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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). |
Just ask: "Help me solve this problem."
A block of mass [katex] m [/katex] is moving on a horizontal frictionless surface with a speed [katex] v_0 [/katex] as it approaches a block of mass [katex] 2m [/katex] which is at rest and has an ideal spring attached to one side.
When the two blocks collide, the spring is completely compressed and the two blocks momentarily move at the same speed, and then separate again, each continuing to move.
A car traveling to the right with a speed v brakes to a stop in a distance d. What is the work done on the car by the frictional force F? (Assume that the frictional force is constant)
A car accelerates uniformly from rest to [katex] 29.4 [/katex] m/s in [katex] 6.93 [/katex] s along a level stretch of road. Ignoring friction, determine the average power in both watts and horsepower ([katex] 1 \text{ horsepower} = 745.7 \text{ Watts} [/katex]) required to accelerate the car if:
A 0.2 kg object is attached to a horizontal spring undergoes SHM with the total energy of 0.4 J. The kinetic energy as a function of position presented by the graph.
A 0.50-kg mass is attached to a spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. What is the total energy of the system?
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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] | |
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] |
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