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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."
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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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