00:01
This is an excellent problem because it essentially walks you through all of the steps that you should typically take when completing a problem.
00:10
I found myself quite glad because by the time i got to part c, i realized just in reading the problem i had already completed parts a and b because those are things you should do whenever you start pretty much any physics problem.
00:26
So it says in part a to sketch the box just after the push and label it with variable.
00:35
So you typically want to start by drawing a diagram of what's going on.
00:40
And so we've got a box on an inclined plane.
00:46
It's inclined at an angle theta.
00:49
It starts at the bottom of the ramp.
00:52
It has a mass m.
00:53
And it is pushed up the ramp so that it has some.
00:57
Some initial velocity be not up the ramp.
01:02
We are also told that there is friction between the box and the ramp, so i usually like to indicate this in diagrams by writing a squiggly line.
01:14
So that's it for part a.
01:16
For part b, you wanna list all of the known variables and their values.
01:21
So i usually like to do this right below the diagram.
01:25
And from the problem we know the mass is 1 .1 kilograms.
01:31
The angle theta is 42 degrees.
01:36
And then we are given both the coefficient of static friction, which is 0 .63, and the coefficient of kinetic friction, which is 0 .33, as well as the initial velocity, which is 9 .6 meters per second.
01:55
We're given one additional piece of information in this problem.
02:01
So i'm going to go ahead and write that down just so that i don't forget it.
02:05
And that is that it will eventually come to a stop.
02:17
Now, for part c, we want to draw a free body diagram for the box as it slides up and write newton's law equations based off of those free body diagrams.
02:30
And so for a free body diagram, i usually wouldn't make it until i know what i'm looking for but if you're doing an inclined plane problem you're most likely going to need newton's second law so it makes sense so our free body diagram here is the box we know it's on earth so there is some weight m g downward it's on an inclined plane so there is a normal force i usually label that as capital in perpendicular to the surface of the inclined plane.
03:09
And as it is sliding up, we know that the force of friction is going to be opposite that motion.
03:20
And because it is moving, it'll be a kinetic friction force.
03:25
And i think that's it.
03:27
I'm additionally going to write on here that my axes look like this where x is down the ramp and y is out of the ramp now we want to write our equations so we'll start with the y direction equations the sum of the forces in the y direction in the y direction the y direction the spox is in equilibrium so the sum of those forces should be equal to zero newtons and the only forces in the y direction are the normal force and the positive the negative y direction and the, excuse me, the y component of the gravitational force in a negative y direction, which will have a magnitude equal to m g cosine of theta.
04:23
In the x direction, we can sum up our forces.
04:28
We are not in equilibrium in the x direction, so that's all going to be equal to mass times acceleration.
04:36
And in the x direction, we have two forces.
04:41
We have the force of kinetic friction in the x direction, which i'm going to go ahead and put in the kinetic friction force equation, which is that it is equal to the coefficient of kinetic friction times the normal force, as well as the x component of gravity in that same direction.
05:03
So plus m g sign of theta and those are our forces for b or sorry for part c so we can move on to part d where we are doing the same thing but drawing the diagram for the box when it is stopped at the top of the incline and we're again going to draw our diagram and write out newton's laws.
05:34
So at this point, again, we can go ahead and add the gravitational force mg and the normal force in.
05:43
Again, our coordinate axes are going to be tilted.
05:49
But this time, when it's at the top, it wants to slide down the ramp.
05:56
So we're going to have a force of friction directed up the ramp.
06:01
And we don't know if the box is going to stay stopped or if it is going to slide back down.
06:11
So i'm going to label this as just a generic force of friction.
06:18
It will, you know, when it stopped there for a split second, it will be static friction force, but it might immediately start falling again.
06:28
So i just want to call this the force of friction.
06:31
So that i can know myself that just because it's static friction force doesn't mean that it's going to stay stationary.
06:42
In the y direction, our forces look exactly the same.
06:47
We have some of the forces equal zero.
06:49
That'll be zero newton's and be equal to the normal force minus mg cosine of theta.
06:57
In the x direction, things are a little bit different because if you look at the x and y components of the gravitational force, you will see that the x component of gravity is opposite in direction to the x direction of friction.
07:16
So the sum of the forces in the x direction, we don't know if it's going to be equal to zero or if it's going to be equal to m .a.
07:26
So we'll just go ahead and write m .a here.
07:29
And our acceleration might be zero.
07:34
And that's going to be equal to this force of friction in the negative x direction plus mg sign of beta...