The equations we'll need to use are:
1. Hooke's Law: F = -kx
2. Energy Conservation: 1/2 kx^2 = mgh = 1/2 mv^2
3. Period of motion: T = 2π√(m/k)
where:
F = force exerted by the spring
k = spring constant
x = displacement from equilibrium
m = mass of the block
g = acceleration due to gravity
h = height of the block above the equilibrium point
v = velocity of the block
T = period of motion
a. When the block is at equilibrium, it is at rest and the net force on it is zero. Therefore, we have:
F = -kx = 0
Solving for x, we get:
x = 0
So the compression of the spring at equilibrium is zero.
b. The maximum compression in the spring occurs when the block is at its maximum displacement from equilibrium. At this point, the block momentarily stops before reversing direction. Using energy conservation, we have:
1/2 kx^2 = 1/2 mv^2
where v = 0 at the maximum compression point. Solving for x, we get:
x = √(2mg/k)
So the maximum compression in the spring is √(2mg/k).
c. The maximum acceleration of the block occurs at the equilibrium point, when the spring is fully compressed and then released. At this point, the net force on the block is equal to the maximum force exerted by the spring. Using Hooke's Law, we have:
F = -kx
At the equilibrium point, x = √(2mg/k), so we have:
F = -2mg
The acceleration of the block is given by:
a = F/m = -2g
So the maximum acceleration of the block is 2g downward.