its mass \( m = 1.0 \) \( \text{kg} \). The moment of inertia for a sphere is \( \frac{2}{5}mR^2 \). What is the speed of the sphere when it reaches the bottom of the plane?
A spring stretches \( 8.0 \) \( \text{cm} \) when a \( 13 \) \( \text{N} \) force is applied. How far does it stretch when a \( 26 \) \( \text{N} \) force is applied?
A net force of \( 8.0 \) \( \text{N} \) accelerates a \( 4.0 \) \( \text{kg} \) body from rest to a speed of \( 5.0 \) \( \text{m s}^{-1} \). Which of the following is equal to the work done by the force?
The elliptical orbit of a comet is shown above. Positions 1 and 2 are, respectively, the farthest and nearest positions to the Sun, and at position 1 the distance from the comet to the Sun is 10 times that at position 2. At position 2, the comet’s kinetic energy is
A horizontal force of \(110 \, \text{N}\) is applied to a \(12 \, \text{kg}\) object, moving it \(6 \, \text{m}\) on a horizontal surface where the kinetic friction coefficient is \(\mu_k = 0.25\). The object then slides up a \(17^\circ\) inclined plane. Assuming the \(110 \, \text{N}\) force is no longer acting on the incline, and the coefficient of kinetic friction there is \(\mu_k = 0.45\), calculate the distance the object will slide on the incline.
The total energy of a system in SHM is given by \( E \). At what displacement from equilibrium is the kinetic energy equal to the potential energy?
A snowboarder starts from rest and slides down a \(32^\circ\) incline that’s \(75 \, \text{m}\) long.
A typical \( 68 \text{-kg} \) person generates a steady mechanical power output of \( 120 \text{ W} \) at the pedals of a bicycle. Approximately how many Calories are “burned” (total metabolic energy expended) when the person rides a bicycle for \( 15 \text{ minutes} \)? A typical energy efficiency for the human body is \( 25\% \), which takes into account the release of thermal energy. Note (\( 1 \text{ Cal} = 4186 \text{ J} \)).
A bullet at speed [katex] v_0 [/katex] trikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height h. Which of the following statements is true?
It takes \(4 \, \text{s}\) for an individual to push a \(70 \, \text{kg}\) box up a \(5 \, \text{m}\) long, \(12^\circ\) ramp. The box starts from rest and achieves a speed of \(2.5 \, \text{m/s}\) at the top. Friction does \(350 \, \text{J}\) of work during its ascent. Calculate the power output of the individual pushing the box.
\(\text{Negative}\)
\(160\,\text{J}\)
\(1.28\times10^{3}\,\text{N/m}\)
\(32\,\text{m/s^{2}}\)
\(1.27\,\text{Hz}\)
\(K(x)=160-640x^{2}\)
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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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