| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[F \Delta x = \frac{1}{2} m v_i^{2}\] | The work done by the constant braking force \(F\) over the stopping distance \(\Delta x\) equals the initial kinetic energy of the car. |
| 2 | \[F = \frac{ \frac{1}{2} m v_i^{2} }{\Delta x}\] | Algebraically solve the above equation for the braking force. |
| 3 | \[F_{10} = \frac{ \frac{1}{2}(1500)(10)^{2} }{6.1}\] | Substitute \(m = 1500\, \text{kg}\), \(v_i = 10\, \text{m/s}\), and \(\Delta x = 6.1\, \text{m}\) for the first data point. |
| 4 | \[F_{10} \approx 1.23 \times 10^{4}\, \text{N}\] | Compute the numerical value of the force for \(10\, \text{m/s}\). |
| 5 | \[F_{20} = \frac{ \frac{1}{2}(1500)(20)^{2} }{23.9} \approx 1.26 \times 10^{4}\, \text{N}\] | Repeat the calculation for \(v_i = 20\, \text{m/s}\) and \(\Delta x = 23.9\, \text{m}\). |
| 6 | \[F_{30} = \frac{ \frac{1}{2}(1500)(30)^{2} }{53.5} \approx 1.26 \times 10^{4}\, \text{N}\] | Repeat the calculation for \(v_i = 30\, \text{m/s}\) and \(\Delta x = 53.5\, \text{m}\). |
| 7 | \[\boxed{F \approx 1.2 \times 10^{4}\, \text{N}}\] | The three values are nearly identical; their average is about \(1.25 \times 10^{4}\, \text{N}\), which corresponds to option (c) \(12,000\, \text{N}\). |
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The maximum energy a bone can absorb without breaking is surprisingly small. Experimental data show that a leg bone of a healthy, \( 80 \) \( \text{kg} \) human can absorb about \( 240 \) \( \text{J} \). From what maximum height could a \( 80 \) \( \text{kg} \) person jump and land rigidly upright on both feet without breaking their legs? Assume that all energy is absorbed by the leg bones in a rigid landing. Express your answer with the appropriate units.
A student is designing an experiment to find the spring constant \( k \) of a spring using only a set of known masses and a stopwatch. Which procedure would work?
An object undergoing simple harmonic motion has a maximum displacement of \(6.2\) \(\text{m}\) at \(t = 0.0\) \(\text{s}\). If the angular frequency of oscillation is \(1.6\) \(\text{rad/s}\), what is the object’s displacement when \(t = 3.5\) \(\text{s}\)?
A rocket of mass \( m \) is launched with kinetic energy \( K_0 \), from the surface of the Earth. How much less kinetic energy does the rocket have at an altitude of two Earth radii? Give your answer in terms of the gravitational constant \( G \), the mass of the Earth \( m_E \), the radius of the Earth \( R_E \), and the mass of the rocket?
An object of mass 2 kg is thrown vertically downwards with an initial kinetic energy of 100 J. What is the distance fallen by the object at the instant when its kinetic energy has doubled?
A box of mass \( 20 \) \( \text{kg} \) moves to the right on a horizontal frictionless surface with a speed of \( 4.0 \) \( \text{m/s} \). The box collides with and remains attached to one end of a spring of negligible mass whose other end is fixed to a wall. After the collision, the spring compresses a maximum distance of \( 0.50 \) \( \text{m} \), and the box then oscillates back and forth.
A rescue helicopter lifts a 79 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted through a distance of 11 m.
A spring in a pogo-stick is compressed \( 12 \) \( \text{cm} \) when a \( 40. \) \( \text{kg} \) girl stands on it. What is the spring constant for the pogo-stick spring?
A pendulum bob of mass m on a cord of length L is pulled sideways until the cord makes an angle \( \theta \) with the vertical. The change in potential energy of the bob during the displacement is:
A child pushes horizontally on a box of mass m with constant speed v across a rough horizontal floor. The coefficient of friction between the box and the floor is µ. At what rate does the child do work on the box?
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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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