00:01
Now here in this question, we look at basically the motion of a box, right, pushed by a spring, by compressed spring, right? as you're in this diagram.
00:13
So according to the question, the spring has a spring constant, which is given by four newton per meter.
00:19
And the box has a friction coefficient with the surface.
00:27
The static coefficient is mule s.
00:29
0 .2.
00:30
And the kinetic 1 .1.
00:32
So first you ask to derive expression for the speed, v of the box, just as it leaves the spring, just as it leaves the spring.
00:43
So what's happening is that if you compress the spring, suppose the natural length of the natural length of the spring is from here to here.
00:54
Now you compress it by, let's say, x.
00:57
And then the spring will push the box, of course, up to this point, right? and then you asked what is the speech of it.
01:09
But actually, first, we have to really just think of which two cases.
01:14
The first is if the compress x and the recoil force, which is k times x, is actually smaller or less, i mean less or equal, the static kinetic, the maximum static friction, which is actually given by muo s times the weight m, g, of the box, right? where m is the mass of the box, g is gravity, acceleration of gravity in earth.
01:46
And then obviously, the box will not move, right? therefore, the speed will be zero.
01:51
In such case, if kx is less than this maximum static friction, then we can be sure that the speed were zero.
01:59
It will not move.
02:00
Right.
02:02
And on the other hand, if the box does have a, if the recoil force is bigger than the static friction force and then this box will actually move...