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

1 | E_{\text{spring}} = \frac{1}{2} k x^2 | The initial energy stored in the spring is given by the spring potential energy formula, where k is the spring constant and x is the compression. |

2 | E_{\text{spring}} = \frac{1}{2} \times 1800 \times 0.3^2 | Substitute k = 1800 \, \text{N/m} and x = 0.3 \, \text{m} into the spring energy formula to calculate the initial energy. |

3 | E_{\text{spring}} = 81 \, \text{J} | Calculate the total energy stored in the spring. |

4 | W_f = f_k \times d | Calculate the work done by friction, where f_k is the kinetic friction force and d is the distance over which the force acts. |

5 | f_k = \mu_k \times m \times g | Kinetic friction force is the coefficient of kinetic friction times the normal force. Here, \mu_k = 0.12 , m = 6 \, \text{kg} , and g = 9.8 \, \text{m/s}^2 . On horizontal surface, normal force equals mg. |

6 | f_k = 0.12 \times 6 \times 9.8 = 7.056 \, \text{N} | Calculate the kinetic friction force on the horizontal surface. |

7 | W_f = 7.056 \times 3.3 = 23.28 \, \text{J} | Calculate the work done by friction over the 3 m horizontal surface and the distance it slides while the spring decompresses (.3 m) for a total distance of 3.3 m. |

8 | W_{f, \text{slope}} = f_{k, \text{slope}} \times d_{\text{slope}} | Work done by friction on the slope, where f_{k, \text{slope}} is the kinetic friction force on the slope and d_{\text{slope}} = 4.5 \, \text{m} . |

9 | f_{k, \text{slope}} = \mu_{k, \text{slope}} \times m \times g \times \cos(\theta) | On the slope, the normal force is reduced by the cosine of the slope angle \theta = 12^\circ . |

10 | W_{f, \text{slope}} = \mu_{k, \text{slope}} \times m \times g \times \cos(\theta) \times d_{\text{slope}} | Substitute equation from step 9 into the equation for work done by the friction of the slope (step 8). |

11 | W_{f, \text{slope}} = \mu_{k, \text{slope}} \times 258.818 \, \text{J} | Plug in given values and simplify the equation for work done by the friction of the slope as much as possible. |

12 | PE_{\text{top}} = mgh = 5 \times 9.8 \times 4.5sin(12^\circ) \approx 55.01 \, \text{J} | Now find the final potential energy at the top of the incline. |

13 | E_{\text{spring}} = W_f + W_{f, \text{slope}} + PE_{\text{top}} | Finally add up all the energy types together. Use the law of conservation of energy which tells us the initial energy from the spring will turn into work done by friction on level ground, work done by friction on the slope, and the potential energy gained traveling up the slope. |

14 | 81 = 23.28 + (\mu_{k, \text{slope}} \times 258.818) + 55.01 | Substitute all numbers found in the previous steps. |

15 | \boxed{\mu_{k, \text{slope}} \approx .01} | Solve for the coefficient of kinetic friction on the slope. |

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

- Statistics

Intermediate

Conceptual

MCQ

A crate rests on a horizontal surface and a woman pulls on it with a 10-N force. No matter what the orientation of the force, the crate does not move. From least to greatest, rank the normal force on the crate.

- Linear Forces, Tension

Intermediate

Mathematical

GQ

- Inclines, Linear Forces

Advanced

Mathematical

FRQ

*A* has a mass of 3.2 kg and block *B* a mass of 2.4 kg. The pulley is frictionless and has no mass.

- Atwood Machine, Linear Forces

Beginner

Conceptual

MCQ

In which direction does the acceleration always face?

- Linear Forces

Beginner

Mathematical

GQ

A 1100 kg car accelerates from 32 m/s to 8.0 m/s in 4.0 sec. What amount of force was needed to slow it down?

- 1D Kinematics, Linear Forces

\mu_{k, \text{slope}} \approx .01

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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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