0 attempts

0% avg

UBQ Credits

Objective: Determine the friction force and the coefficient of kinetic friction for a 25.0 kg box accelerating down a 23.5° incline.

**Finding the Friction Force:**

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

1 | [katex]f_{\text{net}} = ma[/katex] | Net force is equal to mass times acceleration. |

2 | [katex]f_{\text{net}} = mg\sin(\theta) – f_k[/katex] | The net force down the incline is the component of gravity minus kinetic friction force ([katex]f_k[/katex]). |

3 | [katex]ma = mg\sin(\theta) – f_k[/katex] | Substitute step 1 into step 2. |

4 | [katex]f_k = mg\sin(\theta) – ma[/katex] | Rearrange to solve for [katex]f_k[/katex]. |

5 | Substitute values and solve for [katex]f_k[/katex]. <br> [katex]m = 25.0 \text{ kg}[/katex], [katex]g = 9.81 \text{ m/s}^2[/katex], [katex]\theta = 23.5^\circ[/katex], [katex]a = 0.35 \text{ m/s}^2[/katex] | Use given values to calculate [katex]f_k[/katex]. |

Now, let’s calculate the friction force [katex]f_k[/katex]:

Step | Result |
---|---|

6 | [katex] \boxed{f_k \approx 89.04 \text{ N}} [/katex] |

The friction force impeding the box’s motion is approximately [katex]89.04 \text{ N}[/katex].

**Finding the Coefficient of Kinetic Friction:**

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

7 | [katex]\mu_k = \frac{f_k}{N}[/katex] | The coefficient of kinetic friction ([katex]\mu_k[/katex]) is the ratio of the kinetic friction force to the normal force. |

8 | [katex]N = mg\cos(\theta)[/katex] | The normal force is the component of gravity perpendicular to the incline. |

9 | [katex]\mu_k = \frac{f_k}{mg\cos(\theta)}[/katex] | Substitute step 8 into step 7. |

10 | Substitute values and solve for [katex]\mu_k[/katex]. | Use the calculated [katex]f_k[/katex] and given values to find [katex]\mu_k[/katex]. |

Calculate the coefficient of kinetic friction [katex]\mu_k[/katex]:

Step | Result |
---|---|

11 | [katex] \boxed{\mu_k \approx 0.396} [/katex] |

The coefficient of kinetic friction is approximately [katex]0.396[/katex].

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

- Statistics

Intermediate

Mathematical

GQ

Two objects (49.0 and 24.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find the acceleration of the objects and the tension in the string.

- Atwood Machine, 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

Advanced

Mathematical

GQ

A rocket-powered hockey puck has a thrust of 4.40 N and a total mass of 1.00 kg . It is released from rest on a frictionless table, 2.10 m from the edge of a 2.10 m drop. The front of the rocket is pointed directly toward the edge. Assuming that the thrust of the rocket present for the entire time of travel, how far does the puck land from the base of the table?

- Linear Forces, Projectiles

Advanced

Mathematical

GQ

Three blocks of masses 5, 4, and 3 kg are placed side by side in that order. A 25 N force applied on the 5 kg block accelerates all three blocks together to the right. Find the acceleration of the blocks and the normal force the 4 kg block exerts on the 3 kg block.

- Linear Forces, Multi-Body Systems

Advanced

Mathematical

GQ

A car is driving at \(25 \, \text{m/s}\) when a light turns red \(100 \, \text{m}\) ahead. The driver takes an unknown amount of time to react and hit the brakes, but manages to skid to a stop at the red light. If \(\mu_s = 0.9\) and \(\mu_k = 0.65\), what was the reaction time of the driver?

- 1D Kinematics, Friction, Linear Forces

friction = 89 N, µ_{k} = .4

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