To solve part (a), calculate the work performed by the kinetic frictional force acting on the skis.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | KE_i = \frac{1}{2}mv_i^2 | Calculate the initial kinetic energy (KE) using the mass m = 58 \, \text{kg} and the initial velocity v_i = 7.2 \, \text{m/s} . |
| 2 | KE_i = \frac{1}{2} \times 58 \times (7.2)^2 = 1503.36 \, \text{J} | Substitute the values into the kinetic energy formula. |
| 3 | KE_f = \frac{1}{2}mv_f^2 | Calculate the final kinetic energy using the final velocity v_f = 3.8 \, \text{m/s} . |
| 4 | KE_f = \frac{1}{2} \times 58 \times (3.8)^2 = 418.76 \, \text{J} | Substitute the values into the kinetic energy formula. |
| 5 | W_{\text{gravity}} = mgh | Calculate the work done by gravity, where h = d \sin(\theta) is the height gained climbing the incline. d = 2.3 \, \text{m} and \theta = 28^\circ . |
| 6 | h = 2.3 \sin(28^\circ) = 1.08 \, \text{m} | Calculate the vertical height climbed using \sin(28^\circ) \approx 0.4695 . |
| 7 | W_{\text{gravity}} = 58 \times 9.8 \times 1.080 = 614 \, \text{J} | Substitute g = 9.8 \, \text{m/s}^2 . |
| 8 | KE_i = KE_f + W_f + PE | 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 | W_f = 1503.36 \, – \, 418.76 \, – \, 614 | Plug in all values and solve for work done by friction W_f . |
| 10 | W_f = 470.6 \, \text{J} | 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 | W = f_k \times d \times \cos(\theta) | Work done by a force f_k over a distance d , where \theta is the angle between the force and the displacement (which in the case of friction, is 180^\circ). |
| 2 | 470.6 = -f_k \times 2.3 \times \cos(180^\circ) | Substitute the work calculated from part (a) and the distance 2.3 \, \text{m}. \cos(180^\circ) = -1 . |
| 3 | -470.6 = f_k \times 2.3 | Simplify the equation. |
| 4 | f_k = \frac{-470.6}{2.3} | Solve for f_k. |
| 5 | f_k = -204.6 \, \text{N} | The magnitude of the kinetic frictional force. |
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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_1m_2}{r^2} |
| a = \frac{\Delta v}{\Delta t} | f = \mu N |
| R = \frac{v_i^2 \sin(2\theta)}{g} |
| Circular Motion | Energy |
|---|---|
| F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
| a_c = \frac{v^2}{r} | PE = mgh |
| KE_i + PE_i = KE_f + PE_f |
| Momentum | Torque and Rotations |
|---|---|
| p = m v | \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 |
|---|
| F = -k x |
| T = 2\pi \sqrt{\frac{l}{g}} |
| T = 2\pi \sqrt{\frac{m}{k}} |
| Constant | Description |
|---|---|
| g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
| G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
| \mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
| k | Spring constant, in \text{N/m} |
| M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
| M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
| M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| s (Displacement) | \text{meters (m)} |
| v (Velocity) | \text{meters per second (m/s)} |
| a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
| t (Time) | \text{seconds (s)} |
| m (Mass) | \text{kilograms (kg)} |
| Variable | Derived SI Unit |
|---|---|
| F (Force) | \text{newtons (N)} |
| E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
| P (Power) | \text{watts (W)} |
| p (Momentum) | \text{kilogram meters per second (kgm/s)} |
| \omega (Angular Velocity) | \text{radians per second (rad/s)} |
| \tau (Torque) | \text{newton meters (Nm)} |
| I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
| f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | 10^{-12} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
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