# 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). |
Phy can also check your working. Just snap a picture!
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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} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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