00:01
Okay, so we have this small mass that is two kilograms and it's being thrown down the ramp at a speed of five meters per second.
00:09
It's going to run into this spring that has a spring constant of 88 mutons per meter.
00:15
And the ramp itself is at an angle of 37 degrees.
00:18
And we need to figure out how far the spring will compress.
00:26
So to do this, we can use conservation of energy where the initial state of the block is going to be, its potential energy, which will be mg times the height above its resting position it'll get to.
00:40
So that's going to be h, which is our two meters, plus the compression of the spring x, and that's times the sign of our angle of the ramp.
00:53
So this this h plus x times sine of theta, that will give us the height necessary for the potential energy.
01:03
And then it also has some kinetic energy in the form of 1 .5 mb squared because it has an initial velocity of 5 meters per second.
01:12
And then once it reaches the maximum compression, it'll only have the elastic potential energy, the stored potential energy in the spring of 1 .5kx squared.
01:25
And so we need to simplify this equation.
01:30
We're going to turn it into a quadratic and then find the.
01:34
The compression from there.
01:37
Now, first thing i'm going to do is distribute over here so i can have my x kind of by itself.
01:45
So this is going to become mgh sine theta plus mgx sine theta plus one -half mv squared.
02:02
That's going to equal one -half k -x squared.
02:06
And so now, now, we need everything on the right -hand side so i'm just going to subtract all these terms over here i'm also going to multiply through by two to get rid of those halves to make this a little bit easier to manipulate so this is just going to become kx squared like so and then we're going to minus the x term here so once again i've multiplied through by two so this becomes 2mg sine theta x and then these terms are going to be combined together.
02:40
And it's going to be, oops, that should be minus, minus mgh -syne theta minus one -half mv squared.
02:52
At this point, i'm going to plug numbers in to make the quadratic formula work a little bit easier.
02:58
So this becomes 0 equals 88x squared minus 2 times 2 kilograms times 9 .8, so i use 9 .8 for g, times the sign of 37 degrees times x.
03:15
So if you multiply those things together, you get a value of 23 .6.
03:21
And then that's multiplied by x.
03:24
And then once again here, two times 9 .8.
03:29
Oh, i forgot to put the two there.
03:31
So there's another two here.
03:33
So two times two, and then this half should be gone.
03:36
That was close.
03:38
Since i multiplied everything through by two.
03:41
So it's two times two times nine point eight times two times sine of theta...