00:01
In this problem, it is given that a spring that is frictionless has a 9 kilogram mass attached to it and a force of 70 newton stretches the spring 1 .6 meters beyond its natural length.
00:15
And also starting from the equilibrium position, the spring is pushed with an initial velocity of 2 meter per second.
00:26
We are asked to determine the position of the mass after t seconds.
00:34
Now let x of t denotes the position of the mass after t seconds.
00:40
Now it is given that a mass of 9 kilogram is attached to it.
00:47
Let m denotes the mass of the object, then it is 9 kilogram.
00:52
And also if f denotes the force, it is 70 newton.
01:00
And the spring is stretched 1 .6 meters beyond its natural length by the force of 70 newton.
01:10
So we have if k is the spring constant by hook slow, it is the force 70 newton divided by the distance beyond which it is stretched with respect to its natural length, that is 1 .6 newton per meter...