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# Part (a) Free Body Diagram Explanation

Please visualize or draw out the FBD as described:

Gravity ( \vec{mg} ): Acts directly downwards. With a mass of m = 90 + 12 = 102 kg.
Normal Force ( \vec{N} ): Acts perpendicular to the surface of the incline.
Frictional Force ( \vec{f} ): Acts parallel to the incline against the direction of motion during ascent and in the opposite direction of velocity during descent.
Component of Gravitational Force down the incline ( \vec{mg} \sin(\theta) ): Helps in descending and resists during ascending. Here \theta = 30^\circ .

# Part (b) Calculation of Work Done by Friction to Stop the Bicycle

Step Derivation/Formula Reasoning
1 F_f = \mu_k N Frictional force, F_f , is the product of the coefficient of kinetic friction, \mu_k , and the normal force, N .
2 N = mg \cos(\theta) Normal force is the component of the gravitational force perpendicular to the incline.
3 F_f = 0.7 \times 102 \times 9.8 \times \cos(30^\circ) Calculate F_f using \mu_k = 0.7 , m = 102 kg, g = 9.8 \, \text{m/s}^2 , and \theta = 30^\circ .
4 W_f = -F_f d Work done by friction, W_f , is the product of the frictional force and the distance, d , over which it acts, with a negative sign indicating work done against the motion.
5 W_f = -0.7 \times 102 \times 9.8 \times \cos(30^\circ) \times 9 Substitute values to calculate the work done. The distance d = 9 m.
6 W_f \approx -5454 \, \text{J} Calculated work done by friction; it’s negative as it opposes the direction of motion.

# Part (c) Explanation – Difficulty of Traveling Up vs. Down the Incline

Step Derivation/Formula Reasoning
1 F_{\text{gravity, down}} = mg \sin(\theta) Component of gravitational force along the incline that assists in descending and opposes during ascending.
2 F_{\text{friction, up}} = \mu_s N Static friction opposing the upward motion, which is higher due to increased force requirements.
3 F_{\text{net, up}} = F_{\text{gravity, down}} + F_{\text{friction, up}} Sum of forces opposing the ascent, both the gravitational pull back down the incline and the frictional force.
4 F_{\text{net, down}} = F_{\text{gravity, down}} – F_{\text{friction, down}} Net force during descent is reduced because friction (now kinetic) is less than static friction and gravity assists in motion.
5 Comparison The cyclist finds it harder to travel up due to higher net opposing force (more friction and gravity opposing motion).

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1. FBD with 3 forces.
2. W_f \approx -5454 \, \text{J}
3. The cyclist finds it harder to travel up due to higher net opposing force (more friction and gravity opposing motion).

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_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 MotionEnergy
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
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients 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.
kSpring 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
VariableSI 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)}
VariableDerived 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)}
\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.

1. Start with the given measurement: \text{5 km}

2. 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}}

3. 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}

4. 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}

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 v_i is written as u ; sometimes \Delta x is written as s .
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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