00:01
Hello students, in this question there is a spring and when the spring is loaded with a mass m1 is equal to 0 .20 kg, it elongates through a distance x1 is equal to 9 .50 cm.
00:18
Similarly, when the same spring is loaded with a mass m2 is equal to 1 kg, it is getting elongated through a distance x2 is equal to 12 cm.
00:28
So, we have to calculate the spring constant k.
00:33
So, for this we can take the hooke's law which is given by f is equal to k into x and this can be written as mg is equal to k into 9.
00:57
This is for mass m1.
00:59
So for mass m1, this can be written as mg is equal to k into x that is x can be written as 9 .50 cm minus x square.
01:17
Now we can substitute the mass value and the acceleration due to gravity value.
01:22
Therefore, it is 0 .2 kg into 9 .8 meter per second square is equal to k into that is it can be written as 9 .50 k minus x naught k.
01:37
Let this be equation 1.
01:40
Now for mass 2, it can be written as for mass m2, it can be written as 1 kg into 9 .8 meter per second square is equal to k into this is k into 12 cm minus x naught which can be rewritten as 9 .8 kg meter per second square is equal to 12 k minus x naught k.
02:20
Here also it can be written like 1 .96 newton or kg meter per second square is equal to 9 .50 k minus x naught k...