| Derivation or Formula | Reasoning |
|---|---|
| \[E_s = \tfrac12 k x^2\] | Initial spring potential energy when compressed by \(x = 0.3\,\text{m}\). |
| \[W_{f1}=\mu_{1} m g d_{1}\] | Work done by kinetic friction on the horizontal section of length \(d_1 = 3\,\text{m}\) with coefficient \(\mu_1 = 0.12\). |
| \[KE = E_s – W_{f1}\] | Remaining kinetic energy after the cube finishes the horizontal slide. |
| \[W_{g}=m g d_{2} \sin\theta\] | Gravitational work required to lift the cube up the \(12^{\circ}\) slope for a path length \(d_2 = 4.5\,\text{m}\). |
| \[W_{f2}=\mu_{2} m g \cos\theta \, d_{2}\] | Work done by kinetic friction on the slope; \(\mu_2\) is the unknown coefficient. |
| \[E_s – W_{f1}=W_{g}+W_{f2}\] | Energy conservation: the kinetic energy at the foot of the incline is completely used up by gravitational and frictional work on the slope. |
| \[\mu_{2}=\frac{\left(\tfrac12 k x^{2}-\mu_{1} m g d_{1}\right)/(m g d_{2})-\sin\theta}{\cos\theta}\] | Algebraic solution for \(\mu_2\) obtained by substituting the expressions for each work term and dividing by \(m g d_{2}\). |
| \[\mu_{2}=\frac{(81\,\text{J}-21.168\,\text{J})/(58.8\,\text{N}\cdot4.5\,\text{m})-\sin12^{\circ}}{\cos12^{\circ}}\] | Numeric substitution with \(k = 1800\,\text{N/m}\), \(m=6\,\text{kg}\), and \(g = 9.8\,\text{m/s}^2\). |
| \[\mu_{2}=\boxed{0.0188}\] | Simplified result; coefficient of kinetic friction on the slope is approximately \(0.019\). |
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By pressing a painting of mass \( 2.00 \) \( \text{kg} \) against a wall, a man is trying to determine whether it is appropriately positioned. The wall is perpendicular to the pushing force. The coefficient of static friction between the image and the wall is \( 0.660 \). What is the bare minimum pushing force that must be applied?
In the laboratory, you are given a cylindrical beaker containing a fluid and you are asked to determine the density \( \rho \) of the fluid. You are to use a spring of negligible mass and unknown spring constant \( k \) that is attached to a vertical stand.
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?
A cardinal (Richmondena cardinalis) of mass \( 3.80 \times 10^{-2} \) \( \text{kg} \) and a baseball of mass \( 0.150 \) \( \text{kg} \) have the same kinetic energy. What is the ratio of the cardinal’s magnitude \( p_c \) of momentum to the magnitude \( p_b \) of the baseball’s momentum?
A \( 0.30 \text{-kg} \) mass is suspended on a spring. In equilibrium the mass stretches the spring \( 2.0 \) \( \text{cm} \) downward. The mass is then pulled an additional distance of \( 1.0 \) \( \text{cm} \) down and released from rest. Write down its equation of motion.
Which of the following must be true for an object at translational equilibrium?
A student uses a pendulum to determine the acceleration due to gravity, \( g \). They measure the pendulum’s length \( L \) and its period \( T \). Which equation should they use to calculate \( g \)?
An object is projected vertically upward from ground level. It rises to a maximum height \( H \). If air resistance is negligible, which of the following must be true for the object when it is at a height \( H/2 \) ?
A rocket launches upward by expelling exhaust gases downward. This is an illustration of Newton’s ____ Law.
Two objects attract each other gravitationally with a force of \( 2.5 \times 10^{-10} \) \( \text{N} \) when they are \( 0.25 \) \( \text{m} \) apart. Their total mass is \( 4.00 \) \( \text{kg} \). Find their individual masses.
\(0.019\)
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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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