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# Part (a) Free Body Diagram Explanation
Please visualize or draw out the FBD as described:
– Gravity ([katex] \vec{mg} [/katex]): Acts directly downwards. With a mass of [katex] m = 90 + 12 = 102 [/katex] kg.
– Normal Force ([katex] \vec{N} [/katex]): Acts perpendicular to the surface of the incline.
– Frictional Force ([katex] \vec{f} [/katex]): 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 ([katex] \vec{mg} \sin(\theta) [/katex]): Helps in descending and resists during ascending. Here [katex] \theta = 30^\circ [/katex].
# Part (b) Calculation of Work Done by Friction to Stop the Bicycle
Step | Derivation/Formula | Reasoning |
---|---|---|
1 | [katex] F_f = \mu_k N [/katex] | Frictional force, [katex] F_f [/katex], is the product of the coefficient of kinetic friction, [katex] \mu_k [/katex], and the normal force, [katex] N [/katex]. |
2 | [katex] N = mg \cos(\theta) [/katex] | Normal force is the component of the gravitational force perpendicular to the incline. |
3 | [katex] F_f = 0.7 \times 102 \times 9.8 \times \cos(30^\circ) [/katex] | Calculate [katex] F_f [/katex] using [katex] \mu_k = 0.7 [/katex], [katex] m = 102 [/katex] kg, [katex] g = 9.8 \, \text{m/s}^2 [/katex], and [katex] \theta = 30^\circ [/katex]. |
4 | [katex] W_f = -F_f d [/katex] | Work done by friction, [katex] W_f [/katex], is the product of the frictional force and the distance, [katex] d [/katex], over which it acts, with a negative sign indicating work done against the motion. |
5 | [katex] W_f = -0.7 \times 102 \times 9.8 \times \cos(30^\circ) \times 9 [/katex] | Substitute values to calculate the work done. The distance [katex] d = 9 [/katex] m. |
6 | [katex] W_f \approx -5454 \, \text{J} [/katex] | 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 | [katex] F_{\text{gravity, down}} = mg \sin(\theta) [/katex] | Component of gravitational force along the incline that assists in descending and opposes during ascending. |
2 | [katex] F_{\text{friction, up}} = \mu_s N [/katex] | Static friction opposing the upward motion, which is higher due to increased force requirements. |
3 | [katex] F_{\text{net, up}} = F_{\text{gravity, down}} + F_{\text{friction, up}} [/katex] | Sum of forces opposing the ascent, both the gravitational pull back down the incline and the frictional force. |
4 | [katex] F_{\text{net, down}} = F_{\text{gravity, down}} – F_{\text{friction, down}} [/katex] | 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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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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