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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[h = L \sin\theta\] | The vertical drop equals the incline length times \(\sin\theta\). |
| 2 | \[PE_i = m g h\] | Initial potential energy when the rider starts from rest. |
| 3 | \[W_f = -\mu_k m g \cos\theta\, L\] | Work done by kinetic friction acting up the slope. |
| 4 | \[m g h + W_f = \tfrac{1}{2} m v_x^2\] | Conservation of energy: potential energy lost and friction work convert to kinetic energy. |
| 5 | \[v_x = \sqrt{2 g \big(L \sin\theta – \mu_k L \cos\theta\big)}\] | Algebraic solution for \(v_x\) after cancelling mass. |
| 6 | \[v_x = \sqrt{2(9.8)(75)(\sin32^\circ-0.12\cos32^\circ)} \approx 25\,\text{m/s}\] | Numeric substitution of \(g=9.8\,\text{m/s}^2\), \(L=75\,\text{m}\), and \(\theta=32^\circ\). |
| 7 | \[\boxed{v_x \approx 25\,\text{m/s}}\] | Speed at the base of the incline. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[KE = \tfrac{1}{2} m v_x^2\] | Kinetic energy at the start of the flat section. |
| 2 | \[W_f = -\mu_k m g \Delta x\] | Work done by friction on level ground; normal force equals \(mg\). |
| 3 | \[\tfrac{1}{2} m v_x^2 + W_f = 0\] | Energy conservation: all kinetic energy is dissipated by friction. |
| 4 | \[\Delta x = \frac{v_x^2}{2 \mu_k g}\] | Algebraic solution for the stopping distance. |
| 5 | \[\Delta x = \frac{(25\,\text{m/s})^2}{2(0.12)(9.8)} \approx 2.7 \times 10^2\,\text{m}\] | Numeric evaluation using \(v_x\) from part (a). |
| 6 | \[\boxed{\Delta x \approx 2.7 \times 10^2\,\text{m}}\] | Distance travelled on the level surface before stopping. |
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| Speed | \( 10 \, \mathrm{m/s} \) | \( 20 \, \mathrm{m/s} \) | \( 30 \, \mathrm{m/s} \) |
| Braking Distance | \( 6.1 \, \mathrm{m} \) | \( 23.9 \, \mathrm{m} \) | \( 53.5 \, \mathrm{m} \) |
A car of mass \( 1500 \, \mathrm{kg} \) is traveling at one of the speeds listed when the brakes are first applied. Using the data above, what is the magnitude of the average braking force required to stop the car?
A sphere of mass \( M \) and radius \( r \), and rotational inertia \( I \) is released from the top of an inclined plane of height \( h \). The surface has considerable friction. Using only the variables mentioned, derive an expression for the sphere’s center of mass velocity.
An object of mass \( m = 3.0 \) \( \text{kg} \) is attached to one end of a string with negligible mass and length \( L = 0.80 \) \( \text{m} \). The object is released from rest at time \( t = 0 \), when the string is horizontal. At time \( t = t_1 \) the object is at the location shown in the figure, where the string is vertical. Which of the following is most nearly the magnitude of the tension in the string at time \( t = t_1 \)?
How does the time t1 of a block m reaching the bottom of slide 1 compare with t2, the time taken block of mass 2m to reach the end of slide 2 that’s curved? The blocks are released from the same height.
A \(2,000 \, \text{kg}\) car collides with a stationary \(1,000 \, \text{kg}\) car. Afterwards, they slide \(6 \, \text{m}\) before coming to a stop. The coefficient of friction between the tires and the road is \(0.7\). Find the initial velocity of the \(2,000 \, \text{kg}\) car before the collision?
The heaviest train ever pulled by a single engine was over [katex] 2 \, \text{km} [/katex] long. A force of [katex] 1.13 \times 10^5 \, \text{N} [/katex] is needed to get the train to start moving. If the coefficient of static friction is [katex] 0.741 [/katex] and the coefficient of kinetic friction is [katex] .592 [/katex], what is the train’s mass?
A \( 50 \) \( \text{g} \) ice cube can slide up and down a frictionless \( 30^{\circ}\) slope. At the bottom, a spring with spring constant \( 25 \) \( \text{N/m} \) is compressed \( 10 \) \( \text{cm} \) and is used to launch the ice cube up the slope. How high does it go above its starting point? Express your answer with the appropriate units.
A pendulum consists of a mass \( M \) hanging at the bottom end of a massless rod of length \( \ell \) which has a frictionless pivot at its top end. A mass \( m \), moving with velocity \( v \), impacts \( M \) and becomes embedded. In terms of the given variables and constants, what is the smallest value of \( v \) sufficient to cause the pendulum (with embedded mass \( m \)) to swing clear over the top of its arc?
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 block of mass 3.0 kg is hung from a spring, causing it to stretch 12 cm at equilibrium. The 3.0 kg block is then taken off and the spring returns to its original height. Now a 4.0 kg block is placed on the spring and released from rest. How far will the 4.0 kg block fall before its direction is reversed?
\(v_x \approx 25\,\text{m/s}\)
\(\Delta x \approx 2.7\times10^{2}\,\text{m}\)
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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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