| 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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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 \( 10.00 \; \text{cm} \).)
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 \(t = 3.5\) \(\text{s}\)?
Determine the force needed to push a \( 150 \) \( \text{kg} \) body up a smooth \( 30^\circ \) incline with an acceleration of \( 6 \) \( \text{m/s}^2 \).
A block is released from rest and slides down a frictionless ramp inclined at \( 30^\circ \) from the horizontal. When the block reaches the bottom, the block-Earth system has mechanical energy \( \text{E}_i \). The experiment is repeated, but now horizontal and vertical forces of magnitude \( F \) are exerted on the block while it slides, as shown above. When the block reaches the bottom, the mechanical energy of the block-Earth system.
A car is going over the top of a hill whose curvature approximates a circle of radius \( 350 \) \( \text{m} \). At what velocity will the occupants of the car appear to weigh \( 10\% \) less than their normal weight?
A particle of mass \(m\) slides down a fixed, frictionless sphere of radius \(R\), starting from rest at the top.
In terms of \(m\), \(g\), \(R\), and \(\theta\), determine each of the following for the particle while it is sliding on the sphere.
A car slides up a frictionless inclined plane. How does the normal force of the incline on the car compare with the weight of the car?
A \(3300 \, \text{m}\)-high mountain is located on the equator. How much faster does a climber on top of the mountain move than a surfer at a nearby beach? The Earth’s radius is \(6400 \, \text{km}\) and the Earth’s mass is \(5.97 \times 10^{24} \, \text{kg}\).
What force would have to be applied to start a \(12.3 \, \text{kg}\) wood block moving on a surface with a static coefficient of friction of \(0.438\)?
Why do you need to “pump” your legs when you begin swinging on a park swing?
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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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