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Unit 2.6 | Advanced Problem Solving – Objects on Incline Planes

Illustration of a slope with a ball rolling down, annotated with force vectors, friction coefficients, and linear force equations (1)
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Jason Kuma

Writer | Coach | Builder | Fremont, CA

Article Content
This lesson deals with objects on an incline plane. While these can be tricky, I cover a simple way to solve these problems.

As Always be sure to do all the practice questions to fully get the hang of the material.

This course article contains videos that can only be accessed once enrolled in the Learn AP Physics from Scratch Course.

Unit 2 Breakdown

You are on Lesson 6 of 8:

  1. Unit 2.1 | Understanding and Applying Newton’s Law in Depth
  2. Unit 2.2 | Common Linear Forces, Equations, and Misconceptions
  3. Unit 2.3 | Drawing and Understanding Force Body Diagrams
  4. Unit 2.4 | Deriving Equations to Solve Linear Force Problems
  5. Unit 2.5 | Advanced Force Problems – Tension and Elevators
  6. Unit 2.6 | Advanced Force Problems – Ramps (Inclines) [Current Lesson]
  7. Unit 2.7 | Advanced Force Problems – Pulley System
  8. Unit 2.8 | Advanced Force Problems – Multi-Body System

In this lesson: 

  • FBDs for incline problem
  • Dealing with vectors at angles
  • Solving ramp problems
  • Practice ramp problems

Framework

To solve incline problems. We will continue to use the same frame work covered in the previous two lessons.

Note that incline problems are often referred to as ramp problems. Typically there will be an object placed on the ramp. It will be your job to calculate the unknown variable(s), which can be anything from acceleration, mass, to finding the coefficient of friction (µ).

Rather than reading, these next few lesson will use videos to demonstrate problem solving. As always attempt to solve questions listed, before watching the video.

PS – Problem 1 (Easy)

A block with a mass of 10.0 kg lies on a frictionless surface tilted at an angle of 35° to the horizontal.

  1. Determine the acceleration of the block as it slides down the plane ( a = ? m/s2)
  2. Bonus: If the block starts from rest 9.0 m up the plane from its base, what will be the block’s speed when it reaches the bottom of the incline? (v = ? m/s)

PS – Problem 2 (Medium)

A 12 kg box is sliding down an incline where the coefficient of kinetic friction between the box and the incline is 0.15.  Find the angle of the incline where the box slides down with a constant velocity.

PS – Problem 3 (Hard)

A block rests on a flat plane inclined at an angle of 30° with respect to the horizontal. What is the minimum coefficient of friction necessary to keep the block from sliding?

PS – Problem 4 (Mastery)

A penguin slides at a constant velocity of 3.4 m/s down an icy incline. The incline slopes above the horizontal at an angle of 6.2°. At the bottom of the incline, the penguin slides onto a horizontal patch of ice. The coefficient of kinetic friction between the penguin and the ice is the same for the incline as for the horizontal patch. How much time is required for the penguin to slide to a halt after entering the horizontal patch of ice?

Practice Questions

  1. A 42 kg box is being pulled up a rough surfaced incline, with an applied force of 420 N, which makes an angle of 42° with the horizontal.  The coefficient of kinetic friction between the box and the surface is 0.20. What is box’s acceleration?
  2. An 9.2 kg box is sitting in the middle of a horizontal board.  The coefficient of static friction (µs) between the board and the box is 0.46.  The coefficient of kinetic friction is 0.32.  The board is slowly lifted from one end.
    • At what angle will the box start accelerating down the incline? 
    • Once the box is moving, what is its acceleration?
  3. A 19 kg box is sliding down an incline ( µk = 0.15).  Find the angle of the incline where the box will slide down with a constant velocity. 
  4. A man drags a 100 kg crate up the ramp of a truck. The ramp is inclined at 20° and the man pulls at an angle of 30° above the ramp. The coefficient of friction between the crate and the ramp is 0.2.
    • What is the normal force on the crate in terms of the applied force, F?
    • What is the minimum force the man would have to apply to pull the crate up the ramp

Lesson 2.7 Preview

In the next lesson we will move on to solving pulley problems.

While this is confusing for many students, pulleys are simply just tension and weight forces. So we can apply our force problem-solving framework!

As a bonus, I’ll go over a quick shortcut that you can use to solve any pulley question in 15 seconds or less.

Picture of Jason Kuma
Jason Kuma

Writer | Coach | Builder | Fremont, CA

Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy 

Unit 5 – Momentum 

Unit 6 – Torque 

Unit 7 – Oscillations 

Unit 8 – Fluids

Reading Key

LRN
RE
PS
PQ
Black
White
Blue
Orange

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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.

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