00:01
So here we're doing some basic mechanics problems around work, energy, and power for problem one.
00:12
We're told that a box with a weight of 196 newtons slides down an incline.
00:21
When it calls it a 10 meter incline, i'm assuming that the height of the incline is 10 meters.
00:28
Otherwise we can't solve this problem without knowing the angle of incline.
00:34
We're told that the box starts at rest at the top, so its initial speed is zero meters per second, and it reaches the bottom with a speed of nine meters per second.
00:46
And the question is how much mechanical energy is lost as the box slides down the incline.
00:54
So the loss in mechanical energy will be the difference between the, total energy that the box starts with and the amount of energy that the box ends with.
01:04
It starts at the top of the incline with an amount of energy equal to its gravitational potential energy, which is mgh.
01:15
And it's going to end with some kinetic energy once it gets to the bottom, which is where i'm labeling as my zero point for gravitational potential energy.
01:27
And that will be equal to one half mb squared.
01:32
Now, ideally, all of the potential energy would be converted into kinetic energy and we would get zero, but in reality, some of the energy is lost.
01:43
And so the difference between these two will tell us how much is lost.
01:48
To go ahead and plug in, we want to, for m .g plug in the weight of the box, so 196 newtons.
02:00
The height can be plugged in as is 10 meters, and then we're plugging in one half the mass, which we're going to plug in the weight divided by g to get the mass, so 196 newtons divided by 9.
02:17
I use 9 .81 meters per second squared for g.
02:23
And then times the final velocity of nine meters per second squared.
02:30
So that gives me 1 ,960 joules minus 800 and 9 .17 joules, which is 1 .14 times 10 to the third joules.
02:51
The closest answer of the choices is a1 .15, and i think that's close enough that there was probably just a rounding error somewhere along the way that caused that to be off.
03:07
So i would go with answer choice a here.
03:11
For problem two, i'm going to erase each problem as we go along, just so we have a clean slate to start with for each problem and plenty of space to solve it.
03:22
But for problem two, we are thinking about a crate that is thrown upwards.
03:30
It is thrown vertically off of a 300 meter tall building.
03:39
And it has an initial velocity of 15 meters per second.
03:46
And we want to know what is the speed at which it reaches the ground.
03:49
I'm going to make the assumption that this crate is moving as a projectile and assume that only gravity is acting on it so that we can use the kinematic equations substituting in acceleration as negative g.
04:15
And because we're not given any information about how long this crate spends in the air, the amount of time, we're going to use the kinematic equation that does not include time.
04:28
So that tells us that the final speed squared is equal to the initial speed squared.
04:33
And again, we're replacing acceleration with negative g.
04:36
So we get minus 2g.
04:39
And in this case, it's vertical motion, so delta y.
04:43
Where delta y is going to be the vertical displacement, which in this case will be negative 300 meters because it starts from the top of the building, which is 300 meters.
04:54
Goes down to the ground, which is zero.
04:57
So 0 minus 300 is negative 300.
05:00
So at this point, we can plug in the values that we're given, 15 meters per second for the initial speed.
05:10
Again, i use 9 .81 meters per second squared for g, and we're plugging in negative 300 meters for delta to y.
05:20
And we get that b squared is 6 ,11 meters squared per second squared, which means that v will be 78 .2 meters per second or option answer choice c.
05:40
Moving on to part three.
05:44
And i apologize if you hear the barking dogs in the background...