00:01
For this problem, we're considering a mass on an inclined plane.
00:06
And we want to start off by first finding the acceleration of that mass in the absence of friction.
00:14
In any sort of problem like this, we want to start with a free body diagram.
00:18
And for a mass on an inclined plane without friction, the free body diagram should always look like this.
00:24
We've got the normal force out of the plane, the gravitational force into the plane and this is going to be the y component of gravity which is equivalent to m g cosine of theta and then the x component of gravity which will be down the plane and is equivalent to m g sign of theta because again we want our axes to be along the plane and out of the plane where i am denoting positive x is down the incline so to solve a problem like this, we just want to apply newton's second law in the x direction.
01:07
So the sum of the forces in the x direction must equal mass times acceleration in the x direction.
01:14
And in this case, the only force we have is mg sine theta.
01:22
So that means that acceleration must be equal to g times sine of theta, which is 5 .8, meters per second squared.
01:36
For part b, we're now considering kinetic friction.
01:41
So we can add the force of kinetic friction to our free body diagram.
01:48
But kinetic friction is going to be equal to the coefficient of kinetic friction times the normal force.
01:55
And in this case, the normal force is going to be equivalent to the y component of gravity m g cosine theta so again when we set up newton second law in the x direction mass times acceleration must equal m g sine theta in the x sorry in the positive x direction and mu k times the normal force which is m g cosine theta in the y direction.
02:31
And so we can go ahead and solve.
02:34
Acceleration is going to be equal to g times sine of theta minus mu k cosine of theta.
02:46
And that is going to be equivalent to 3 .1 meters per second squared.
02:55
For part c, on top of that, we are now considering a spring.
03:03
So i'm going to draw a new free body diagram here.
03:08
Again, we still have the x component of the gravitational force, mg sine theta, the y component of the gravitational force, mg cosine theta, the normal force n.
03:26
But then in the negative x direction, we have both the force of static friction because it is not moving, as well as the force from the spring.
03:44
Actually, i'm going to subscript this fe for elastic.
03:51
So the force of static friction, i'm going to assume we're in the case of maximum static friction, which means that this is going to be mu sub s times in, and the force for a spring is going to always be minus kx.
04:12
And remember in part c or part three, they told us that x is equal to 0 .15 meters.
04:23
So again, we're going to set up newton's second law in the x direction, but in this case, nothing is moving, so the sum of the forces must be equal to zero.
04:36
And we've got additional forces here.
04:40
We've got mg sine theta like before.
04:45
But now we have minus mu sub s, m g cosine theta.
04:51
That's the static friction force minus kx.
04:55
That is the spring force.
04:58
So to solve, we want to solve for k, the spring force.
05:04
And when you solve, you would get that k is equal to the mass, which is 9 .2 times g, which is 9 .8, all divided by the extension, x, which is 0 .15, times what we had before, which is sign of theta, theta again being 36 degrees minus.
05:37
And now we're using the coefficient of static friction.
05:41
So 0 .363 times the cosine of the angle 36.
05:53
And you should get that k is equal to about 177 newton's per meter.
06:03
In that case.
06:05
Okay, i'm going to erase all of this to make space for the last one.
06:10
So in the last one, we've added an additional block to the inclined plane...