Incomplete question.The Complete question is here
A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.
a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.
b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.
Answer:
(a)ω = 1 rad/s
(b)t = 2.41 s
Explanation:
(a) initial angular momentum = final angular momentum
0 = L for disk + L............... for runner
0 = Iω² - mv²r ...................they're opposite in direction
0 = (MR²/2)(ω²) - mv²r
................where is ω is angular speed which is required in part (a) of question
0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)
0=200ω²-200
200=200ω²
ω = 1 rad/s
b.)
lets assume the "starting point" is a point marked on the disk.
The person's angular speed is
v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s
As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.
(angle) + (angle disk turns) = 2π
(1.6 rad/s)(t) + ωt = 2π
t[1.6 rad/s + 1 rad/s] = 2π
t = 2.41 s
Ok, now what do you want to know about it?
We can solve for the acceleration by using a kinematic equation. First we should identify what we know so we can choose the correct equation.
We are given an original velocity of 24 m/s, a final velocity of 0 m/s, and a time of 6 s. We and looking for acceleration (a) in m/s^2.
The following equation has everything we need:

So plug in the known values and solve for a:
0 = 24 + 6a
-24 = 6a
a = -4 m/s^2
Answer:
k = 
b = 
t = 
Solution:
As per the question:
Mass of the block, m = 1000 kg
Height, h = 10 m
Equilibrium position, x = 0.2 m
Now,
The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:

where
u = initial velocity = 0
g = 10
Thus

Force on the mass is given by:
F = mg = 
Also, we know that the spring force is given by:
F = - kx
Thus

Now, to find the damping constant b, we know that:
F = - bv

Now,
Time required for the platform to get settled to 1 mm or 0.001 m is given by:
