Login

AP Physics

Unit 2 - Linear Forces

13357 views

Advanced

Conceptual

FRQ

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 whether it takes longer for the hockey puck to slide up or down a distance on an incline with kinetic friction , given an initial speed of 5 m/s.

Analysis for Upward Motion:

Step Formula Derivation Reasoning
1 [katex]a_{\text{up}} = -g(\sin(\theta) + \mu_k \cos(\theta))[/katex] Acceleration is due to gravity and kinetic friction opposing the motion.
2 [katex]v^2 = u^2 + 2a_{\text{up}}d[/katex] Use the kinematic equation for final velocity. v m/s at the top of incline.
3 [katex]t_{\text{up}} = \frac{v – u}{a_{\text{up}}}[/katex] Time taken is found using the kinematic equation for time.
4 Substitute and solve for . [katex]u = 5 \text{ m/s}[/katex], [katex]\theta = 30^\circ[/katex], [katex]\mu_k = 0.4[/katex] Calculate time up.

Analysis for Downward Motion:

Step Formula Derivation Reasoning
5 [katex]a_{\text{down}} = g(\sin(\theta) – \mu_k \cos(\theta))[/katex] Acceleration is due to gravity assisted by friction.
6 [katex]t_{\text{down}} = \sqrt{\frac{2d}{a_{\text{down}}}}[/katex] Time taken using the kinematic equation for constant acceleration.
7 Substitute and solve for .  [katex]u = 0 \text{ m/s}[/katex] at the peak, [katex]\theta = 30^\circ[/katex], [katex]\mu_k = 0.4[/katex] Calculate time down.

Calculating the time for both upward and downward motions:

Step Result
8 [katex] t_{\text{up}} \approx 0.60 \text{ s} [/katex]
9 [katex] t_{\text{down}} \approx 1.15 \text{ s} [/katex]

The time taken for the hockey puck to move up the distance is approximately 0.60 seconds, while the time to slide down the same distance is approximately 1.15 seconds. Therefore, it takes longer for the puck to move down the distance than to move up.

This result may seem counterintuitive, but it’s important to note that the initial conditions (starting speed and angle of incline) and the presence of friction significantly influence the motion. The initial upward speed allows the puck to cover the upward distance quickly, while friction continuously slows it down during both upward and downward motion, affecting the total time for each path.

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

Beginner
Mathematical
MCQ

The block is moving horizontally at a constant velocity. There are two applied forces on the object as shown in the image. In which direction is the friction force acting on the object?

  • Linear Forces
Solve
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
Solve
Intermediate
Mathematical
GQ

A skateboarder, with an initial speed of \( 20.0 \, \text{m/s} \), rolls virtually friction-free down a straight incline of length \( 18 \, \text{m} \) in \( 3.3 \, \text{s} \). At what angle is the incline oriented above the horizontal?

  • Inclines, Linear Forces
Solve
Intermediate
Mathematical
GQ

The cart with mass \( M = 3 \, \text{kg} \) is pulled by a massless string and moving on a horizontal track. A weight with mass \( m = 1 \, \text{kg} \) is hung from the other end of the string through a pulley system. Due to the gravitational force acting on the weight of mass \( m \), the cart is accelerated to the left. Find the tension in the string.

  • Linear Forces, Tension
Solve
Beginner
Mathematical
GQ

An airplane of weight \( W \) is flying horizontally with constant velocity. The total forward thrust of the engines is \( 3W \). What is the magnitude of the force of the air on the plane in terms of \( W \)

  • Air Resistance, Linear Forces
Solve
Discussion Threads
Login to Discuss

Leave a Reply

It takes less time to slide up. This is because there is a greater net force on the way up (gravity and friction work together to slow the block down quicker).

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

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