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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.

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1. [katex] W_f = 470.6 \, \text{J}[/katex]
2. [katex] f_k = -204.6 \, \text{N}[/katex]

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KinematicsForces
[katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex][katex]F = ma[/katex]
[katex]v = v_i + at[/katex][katex]F_g = \frac{G m_1m_2}{r^2}[/katex]
[katex]a = \frac{\Delta v}{\Delta t}[/katex][katex]f = \mu N[/katex]
[katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex]
Circular MotionEnergy
[katex]F_c = \frac{mv^2}{r}[/katex][katex]KE = \frac{1}{2} mv^2[/katex]
[katex]a_c = \frac{v^2}{r}[/katex][katex]PE = mgh[/katex]
[katex]KE_i + PE_i = KE_f + PE_f[/katex]
MomentumTorque and Rotations
[katex]p = m v[/katex][katex]\tau = r \cdot F \cdot \sin(\theta)[/katex]
[katex]J = \Delta p[/katex][katex]I = \sum mr^2[/katex]
[katex]p_i = p_f[/katex][katex]L = I \cdot \omega[/katex]
Simple Harmonic Motion
[katex]F = -k x[/katex]
[katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex]
[katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex]
ConstantDescription
[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
VariableSI 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]
VariableDerived 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]\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.

1. Start with the given measurement: [katex]\text{5 km}[/katex]

2. 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]

3. 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]

4. 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]

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity [katex] v_i [/katex] is written as [katex] u [/katex]; sometimes [katex] \Delta x [/katex] is written as [katex] s [/katex].
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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