00:01
Hi there.
00:02
So for this problem, we have a block with a mass m that is equal to 3 .5 kilograms.
00:14
And it's attached to a spring with a spring constant that is also given.
00:19
That is spreebs constant is equal to 570 newton's per meter.
00:28
And it is initially at rest on an inclined plane that is at an angle.
00:33
That is also given, that angle in here is equal to 26 degrees with respect to the, as you can see from the figure, with respect to the horizontal.
00:57
And the coefficient of kinetic friction between the block and the plane is given.
01:05
That coefficient of kinetic friction is equal to 0 .12.
01:12
In the position where the spring is compressed by a distance d and that distance d is equal to 0 .18 meters.
01:25
Now the mass is at its lowest position and the spring is compressed the maximum amount.
01:32
So we need to take the initial gravitational energy of the block as zero.
01:37
So the condition that we have for this problem is the one that is shown here in this figure.
01:44
Now, as we are told the initial gravitational energy of the block is zero.
01:54
So at this position, so you can see, gravitational potential energy is zero.
02:02
So the first question for this problem is, what is the blob's initial mechanical energy? now, we know that mechanical energy contains all forms of energy.
02:18
However, in this case, because we are told that initial mechanical energy is equal, the initial gravitational energy is equal to zero.
02:27
And the system is addressed.
02:31
So that means also that the initial kinetic energy is equal to zero.
02:36
So since we have those two equals to zero, the only one that is different from zero is the initial, potential energy due to the spring.
02:47
And that is defined as 1 divided by 2 times the spring's constant times the distance d square.
02:55
Well, in this case, we call it the distance d that this spring has been compressed.
03:02
And then we just need to simply substitute those valleys in there.
03:06
So we will have 1 divided by 2 times the spring's constant that is 570 newton per meter.
03:14
This times the distance d that we are given, which is equal to 0 .18 meters and that to the square.
03:24
So using our calculator, we obtain a value of 9 .234 joules.
03:39
So that's a solution for the first question of this problem.
03:42
Let's call this the equation a and the question a.
03:47
And now for part b of this problem, we are told that if the spring pushes the block up the incline, what distance l in meters while the block travel before coming to rest? now we are told that the spring remains attached to both the block and the fits wall throughout its motion.
04:13
So with that said, we need to calculate that distance at which this mass is at rest.
04:25
So with that said, the condition that we have in here is that, well, let me just draw the situation in here.
04:35
Let's say that let's extend a little bit this, just like in here.
04:43
So let's say that the block ends at this point.
04:55
So what we need to calculate is the distance, this distance, l from here to here.
05:12
This distance l is the one that we want in this case.
05:17
Now, we know that essentially the block is going to have.
05:26
So initial speed due to the spring's energy that this brings potential energy that is transformed into kinetic energy.
05:48
Then due to friction, that energy is going to be transforming into other type of energy, for example, thermal energy, until the block is.
06:01
Addressed.
06:03
So what we need to consider in this case are those two forces.
06:08
We know that the change, well, the war done by the forces in this case, that work is just simply the total war done by all of the forces in here.
06:21
We need to consider all of the forces.
06:24
And it's going to be equal to the change in the kinetic energy plus the change in the botanitary energy.
06:30
Plus the change in the potential energy.
06:35
So having that in mind, let me just draw all of the forces that are acting on this block.
06:42
So we have the normal force that is always perpendicular to the surface.
06:48
And we have the weight of this mass that is always downward.
06:53
So as you can see from this, it meets an angle with the vertical in that angle is the angle theta.
07:00
We also have the spring's, yeah, the spring's force to the right, but also we have some frictional force that opposes the motion of this, just like this.
07:28
Well, let's call this the spring force and this one in here as the frictional force.
07:35
Let's put it in here to differentiate.
07:39
Okay, so those are the, well, in this case, are four forces acting on this.
07:46
So what we need to do is to calculate the total work done by these forces and then just simply see what we obtain from this for the length.
08:02
Now, we know that the initial kinetic energy is zero and the final kinetic energy is going to be also equals to zero because there is no we want the condition of which the block is addressed.
08:17
So this is going to be zero.
08:20
But this in here is going to be different from zero because we are going to have the final gravitational potential energy.
08:28
And we know that the gravitational potential energy depends on the altitude, depends on the yes, on the altitude.
08:38
So that will be the mass times the acceleration due to gravity times the height.
08:42
Now, by trigonometry, we have in here some triangle.
08:46
So let's say that this is to a height age.
08:51
So since we want the length l, and we know this angle right here, what we can do is to use the sign of theta, because the sign of theta is going to be h divided by l.
09:07
So to obtain l, we just need to divide h by, well, in this case, well, we need to define l.
09:20
So we need to write this in terms of l.
09:23
So let me just see.
09:25
That will be, yes, it's going to be l is going to be h divided by the sign of theta, but we want h just to substitute in here.
09:37
So that will be just simply l, sign of theta...