**First find the pushing force, which equal to the maximum force of static friction.**

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

1 | F_{\text{static max}} = \mu_{s} F_{\text{normal}} | Maximum static friction force, where \mu_{s} is the coefficient of static friction. |

2 | F_{\text{normal}} = mg | Normal force equals weight for horizontal motion. |

3 | F_{\text{static max}} = \mu_{s} mg | Substituting F_{\text{normal}}. |

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

1 | F_{\text{net}} = F_{\text{push}} – F_{\text{kinetic}} | Net force is the difference between pushing force and kinetic friction. |

2 | F_{\text{push}} = \mu_{s} F_{\text{normal}} | Pushing force equal to maximum static friction. |

3 | F_{\text{kinetic}} = \mu_{k} F_{\text{normal}} | Kinetic friction force. |

4 | F_{\text{net}} = \mu_{s} mg – \mu_{k} mg | Substituting values for F_{\text{push}} and F_{\text{kinetic}}. |

5 | F_{\text{net}} = \mu_{s} mg – \mu_{k} mg = ma | Set net force equation equal to ma. Then solve for a (acceleration) |

6 | a = 1 m/s^{2} |
Plug in values and solve |

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

1 | v = u + \frac{F_{\text{net}}}{m}t | Kinematic equation for velocity. Note F/m is the acceleration (1 m/s^{2} ) from above. |

2 | v = 0 + \frac{(\mu_{s} – \mu_{k}) mg}{m} \times 5, \text{s} | Initial velocity u = 0; substituting F_{\text{net}} and t. |

3 | v = 5 m/s | Plug in values and find the velocity at the end of the 5 second push. |

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

1 | 0 = v – \mu_{k} gt_{\text{stop}} | Final velocity 0 when stopped, v is final speed after push (5 m/s as found above). |

2 | t_{\text{stop}} = \frac{v}{\mu_{k} g} | Solving for time to stop t_{\text{stop}}. |

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

1 | t_{\text{total}} = t_{\text{push}} + t_{\text{stop}} | Sum of time with force applied and time to stop. |

Let’s calculate t_{\text{total}}. The calculations yield the following results:

- Final speed at the end of 5 seconds: \boxed{4.905, \text{m/s}}
- Time to stop due to kinetic friction: \boxed{1.667, \text{seconds}}
- Total time the block is in motion: \boxed{6.667, \text{seconds}}

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

- Statistics

Intermediate

Conceptual

MCQ

While flying horizontally in an airplane, you notice that a string dangling from the overhead luggage compartment hangs at rest at 15° away from the vertical toward the back of the plane. Using this observation, you can conclude that the airplane is:

- Linear Forces

Intermediate

Mathematical

GQ

A 135.0 N force is applied to a 30.0 kg box at 42 degree angle to the horizontal. If the force of friction is 85.0, what is the net force and acceleration? If the object starts from rest, how far has it traveled in 3.3 sec?

- 1D Kinematics, 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

MCQ

A constant force of 8.0 N is exerted on a 16 kg object initially at rest. How much speed will the object gain after 4 seconds?

- Linear Forces

Intermediate

Conceptual

MCQ

The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is

- Elevators, Linear Forces

6.67 seconds

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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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