Step | Derivation/Formula | Reasoning |
---|---|---|

1 | KE_{\text{initial}} = \frac{1}{2} m v^2 | Calculate the initial kinetic energy of the vehicle using its speed before it halts. Here v = 12.3 \text{ m/s}. |

2 | W_{\text{friction}} = f_k \cdot d = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d | The work done by friction, where \mu_k = 0.650 is the coefficient of kinetic friction, g = 9.8 \text{ m/s}^2 is acceleration due to gravity, \theta = 18^\circ, and d is the distance the vehicle slides. |

3 | W_{\text{gravity}} = m \cdot g \cdot \sin(\theta) \cdot d | The work done by gravity while the vehicle moves down the incline. |

4 | KE_{\text{initial}} = W_{\text{friction}} + W_{\text{gravity}} | By the work-energy principle, the initial kinetic energy is converted into work done against friction plus the work done by gravity. |

5 | \frac{1}{2} m v^2 = \mu_k mg\cos(\theta)d + mg\sin(\theta)d | Substitute expressions from steps 1, 2, and 3 into the work-energy equation. |

6 | d = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) + g \sin(\theta)} | Solve for d, distance the vehicle slides. Notice that mass m cancels out. |

7 | d = \frac{\frac{1}{2} (12.3)^2}{0.650 \times 9.8 \times \cos(18^\circ) + 9.8 \times \sin(18^\circ)} | Substitute numerical values for v, \mu_k, g, and \theta to find the value of d that represents the distance the vehicle slides until it stops. |

8 | d = 8.32 \,\text{m} |
Calculated value. |

To solve part b, look at the equation derived in step 6 of part a. Notice that the angle and distance traveled up the incline are inversely proportional. This means the greater the angle \theta the shorter the distance traveled.

Step | Derivation/Formula | Reasoning |
---|---|---|

1 | d' = \frac{\frac{1}{2} v^2}{\mu_k g \cos(27^\circ) + g \sin(27^\circ)} | Recalculate the distance with the increased angle of 1.5 \times 18^\circ = 27^\circ. |

2 | d' = \frac{d}{\cos(27^\circ) + \tan(27^\circ) \cdot \sin(27^\circ)} | Using the previous formula of d, we express the new sliding distance d' in terms of the old distance d. |

3 | d' = 7.41 \, \text{meters} | At an angle of 27^\circ the vehicale would slide up 7.41 meters, which is less than the orginal dsitance of 8.32 meters. |

3 | \frac{d'}{d} | The ratio \frac{d'}{d} shows how much further the vehicle would slide relative to d. |

4 | \frac{7.41}{8.32} = .89 | Thus the new distance d' = .89d |

Part (c):

Step | Derivation/Formula | Reasoning |
---|---|---|

1 | W_{\text{gravity,up}} = -m \cdot g \cdot \sin(\theta) \cdot d | The work done by gravity as the vehicle slides up the incline is negative since gravity opposes the motion. |

2 | \frac{1}{2} m v^2 = \mu_k m g \cos(\theta) d – m g \sin(\theta) d | Work-energy principle applied while moving up. The kinetic energy has to overcome both friction and an upward gravity force. |

3 | d_{\text{up}} = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) – g \sin(\theta)} | Solve for the distance the vehicle would slide up the incline. |

4 | d_{\text{up}} < d |
The distance d_{\text{up}} will be lesser than d since gravity now acts against the motion, reducing the sliding distance relative to sliding down. |

Phy can also check your working. Just snap a picture!

- Statistics

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

Advanced

Mathematical

GQ

An object of unknown mass is acted upon by multiple forces:

- 100 N to the right at 20°
- 400 N to the left at 40° below the horizontal
- 500 N to the right at 10° below horizontal.

The coefficients of friction are μ_{s}=0.6 and μ_{k}=0.2. Starting from rest, the object travels 10 meters in 4.5 seconds. What is the mass of the unknown object?

- 1D Kinematics, Friction, Linear Forces

Advanced

Mathematical

GQ

A 1000 kg car is traveling east at 20m/s when it collides perfectly inelastically with a northbound 2000 kg car traveling at 15m/s. If the coefficient of kinetic friction is 0.9, how far, and at what angle do the two cars skid before coming to a stop?

- 1D Kinematics, Friction, Momentum

Intermediate

Conceptual

MCQ

Why do raindrops fall with constant speed during the later stages of their descent?

- Linear Forces

Intermediate

Mathematical

GQ

A sled moves with constant speed down a sloped hill. The angle of the hill with respect to the horizontal is 10.0°. What is the coefficient of kinetic friction between the sled and the hill’s surface?

- Inclines, Linear Forces

- d = 8.32 \,\text{meters}
- d' = .89d

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

Conversion Example

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 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - 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 .
- Bookmark questions that you can’t solve so you can come back to them later.
- 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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