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

1 | U_{\text{spring}} = \frac{1}{2} k x^2 | The potential energy stored in the compressed spring, where k is the spring constant and x is the compression distance. |

2 | W_{\text{friction}} = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d | Work done against friction while moving up the incline: \mu_k is the coefficient of kinetic friction, m is mass, g is acceleration due to gravity, \theta is the angle of incline, and d is the distance traveled up the incline. |

3 | PE = m \cdot g \cdot h | The gravitational potential energy gained by the block at the end of the incline. Here, h is the vertical height, calculated further in Step 5. |

4 | U_{\text{spring}} = W_{\text{friction}} + PE | Conservation of mechanical energy: The total mechanical energy (spring potential energy) initially equals the sum of the work done against friction and the gravitational potential energy at the end. |

5 | h = d \cdot \sin(\theta) | Using the distance d traveled up the incline and the angle \theta to calculate the height h . |

6 | Substituting h , W_{\text{friction}} , and PE in step 4: | \frac{1}{2} k x^2 = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d + m \cdot g \cdot d \cdot \sin(\theta) |

7 | Simplifying and solving for x : | x^2 = \frac{2(\mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d + m \cdot g \cdot d \cdot \sin(\theta))}{k} x = \sqrt{\frac{2d(mg(\mu_k \cdot \cos(\theta) + \sin(\theta)))}{k}} |

8 | x = \sqrt{\frac{2 \cdot 55 \cdot (4 \cdot 9.8 \cdot (0.25 \cdot \cos(30^\circ) + \sin(30^\circ)))}{800}} | Substitute values for d = 55 \, \text{m} , m = 4 \, \text{kg} , g = 9.8 \, \text{m/s}^2 , \mu_k = 0.25 , \theta = 30^\circ , and k = 800 \, \text{N/m} to find x . |

9 | x = \boxed{1.965\, m} | Final value. |

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- Statistics

Intermediate

Conceptual

MCQ

An object is projected vertically upward from ground level. It rises to a maximum height H . If air resistance is negligible, which of the following must be true for the object when it is at a height H/2 ?

- 1D Kinematics, Energy

Intermediate

Proportional Analysis

MCQ

Two balls are dropped from the roof of a building. One ball has twice as massive as the other and air resistance is negligible. Just before hitting the ground, the more massive ball has ball ____ the kinetic energy of the less massive ball.

- Energy

Advanced

Mathematical

FRQ

Refer to the diagram above and solve all equations in-terms of R, M, k, and constants.

- Circular Motion, Energy, Momentum

Advanced

Mathematical

GQ

A 2,000 kg car collides with a stationary 1,000 kg car. Afterwards, they slide 6 m before coming to a stop. The coefficient of friction between the tires and the road is 0.7. Find the initial velocity of the 2,000 kg car before the collision?

- 1D Kinematics, Energy, Linear Forces, Momentum

Advanced

Mathematical

FRQ

The diagram above shows a marble rolling down an incline, the bottom part of which has been bent into a loop. The marble is released from point A at a height of 0.80 m above the ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is 0.35 m. The mass of the marble is 0.050 kg. Friction forces and any gain in kinetic energy due to the rotating of the marble can be ignored. When answering the following questions, consider the marble when it is at point C.

- Circular Motion, Energy

Beginner

Mathematical

MCQ

A forward horizontal force of 12 N is used to pull a 240 N crate at constant velocity across a horizontal floor. The coefficient of friction is

- Friction

Advanced

Mathematical

FRQ

A 90 kg individual is cycling up a hill inclined at 30 degrees on a 12 kg bicycle. The hill is quite steep, and the coefficient of static friction is 0.85. The cyclist ascends 12 meters up the hill and then pauses at the summit. If they then start descending from the peak at rest and travel 9 meters before firmly applying the brakes, causing the wheels to lock.

- Energy

Intermediate

Conceptual

MCQ

You kick a ball straight up. Compare the sign of the work done by gravity on the ball while it goes up with the sign of the work done by gravity while it goes down.

- Energy

Advanced

Mathematical

GQ

A horizontal 300 N force pushes a 40 kg object across a horizontal 10 meter frictionless surface. After this, the block slides up a 20° incline. Assuming the incline has a coefficient of kinetic friction of 0.4, how far along the incline with the object slide?

- 1D Kinematics, Friction, Inclines, Linear Forces

Intermediate

Mathematical

GQ

A 0.5 kg cart, on a frictionless 2 m long table, is being pulled by a 0.1 kg mass connected by a string and hanging over a pulley. The system is released from rest. After the hanging mass falls 0.5 m, calculate the speed of the cart on the table. Use ONLY forces and energy.

- Energy

x = \boxed{1.965\, m}

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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

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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