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

1 | F = ma | The equation relates force, mass, and the acceleration of the object, where F is the force, m is the mass, and a is the acceleration. |

2 | a = \frac{F}{m} | Rearrange the formula to solve for acceleration a . |

3 | a = \frac{8.0 \, \text{N}}{16 \, \text{kg}} | Substitute the values F = 8.0 \, \text{N} and m = 16 \, \text{kg} . |

4 | a = 0.5 \, \text{m/s}^2 | Calculate the acceleration. |

5 | v = u + at | Use the equation of motion where v is the final velocity, u is the initial velocity (0 m/s since the object is initially at rest), a is the acceleration, and t is the time. |

6 | v = 0 \, \text{m/s} + (0.5 \, \text{m/s}^2)(4 \, \text{s}) | Substitute a = 0.5 \, \text{m/s}^2 and t = 4 \, \text{s} into the equation of motion. |

7 | v = 2 \, \text{m/s} | Calculate the final velocity after 4 seconds. |

8 | \textbf{2 m/s} |
The change in speed of the object after 4 seconds is 2 m/s, hence the correct answer is (b) 2 m/s. |

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- Statistics

Advanced

Mathematical

FRQ

A hungry bear weighing 700 N walks out on a beam in an attempt to retrieve a basket of goodies hanging at the end of the beam. The beam is uniform, weighs 200 N, and is 6.00 m long. The goodies weigh 80 N.

- Linear Forces, Rotational Motion, Torque

Advanced

Mathematical

MCQ

Block m_{1} is stacked on top of block m2. Block m_{2} is connected by a light cord to block m_{3}, which is pulled along a frictionless surface with a force F as shown in the diagram above. Block m_{1} is accelerated at the same rate as block m_{2} because of the frictional forces between the two blocks. If all three blocks have the same mass m, what is the minimum coefficient of static friction between block m_{1} and block m_{2}?

- Linear Forces, Multi-Body Systems

Intermediate

Mathematical

FRQ

A train consists of 50 cars, each of which has a mass of 6.1 x 10^{3} kg. The train has an acceleration of 8.0 × 10^{-2} m/s^{2}?. Ignore friction and determine the tension in the coupling at the following places:

- Linear Forces, Multi-Body Systems

Intermediate

Mathematical

GQ

A person is trying to judge whether a picture (mass = 1.42 kg) is properly positioned by temporarily pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between the picture and the wall is 0.62. What is the minimum amount of pressing force that must be used?

- Linear Forces

Advanced

Mathematical

GQ

A spring launches a 4 kg block across a frictionless horizontal surface. The block then ascends a 30° incline with a kinetic friction coefficient of 0.25, stopping after 55 m on the incline. If the spring constant is 800 N/m, find the initial compression of the spring. Disregard friction while in contact with the spring.

- Energy, Friction, Springs

Intermediate

Mathematical

GQ

Determine the force needed to push a 150 kg body up a smooth 30° incline with an acceleration of 6 m/s^{2}.

- Inclines, Linear Forces

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

Intermediate

Mathematical

FRQ

Friction provides the force needed for a car to travel around a flat, circular race track. Answer the following:

- Centripetal Acceleration, Circular Motion, Friction

Intermediate

Mathematical

GQ

A 2.0 kg wood box slides down a vertical wood wall while you push on it at a 45 ° angle. The coefficient of kinetic friction of wood µ_{k} = 0.200. What magnitude of force should you apply to cause the box to slide down at a constant speed?

- Linear Forces

Intermediate

Mathematical

GQ

The heaviest train ever pulled by a single engine was over 2 \, \text{km} long. A force of 1.13 \times 10^5 \, \text{N} is needed to get the train to start moving. If the coefficient of static friction is 0.741 and the coefficient of kinetic friction is .592 , what is the train’s mass?

- Friction, Linear Forces

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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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