For more information, please see full course syllabus of AP Physics 1 & 2

For more information, please see full course syllabus of AP Physics 1 & 2

## Discussion

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

## Share this knowledge with your friends!

## Copy & Paste this embed code into your website’s HTML

Please ensure that your website editor is in text mode when you paste the code.(In Wordpress, the mode button is on the top right corner.)

- - Allow users to view the embedded video in full-size.

*Since this lesson is not free, only the preview will appear on your website.*

### Ramps and Inclines

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Question 1 0:18
- Question 2 1:01
- Question 3 2:50
- Question 4 3:11
- Question 5 5:08

### AP Physics 1 & 2 Exam Online Course

### Transcription: Ramps and Inclines

*Hi folks and welcome back to Educator.com.*0000

*In this mini-lesson we are going to go over page 1 of the worksheet on ramps and inclines.*0002

*You can find the link to that down below and I recommend you take a minute, print it out, see if you can answer it and then come back and we will check our answers, so take a second and we will go on to problem 1 here.*0008

*Number 1 -- The diagram at right represents a block at rest on an incline.*0019

*Which diagram below best represents the forces acting on the block?*0023

*Right away I know I am going to have the blocks weight down and we are going to have the normal force, which is always perpendicular to the surface in that direction and if the block wants to slide down the ramp, then I would expect to see force of friction acting up the ramp.*0028

*Of these choices, the one that matches on our free-body diagrams (FBD) here, is Number 4, weight down with a normal force perpendicular to the ramp, and friction acting to oppose the motion of the object up the ramp.*0046

*Number 2 -- The diagram below shows a 50 kg crate on a frictionless plane at angle θ to the horizontal.*0062

*The crate is pushed at constant speed up the incline from point (A) to point (B) by force (F).*0069

*If angle θ were increased, what would be the effect on the magnitude of force (F) and the total work done on the crate as it is moved from (A) to (B).*0074

*Now this is a very interesting question because it is somewhat vague about how this would happen.*0083

*For example, if you are going from (A) to (B), you are going to do the same amount of work, assuming that (B) is at the same height because it is a frictionless plane.*0087

*Your difference in energy says that you must have some potential energy here and that change in potential energy depends only on the height, not the route that you take to get there, assuming that it is frictionless.*0098

*Of course as the angle θ increases, you are going to see that you are going to have to push harder, you are going to have to apply more force to get it up the ramp, but what it does not make clear is as angle θ increases, are you actually going to a higher (B)?*0110

*If that is the case, then you would have to do more work.*0126

*If instead, it is just saying that you have a shorter ramp, another way of getting to the same height, change the angle, but (B) is still at the same height, a different angle there, then you would be doing the same work, but regardless, the force is going up.*0129

*There are a couple of answers here depending on how you looked at the question that could be correct.*0143

*Instead of worrying about exactly which answer is correct, make sure you understand the basic concept here, because as θ goes up, force must increase of course, and if you consider that (B) was getting to a higher point, you would have to do more work.*0148

*If you thought (B) was at the same point and you just shortened up your ramp, then you would have the same amount of work done.*0162

*Number 3 -- In the diagram below, a 10 kg block is at rest on a plane inclined at 15 degrees to the horizontal.*0170

*As the angle of the incline is increased, what will happen to the mass of the block?*0177

*Another trick question there -- the mass of the block is not going to change, the amount of stuff it is made up of is going to remain constant.*0182

*Number 4 -- A block weighing 10 N is on a ramp, inclined at 30 degrees to the horizontal.*0192

*A 3 N force of friction (Ff) acts on the block as it is pulled up the ramp at constant velocity, so acceleration equals 0, all balanced forces, with (Ff), which is parallel to the ramp as shown in the diagram.*0198

*What is the magnitude of force (F)?*0210

*Let us start with our FBD here. We have weight; we have (F) up the ramp; we have force of friction and our normal force.*0213

*But notice that these are not all lined up with an axis, so I am going to redraw this with my axis this way and now I have (F) acting up the ramp.*0226

*I have the normal force lined up with an axis.*0239

*I have the force of friction and this force of gravity -- this weight I am going to break into two components, a portion that is perpendicular to the ramp (mg perpendicular) and a portion parallel with the ramp (mg parallel), which is equal to mg-sin(θ).*0243

*If we want the magnitude of the force (F), we need to realize that if it is moving at constant velocity, all the forces must balance out, so those forces in the x-direction must be equal.*0260

*That tells me that (F) must be equal to mg-sin(θ) plus the force of friction, where mg-sin(θ) is going to be (mg), its weight...*0272

*...so 10 N × sin(30 degrees) + 3 N (force of friction), is 10 sin(30) = 5 N + 3 N, therefore our applied force must be choice 2, 8 N.*0284

*Let us take a look at one more here.*0305

*The diagram below shows a 1 × 10 ^{5} N truck at rest on a hill that makes an angle of 8 degrees with a horizontal.*0308

*What is the component of the truck's weight parallel to the hill?*0317

*We are looking for the component (mg) parallel to the hill, which is (mg) times the sine of that angle.*0320

*If its weight (mg) is 1 × 10 ^{5}, so 10^{5} N × sin(8 degrees)... is about 13,900 N or about 1.4 × 10^{4} N, so the correct answer there is Number 3.*0329

*That concludes page 1 of the ramps and inclines worksheet.*0357

*If this went great -- Wonderful! -- Keep moving on and if it did not, now would be a great time to go back and review that full lecture from previously here on the Educator.com site.*0361

*Thanks so much for your time and make it a great day!*0371

## Start Learning Now

Our free lessons will get you started (Adobe Flash

Sign up for Educator.com^{®}required).Get immediate access to our entire library.

## Membership Overview

Unlimited access to our entire library of courses.Learn at your own pace... anytime, anywhere!