| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(KE_{\text{initial}} = \frac{1}{2} m v^2\) | Calculate the initial kinetic energy of the vehicle using its speed before it halts. Here \(v = 12.3 \text{ m/s}\). |
| 2 | \(W_{\text{friction}} = f_k \cdot d = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d\) | The work done by friction, where \( \mu_k = 0.650\) is the coefficient of kinetic friction, \( g = 9.8 \text{ m/s}^2\) is acceleration due to gravity, \( \theta = 18^\circ\), and \(d\) is the distance the vehicle slides. |
| 3 | \(W_{\text{gravity}} = m \cdot g \cdot \sin(\theta) \cdot d\) | The work done by gravity while the vehicle moves down the incline. |
| 4 | \(KE_{\text{initial}} = W_{\text{friction}} + W_{\text{gravity}}\) | By the work-energy principle, the initial kinetic energy is converted into work done against friction plus the work done by gravity. |
| 5 | \(\frac{1}{2} m v^2 = \mu_k mg\cos(\theta)d + mg\sin(\theta)d\) | Substitute expressions from steps 1, 2, and 3 into the work-energy equation. |
| 6 | \(d = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) + g \sin(\theta)}\) | Solve for \(d\), distance the vehicle slides. Notice that mass \(m\) cancels out. |
| 7 | \(d = \frac{\frac{1}{2} (12.3)^2}{0.650 \times 9.8 \times \cos(18^\circ) + 9.8 \times \sin(18^\circ)}\) | Substitute numerical values for \(v\), \(\mu_k\), \(g\), and \(\theta\) to find the value of \(d\) that represents the distance the vehicle slides until it stops. |
| 8 | \(d = 8.32 \,\text{m}\) | 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 \( \theta \) the shorter the distance traveled.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(d’ = \frac{\frac{1}{2} v^2}{\mu_k g \cos(27^\circ) + g \sin(27^\circ)}\) | Recalculate the distance with the increased angle of \(1.5 \times 18^\circ = 27^\circ\). |
| 2 | \(d’ = \frac{d}{\cos(27^\circ) + \tan(27^\circ) \cdot \sin(27^\circ)}\) | Using the previous formula of \(d\), we express the new sliding distance \(d’\) in terms of the old distance \(d\). |
| 3 | \(d’ = 7.41 \, \text{meters}\) | At an angle of \(27^\circ\) the vehicale would slide up 7.41 meters, which is less than the orginal dsitance of 8.32 meters. |
| 3 | \(\frac{d’}{d}\) | The ratio \(\frac{d’}{d}\) shows how much further the vehicle would slide relative to \(d\). |
| 4 | \(\frac{7.41}{8.32} = .89\) | Thus the new distance \(d’ = .89d\) |
Part (c):
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(W_{\text{gravity,up}} = -m \cdot g \cdot \sin(\theta) \cdot d\) | The work done by gravity as the vehicle slides up the incline is negative since gravity opposes the motion. |
| 2 | \(\frac{1}{2} m v^2 = \mu_k m g \cos(\theta) d – m g \sin(\theta) d\) | Work-energy principle applied while moving up. The kinetic energy has to overcome both friction and an upward gravity force. |
| 3 | \(d_{\text{up}} = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) – g \sin(\theta)}\) | Solve for the distance the vehicle would slide up the incline. |
| 4 | \(d_{\text{up}} < d\) | The distance \(d_{\text{up}}\) will be lesser than \(d\) since gravity now acts against the motion, reducing the sliding distance relative to sliding down. |
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When a skier skis down a hill, the normal force exerted on the skier by the hill is
What is the weight of a person who has a mass of \(75 \, \text{kg}\)?
A \( 1.5 \; \text{kg} \) mass attached to a spring with a force constant of \( 20.0 \; \text{N/m} \) oscillates on a horizontal, frictionless track. At \( t = 0 \), the mass is released from rest at \( x = 10.0 \; \text{cm} \). (That is, the spring is stretched by \( 10.00 \; \text{cm} \).)
A skier with a mass of \(58 \, \text{kg}\) glides up a snowy incline that forms an angle of \(28^\circ\) with the horizontal. The skier initially moves at a speed of \(7.2 \, \text{m/s}\). After traveling a distance of \(2.3 \, \text{m}\) up the slope, the skier’s speed reduces to \(3.8 \, \text{m/s}\).
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.
A ball falls straight down through the air under the influence of gravity. There is a retarding force \(F\) on the ball with magnitude given by \(F=bv\), where \(v\) is the speed of the ball and \(b\) is a positive constant. The ball reaches a terminal velocity after a time \(t\). The magnitude of the acceleration at time \(t/2\) is
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List at least 2 everyday forces that are not conservative, and explain why they aren’t.
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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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