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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]KE_{\text{initial}} = \frac{1}{2} m v^2[/katex] | Calculate the initial kinetic energy of the vehicle using its speed before it halts. Here [katex]v = 12.3 \text{ m/s}[/katex]. |
| 2 | [katex]W_{\text{friction}} = f_k \cdot d = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d[/katex] | The work done by friction, where [katex] \mu_k = 0.650[/katex] is the coefficient of kinetic friction, [katex] g = 9.8 \text{ m/s}^2[/katex] is acceleration due to gravity, [katex] \theta = 18^\circ[/katex], and [katex]d[/katex] is the distance the vehicle slides. |
| 3 | [katex]W_{\text{gravity}} = m \cdot g \cdot \sin(\theta) \cdot d[/katex] | The work done by gravity while the vehicle moves down the incline. |
| 4 | [katex]KE_{\text{initial}} = W_{\text{friction}} + W_{\text{gravity}}[/katex] | By the work-energy principle, the initial kinetic energy is converted into work done against friction plus the work done by gravity. |
| 5 | [katex]\frac{1}{2} m v^2 = \mu_k mg\cos(\theta)d + mg\sin(\theta)d[/katex] | Substitute expressions from steps 1, 2, and 3 into the work-energy equation. |
| 6 | [katex]d = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) + g \sin(\theta)}[/katex] | Solve for [katex]d[/katex], distance the vehicle slides. Notice that mass [katex]m[/katex] cancels out. |
| 7 | [katex]d = \frac{\frac{1}{2} (12.3)^2}{0.650 \times 9.8 \times \cos(18^\circ) + 9.8 \times \sin(18^\circ)}[/katex] | Substitute numerical values for [katex]v[/katex], [katex]\mu_k[/katex], [katex]g[/katex], and [katex]\theta[/katex] to find the value of [katex]d[/katex] that represents the distance the vehicle slides until it stops. |
| 8 | [katex]d = 8.32 \,\text{m}[/katex] | Calculated value. |
To solve part b, look at the equation derived in step 6 of part a. Notice that the angle and distance traveled up the incline are inversely proportional. This means the greater the angle [katex] \theta [/katex] the shorter the distance traveled.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]d’ = \frac{\frac{1}{2} v^2}{\mu_k g \cos(27^\circ) + g \sin(27^\circ)}[/katex] | Recalculate the distance with the increased angle of [katex]1.5 \times 18^\circ = 27^\circ[/katex]. |
| 2 | [katex]d’ = \frac{d}{\cos(27^\circ) + \tan(27^\circ) \cdot \sin(27^\circ)}[/katex] | Using the previous formula of [katex]d[/katex], we express the new sliding distance [katex]d'[/katex] in terms of the old distance [katex]d[/katex]. |
| 3 | [katex]d’ = 7.41 \, \text{meters}[/katex] | At an angle of [katex]27^\circ[/katex] the vehicale would slide up 7.41 meters, which is less than the orginal dsitance of 8.32 meters. |
| 3 | [katex]\frac{d’}{d}[/katex] | The ratio [katex]\frac{d’}{d}[/katex] shows how much further the vehicle would slide relative to [katex]d[/katex]. |
| 4 | [katex]\frac{7.41}{8.32} = .89[/katex] | Thus the new distance [katex]d’ = .89d[/katex] |
Part (c):
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]W_{\text{gravity,up}} = -m \cdot g \cdot \sin(\theta) \cdot d[/katex] | The work done by gravity as the vehicle slides up the incline is negative since gravity opposes the motion. |
| 2 | [katex]\frac{1}{2} m v^2 = \mu_k m g \cos(\theta) d – m g \sin(\theta) d[/katex] | Work-energy principle applied while moving up. The kinetic energy has to overcome both friction and an upward gravity force. |
| 3 | [katex]d_{\text{up}} = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) – g \sin(\theta)}[/katex] | Solve for the distance the vehicle would slide up the incline. |
| 4 | [katex]d_{\text{up}} < d[/katex] | The distance [katex]d_{\text{up}}[/katex] will be lesser than [katex]d[/katex] since gravity now acts against the motion, reducing the sliding distance relative to sliding down. |
Just ask: "Help me solve this problem."
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 horizontal spring with spring constant 162 N/m is compressed 50 cm and used to launch a 3 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the rough surface is 0.2. Find the total distance the box travels before stopping.
Two masses, \( m_1 \) and \( m_2 \), are connected by a cord and arranged as shown in the diagram, with \( m_1 \) sliding along a frictionless surface and \( m_2 \) hanging from a light, frictionless pulley. What would be the mass of the falling mass, \( m_2 \), if both the sliding mass, \( m_1 \), and the tension, \( T \), in the cord were known?
A \(1509 \, \text{g}\) wood block is being pulled by the force meter at a constant velocity. Using the graph above, find:
The gravitational force that the Moon exerts on Earth is often cited as the source of the tides we witness. However, the gravitational force the Sun exerts on Earth is over \(100\) times greater than the force the Moon exerts on Earth.
Why is the force from the Moon credited for the tides, and not the force from the Sun?
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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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