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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."

- Statistics

Intermediate

Mathematical

GQ

A big bird has a mass of about 0.021 kg. Suppose it does 0.36 J of work against gravity, so that it ascends straight up with a net acceleration of 0.625 m/s^{2}. How far up does it move?

- Energy

Advanced

Mathematical

GQ

A 81 kg student dives off a 45 m tall bridge with an 18 m long bungee cord tied to his feet and to the bridge. You can consider the bungee cord to be a flexible spring. What spring constant must the bungee cord have for the student’s lowest point to be 2.0 m above the water?

- Energy, Springs

Intermediate

Mathematical

FRQ

The launching mechanism of a toy gun consists of a spring with an unknown spring constant, \( k \). When the spring is compressed \( 0.120 \, \text{m} \) vertically, a \( 35.0 \, \text{g} \) projectile is able to be fired to a maximum height of \( 25 \, \text{m} \) above the position of the projectile when the spring is compressed. Assume that the barrel of the gun is frictionless.

- Energy, Springs

Beginner

Mathematical

FRQ

If a small motor does 520 J of work to move a toy car 260 meters in a time of 37 seconds.

- Energy

Intermediate

Conceptual

MCQ

Two blocks of ice, one five times as heavy as the other, are at rest on a frozen lake. A person then pushes each block the same distance d. Ignore friction and assume that an equal force *F* is exerted on each block. Which of the following statements is true about the kinetic energy of the heavier block after the push?

- Energy

- FBD with 3 forces.
- [katex] W_f \approx -5454 \, \text{J} [/katex]
- 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

Conversion Example

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 |

- 1. Some answers may vary by 1% due to rounding.
- Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
- Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
- Bookmark questions you can’t solve to revisit them later
- 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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