00:01
For spring system we know that f equals to kx.
00:08
So from here, if f is represented at y -axis, so we can write k -x where x is the elongation represented at x -axis.
00:23
So k will be equal to y -y -x, and this is equal to tan theta, that is this slope.
00:35
Now if we take, let's suppose this particular mark.
00:42
So here the value of y is, so this y can be represented here or the coordinate 4 newton and x is 0 .2.
00:56
So k will be equals to y upon x, that is 4 divided by 0 .2.
01:01
So that is equals to 80 newton per meter.
01:12
Sorry, here it will be 20 newton.
01:15
Per meter because this will become 40 by 2 and this that comes equals to 20 newton per meter.
01:24
So the value of spring constant will be 20 newton per meter.
01:28
Similarly if we take even this particular point here it is 8 newton divided by 0 .4.
01:36
So here k will be equals to 8 divided by 0 .4 that is equals to 20 newton per meter.
01:43
So this is the value of spring constant.
01:46
That is k equals to 20 newton per meter.
01:50
Now coming to the second part of the question, that is given as a mass of 0 .25 kg has been hung from the spring.
02:03
We have to calculate the extension of the spring.
02:08
Now we know that f equals to kx where f is, let's suppose this is the spring, this is the mass air force will be acting m g so we can write 0 .25 into g where g is the oscillation due to gravity k we already found 20 so we can calculate x that is the change in the extension of the length that will be equals to 0 .25 into 9 .8 divided by 20 so x will be equals to 0 .12 meter or 12 so this will be the change in the length of the spring or extension of the spring...
