In order to determine the acceleration of the block, use the following formula:

Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:

Then, you have:

by solving for a, you obtain:

In this case, you have:
k: spring constant = 100N/m
m: mass of the block = 200g = 0.2kg
x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m
Replace the previous values of the parameters into the expression for a:

Hence, the acceleration of the block is 10 m/s^2
I'm pretty sure the answer is d.The weight of the book and the table's upward force on the book are equal in magnitude but opposite in direction.
Answer:
Ф = 239.73 rad
Explanation:
α = 12 + 15×t
W = ∫α×dt
= ∫(12 + 5×t)×dt
= 12×t + 2.5×t^2
then:
Ф = ∫W×dt
= ∫(12×t + 2.5×t^2)dt
= 6×t^2 + 5/6×t^3
therefore the angle at t = 4.88s is:
Ф = 6×(4.88)^2 + 5/6×(4.88)^3
= 239.73 rad
Simply, apply the formula

and insert the values of m = mass, v = velocity and E = Energy.
The result will be

, m = 1 kg