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To solve part (a), calculate the work performed by the kinetic frictional force acting on the skis.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]KE_i = \frac{1}{2}mv_i^2[/katex] | Calculate the initial kinetic energy (KE) using the mass [katex] m = 58 \, \text{kg} [/katex] and the initial velocity [katex] v_i = 7.2 \, \text{m/s} [/katex]. |
| 2 | [katex]KE_i = \frac{1}{2} \times 58 \times (7.2)^2 = 1503.36 \, \text{J}[/katex] | Substitute the values into the kinetic energy formula. |
| 3 | [katex]KE_f = \frac{1}{2}mv_f^2[/katex] | Calculate the final kinetic energy using the final velocity [katex] v_f = 3.8 \, \text{m/s} [/katex]. |
| 4 | [katex]KE_f = \frac{1}{2} \times 58 \times (3.8)^2 = 418.76 \, \text{J}[/katex] | Substitute the values into the kinetic energy formula. |
| 5 | [katex]W_{\text{gravity}} = mgh [/katex] | Calculate the work done by gravity, where [katex] h = d \sin(\theta) [/katex] is the height gained climbing the incline. [katex] d = 2.3 \, \text{m} [/katex] and [katex] \theta = 28^\circ [/katex]. |
| 6 | [katex]h = 2.3 \sin(28^\circ) = 1.08 \, \text{m}[/katex] | Calculate the vertical height climbed using [katex] \sin(28^\circ) \approx 0.4695 [/katex]. |
| 7 | [katex]W_{\text{gravity}} = 58 \times 9.8 \times 1.080 = 614 \, \text{J}[/katex] | Substitute [katex] g = 9.8 \, \text{m/s}^2 [/katex]. |
| 8 | [katex] KE_i = KE_f + W_f + PE [/katex] | Place all energy transformation into a single conservation of energy equation: The initial kinetic energy transforms into the final kinetic energy, work done by friction, and the potential energy of the skier. |
| 9 | [katex]W_f = 1503.36 \, – \, 418.76 \, – \, 614 [/katex] | Plug in all values and solve for work done by friction [katex] W_f [/katex]. |
| 10 | [katex] W_f = 470.6 \, \text{J}[/katex] | The negative sign indicates that the work done by friction is in the direction opposite to the motion. |
To solve part (b), determine the magnitude of the kinetic frictional force.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]W = f_k \times d \times \cos(\theta)[/katex] | Work done by a force [katex] f_k [/katex] over a distance [katex] d [/katex], where [katex] \theta [/katex] is the angle between the force and the displacement (which in the case of friction, is [katex]180^\circ[/katex]). |
| 2 | [katex] 470.6 = -f_k \times 2.3 \times \cos(180^\circ)[/katex] | Substitute the work calculated from part (a) and the distance [katex]2.3 \, \text{m}[/katex]. [katex] \cos(180^\circ) = -1 [/katex]. |
| 3 | [katex] -470.6 = f_k \times 2.3[/katex] | Simplify the equation. |
| 4 | [katex]f_k = \frac{-470.6}{2.3}[/katex] | Solve for [katex]f_k[/katex]. |
| 5 | [katex]f_k = -204.6 \, \text{N}[/katex] | The magnitude of the kinetic frictional force. |
Just ask: "Help me solve this problem."
Two blocks, [katex] m_2 > m_1 [/katex], having the same kinetic energy, move from a frictionless surface onto a surface having friction coefficient [katex] \mu_k [/katex]. Which block will travel further before stopping.
A bullet of mass 0.0500 kg traveling at 50.0 m/s is fired horizontally into a wooden block suspended from a long rope. The mass of the wooden block is 0.300 kg and it is initially at rest. The collision is completely inelastic and after impact the bullet+ wooden block move together until the center of mass of the system rises a vertical distance h above its initial position.
A horizontal uniform meter stick of mass 0.2 kg is supported at its midpoint by a pivot point. A mass of 0.1 kg is attached to the left end of the meter stick, and another mass of 0.15 kg is attached to the right end of the meter stick. The meter stick is free to rotate in the horizontal plane around the pivot point. What is the tension in the string supporting the left end of the meter stick?
One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance [katex] h [/katex] above the floor. A block of mass M is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance [katex] x [/katex]. The block is released and strikes the floor a horizontal distance [katex] D [/katex] from the edge of the table. Air resistance is negligible.
Derive an expressions for the following quantities only in terms of [katex] M, x, D, h, [/katex] and any constants.
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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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