0 attempts

0% avg

UBQ Credits

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

1 | [katex] F_{\text{gravity}} = mg [/katex] | Calculate the gravitational force acting on the crate. This uses the mass of the crate [katex]m = 25.0 \, \text{kg}[/katex] and the acceleration due to gravity [katex]g = 9.8 \, \text{m/s}^2[/katex]. |

2 | [katex] F_{\text{gravity}} = 25.0 \times 9.8 = 245 \, \text{N} [/katex] | Performing the calculation gives the gravitational force acting on the crate. |

3 | [katex] F_{\text{gravity, parallel}} = F_{\text{gravity}} \sin(\theta) [/katex] | Determine the component of the gravitational force that acts parallel to the slope. Here, [katex]\theta = 38.0^\circ[/katex]. |

4 | [katex] F_{\text{gravity, parallel}} = 245 \sin(38.0^\circ) = 150.8 \, \text{N} [/katex] | Calculation of the parallel component of gravitational force using the angle’s sine value. |

4.5 | [katex] F_{\text{gravity, parallel}} = F_{\text{friction needed}} [/katex] | Answer – [katex] \boxed{150.8 \, \text{N}} [/katex]. This is the total force frictional force needed to hold the box in place. To verify the static friction can provide enough force we can continue calculate the MAX static friction force. This number should be greater than [katex] 150.8 \, \text{N} [/katex]. |

5 | [katex] f_{max} = \mu_s F_{\text{normal}} [/katex] | Maximum force of static friction that can be applied. |

6 | [katex] F_{\text{normal}} = F_{\text{gravity}} \cos(\theta) + F_{\text{applied}} [/katex] | The normal force is affected by both the gravitational perpendicular force and the additional applied force of 59 N, which acts perpendicular to the slope, reducing the normal force. |

6 | [katex] F_{\text{normal}} = 245 \cos(38.0^\circ) + 59 = 252.1 \, \text{N} [/katex] | Calculation of the normal force considering the cosine of the incline angle and the applied force. |

7 | [katex] f_s = \mu_s F_{\text{normal}} [/katex] | Calculation of the max static friction force, where [katex]\mu_s[/katex] is the coefficient of static friction between the crate and the pavement. |

8 | [katex] f_s = 0.599 \times 252.1 = 151 \, \text{N} [/katex] | The magnitude of the maximum frictional force is [katex]151 \, \text{N}[/katex] which is greater than the [katex] 150.8 \, \text{N} [/katex]. to keep it in place. Thus the scenario is possible. |

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

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

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

Intermediate

Mathematical

FRQ

A 10kg box is pushed to the right by an unknown force at an angle of 25° below the horizontal while a friction force of 50 N acts on the box as well. The box accelerates from rest and travels a distance of 4 m where it is moving at 3 m/s. Solve the following without the use of energy.

- 1D Kinematics, Linear Forces

Intermediate

Conceptual

MCQ

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

- Linear Forces

[katex] 150.8 \, \text{N}} [/katex].

By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

Kinematics | Forces |
---|---|

[katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |

[katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |

[katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |

[katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |

Circular Motion | Energy |
---|---|

[katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |

[katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |

[katex]KE_i + PE_i = KE_f + PE_f[/katex] |

Momentum | Torque and Rotations |
---|---|

[katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |

[katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |

[katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |

Simple Harmonic Motion |
---|

[katex]F = -k x[/katex] |

[katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |

[katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |

Constant | Description |
---|---|

[katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |

[katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |

[katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |

[katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |

[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |

[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |

[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |

Variable | SI Unit |
---|---|

[katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |

[katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |

[katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |

[katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |

[katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |

Variable | Derived SI Unit |
---|---|

[katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |

[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |

[katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |

[katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |

[katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |

[katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |

[katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |

[katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |

General Metric Conversion Chart

Conversion Example

Example of using unit analysis: Convert 5 kilometers to millimeters.

Start with the given measurement:

`[katex]\text{5 km}[/katex]`

Use the conversion factors for kilometers to meters and meters to millimeters:

`[katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]`

Perform the multiplication:

`[katex]\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}[/katex]`

Simplify to get the final answer:

`[katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]`

Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|

Pico- | p | [katex]10^{-12}[/katex] | 0.000000000001 |

Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |

Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |

Milli- | m | [katex]10^{-3}[/katex] | 0.001 |

Centi- | c | [katex]10^{-2}[/katex] | 0.01 |

Deci- | d | [katex]10^{-1}[/katex] | 0.1 |

(Base unit) | – | [katex]10^{0}[/katex] | 1 |

Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |

Hecto- | h | [katex]10^{2}[/katex] | 100 |

Kilo- | k | [katex]10^{3}[/katex] | 1,000 |

Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |

Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |

Tera- | T | [katex]10^{12}[/katex] | 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 [katex] v_i [/katex] is written as [katex] u [/katex]; sometimes [katex] \Delta x [/katex] is written as [katex] s [/katex].
- 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!

The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.

per month

Billed Monthly. Cancel Anytime.

Trial –> Phy Pro

- Unlimited Messages and Images
- Unlimited UBQ Credits
- 157% Better than GPT
- 30 --> 300 Word Input
- 3 --> 15 MB Image Size Limit
- 1 --> 3 Images per Message
- All Smart Actions
- Mobile Snaps
- Focus Mode
- No Ads

A quick explanation

UBQ credits are specifically used to grade your FRQs and GQs.

You can still view questions and see answers without credits.

Submitting an answer counts as 1 attempt.

Seeing answer or explanation counts as a failed attempt.

Lastly, check your average score, across every attempt, in the top left.

MCQs are 1 point each. GQs are 1 point. FRQs will state points for each part.

Phy can give partial credit for GQs & FRQs.

Phy sees everything.

It customizes responses, explanations, and feedback based on what you struggle with. Try your best on every question!

Understand you mistakes quicker.

For GQs and FRQs, Phy provides brief feedback as to how you can improve your answer.

Aim to increase your understadning and average score with every attempt!

10 Free Credits To Get You Started

*Phy Pro members get unlimited credits