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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ \Delta x = v_i t + \frac{1}{2} a t^2 \] | This is the kinematic equation for constant acceleration along the path (with \(\Delta x\) representing the displacement along the incline). |
| 2 | \[ a = -g \sin(\theta) \] | Since the skateboarder is moving up the incline (against gravity), the acceleration along the ramp is the component of gravitational acceleration in the opposite direction, where \(g=9.8\,\text{m/s}^2\). |
| 3 | \[ \Delta x = v_i t – \frac{1}{2} g \sin(\theta) t^2 \] | Substitute the expression for \(a\) into the displacement equation, keeping the sign convention consistent (upward as positive). |
| 4 | \[ 18 = 20(3.3) – \frac{1}{2}(9.8)\sin(\theta)(3.3)^2 \] | Insert the given values: \(\Delta x=18\,\text{m}\), \(v_i=20.0\,\text{m/s}\), and \(t=3.3\,\text{s}\). |
| 5 | \[ 18 = 66 – 4.9(10.89)\sin(\theta) \] | Calculate \(20 \times 3.3=66\) and note that \(\frac{1}{2} \times 9.8=4.9\) and \((3.3)^2=10.89\). |
| 6 | \[ 4.9(10.89)=53.361 \quad \Rightarrow \quad 18 = 66 – 53.361\,\sin(\theta) \] | Multiply \(4.9\) and \(10.89\) to simplify the equation. |
| 7 | \[ 53.361\,\sin(\theta) = 66 – 18 = 48 \quad \Rightarrow \quad \sin(\theta)=\frac{48}{53.361}\approx0.900 \] | Rearrange the equation to solve for \(\sin(\theta)\). |
| 8 | \[ \theta = \arcsin(0.900) \approx 64^\circ \] | Take the inverse sine of \(0.900\) to find the angle. This yields an approximate angle of \(64^\circ\) above the horizontal. |
| 9 | \[ \boxed{\theta \approx 64^\circ} \] | This is the final answer for the orientation angle of the incline. |
Just ask: "Help me solve this problem."
Block m1 is stacked on top of block m2. Block m2 is connected by a light cord to block m3, which is pulled along a frictionless surface with a force F as shown in the diagram above. Block m1 is accelerated at the same rate as block m2 because of the frictional forces between the two blocks. If all three blocks have the same mass m, what is the minimum coefficient of static friction between block m1 and block m2?
Which of the following statements about the acceleration due to gravity is TRUE?
A “doomsday” asteroid with a mass of \( 1010 \, \text{kg} \) is hurtling through space. Unless the asteroid’s speed is changed by about \( 0.20 \, \text{cm/s} \), it will collide with Earth and cause tremendous damage. Researchers suggest that a small “space tug” sent to the asteroid’s surface could exert a gentle constant force of \( 2.5 \, \text{N} \). For how long must this force act?
When a basketball is dropped to the pavement, it bounces back up. Is a force needed to make it bounce back up? If so, what exerts the force?
A block of mass \( m \), acted on by a force \( F \) directed horizontally, slides up an inclined plane that makes an angle \( \theta \) with the horizontal. The coefficient of sliding friction between the block and the plane is \( \mu \).
\( \theta \approx 64^\circ} \)
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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] |
General Metric Conversion Chart
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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