Login

AP Physics

Unit 2 - Linear Forces

14415 views

Intermediate

Mathematical

GQ

You're a Pro Member

Supercharge UBQ

Go Pro

0 attempts

0% avg

UBQ Credits

Verfied Answer
Verfied Explanation 0 likes
Written by Phy
0

Objective: Determine the maximum angle of a hill on which a car can be parked without sliding, given the coefficient of static friction between the tires and the pavement.

Step Formula Derivation Reasoning
1 [katex]f_s = \mu_s N[/katex] The maximum static friction force ([katex]f_s[/katex]) is equal to the coefficient of static friction ([katex]\mu_s[/katex]) times the normal force ([katex]N[/katex]).
2 [katex]N = mg \cos(\theta)[/katex] The normal force is the component of the gravitational force perpendicular to the slope, where [katex]m[/katex] is the mass of the car, [katex]g[/katex] is the acceleration due to gravity, and [katex]\theta[/katex] is the angle of the hill.
3 [katex]f_s = mg \sin(\theta)[/katex] The force down the slope is the component of gravity parallel to the slope.
4 Set [katex]f_s[/katex] from step 1 equal to [katex]f_s[/katex] from step 3. <br> [katex]\mu_s mg \cos(\theta) = mg \sin(\theta)[/katex] Equate the maximum static friction force to the gravitational force down the slope.
5 Simplify and solve for [katex]\theta[/katex]. <br> [katex]\mu_s \cos(\theta) = \sin(\theta) \Rightarrow \tan(\theta) = \mu_s[/katex] <br> [katex]\theta = \arctan(\mu_s)[/katex] Divide both sides by [katex]mg \cos(\theta)[/katex] and rearrange to solve for [katex]\theta[/katex].
6 [katex]\theta = \arctan(0.85)[/katex] Substitute [katex]\mu_s = 0.85[/katex] into the equation.

Now, calculate the angle [katex]\theta[/katex]:

Step Result
7 [katex] \boxed{\theta \approx 40.36^\circ} [/katex]

The maximum angle of a hill on which a car can be parked without sliding, given the coefficient of static friction, is approximately [katex]40.36^\circ[/katex].

Need Help? Ask Phy To Explain

Just ask: "Help me solve this problem."

Just Drag and Drop!
Quick Actions ?
×

Topics in this question

Join 1-to-1 Elite Tutoring

  • Statistics

See how Others Did on this question | Coming Soon

Intermediate
Mathematical
FRQ

A horizontal, uniform board of weight \( 125 \, \text{N} \) and length \( 4 \, \text{m} \) is supported by vertical chains at each end. A person weighing \( 500 \, \text{N} \) is hanging from the board. The tension in the right chain is \( 250 \, \text{N} \).

  • Linear Forces, Tension
Solve
Advanced
Mathematical
MCQ

A piece of metal of weight\(W\) is suspended by two identical strings. Each string passes through a pulley and is attached to a block of mass \(m\) . The system is in equilibrium.What must be true for \(m\) such that the two strings attached to the piece of metal are almost horizontal.

  • Linear Forces, Tension
Solve
Intermediate
Mathematical
GQ

A crane’s trolley at point \( P \) moves for a few seconds to the right with constant acceleration, and the \( 870 \, \text{kg} \) load hangs on a light cable at a \( 5^\circ \) angle to the vertical as shown. What is the acceleration of the trolley and load?

  • Inclines, Linear Forces, Tension
Solve
Advanced
Mathematical
GQ

An object of unknown mass is acted upon by multiple forces:

  • 100 N to the right at 20°
  • 400 N to the left at 40° below the horizontal
  • 500 N to the right at 10° below horizontal.

The coefficients of friction are μs=0.6 and μk=0.2. Starting from rest, the object travels 10 meters in 4.5 seconds. What is the mass of the unknown object?

  • 1D Kinematics, Friction, Linear Forces
Solve
Advanced
Conceptual
MCQ

A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant (an approximation), when would the tension in the cord connected to the ball be greatest?

  • Circular Motion, Tension
Solve
Discussion Threads
Login to Discuss

Leave a Reply

~ 40°

Nerd Notes

Discover the world's best Physics resources

Continue with

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.

Error Report

Sign in before submitting feedback.

Sign In

Sign In to View Your Questions

Sign In

Share This Question

Enjoying UBQ? Share the 🔗 with friends!

Link Copied!
Made By Nerd-Notes.com
Download Equation Sheet
KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[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
VariableSI 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]
VariableDerived 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

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

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. 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]

  3. 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]

  4. 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]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

–

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

  1. 1. Some answers may vary by 1% due to rounding.
  2. Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
  3. Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
  4. Bookmark questions you can’t solve to revisit them later
  5. 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

Phy Pro

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

$11.99

per month

Subscribe

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

📚 Predict Your AP Physics Exam Score

Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.

Calculate

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

You can close this ad in 7 seconds.

Ads display every few minutes. Upgrade to Phy Pro to remove ads.

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

How to UBQ

A quick explanation

1. UBQ Credits

Credits are used to grade your FRQs and GQs. Pro users get unlimited credits. 

2. Auto Track Stats

Submitting counts as 1 attempt. 

Viewing answers or explanations count as a failed attempts.

3. Point System

Phy gives partial credit if needed

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

Phy helps customize your learning experience.

4. Question Explantions

Phy customizes problem explanations based on what you struggle with. Just hit the explanation button to see.

Phy gives feedback.

Understand you mistakes quicker.

4. Feedback

Phy automatically provides feedback so you can improve your responses. 

10 Free Credits To Get You Started 

Start Solving

By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data. 

Jason here! Feeling uneasy about your next physics test? We will help boost your grade in just two hours.

Book a Free Lesson

NEW! PHY instantly solves any question

Try Now

NEW! Elite Members get 40% off Physics Tutoring

Join Now
We use site cookies to improve your experience. By continuing to browse on this website, you accept the use of cookies as outlined in our privacy policy.
I Accept