You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it at position y0. Imagine that you slowly pull down with a spring scale so the block is now at position y bottom, below the equilibrium position y0 where it was hanging at rest. The scale reading when you let go of the block is 3.0 N.
a. Where was the block when you let go? Assume y0 is the equilibrium position of the block and that "down" is a positive direction.
b. Determine the work you did stretching the spring.
Express your answer to two significant figures and include the appropriate units.
c. What was the energy of the spring-Earth system when you let go (assume that zero potential energy corresponds to the equilibrium position of the block)?
Express your answer to two significant figures and include the appropriate units.
d. How far will the block rise after you release it?
Express your answer to two significant figures and include the appropriate units.

a) Δy = 0.144 m

b) W = 0.145 J

c) Us = 0.32 J

d) ymax = 0.144 m

Explanation:

a) First let's find the spring constant using Hooke's Law

F = k*Δy   ⇒  k = F/Δy

where

F = m*g = 0.1 kg*9.81 m/s² = 0.981 N

and  Δy = 0.07 m. Hence

k = 0.981 N/0.07 m = 14.014 N/m ≈ 14 N/m

In order to find the position of the block when we let it go, we need to find the force that caused this expansion in the spring, we know that the reading of the scale was 3 N and this reading includes the force we want to find and the weight of the block, therefore:

f = 3 N - F = 3 N - 0.981 N = 2.019 N

Now that we have found the force we can use Hooke's Law in order to find the position of the block

f = k*Δy   ⇒   Δy = f/k

⇒   Δy = 2.019 N/14 N/m

⇒   Δy = 0.144 m

b) First, notice that there are two kind of potential energy: the potential energy in the spring and the potential energy due to the gravitational field:

W = ΔU

W = ΔUs + ΔUg

W = (Usf - Usi) + (Ugf - Ugi)

Notice that

Us = 0.5*k*y²

where

yf = 0.07 m + 0.144 m = 0.214 m  and

yi = 0.07 m

and we will take the zero level to be the equilibrium position where the block was hanging at rest. Hence

W = 0.5*k*(yf² - yi²) + m*g*(0 - Δy)

⇒ W = 0.5*14 N/m*((0.214 m)² - (0.07 m)²) + (0.1 kg)*(9.81 m/s²)*(0 - 0.144 m)

⇒ W = 0.145 J

c) When we let the block go the spring was stretched by

y = 0.07 m + 0.144 m = 0.214 m

Therefore:

Us = 0.5*k*y²

⇒ Us = 0.5*14 N/m*(0.214 m)²

⇒ Us = 0.32 J

d) Because the position that we pulled the block to it is considered as the amplitude for the vibrational motion that will happen after we release the block, then the maximum height the particle will reach above the equilibrium position is

ymax = Δy = 0.144 m

their crystal patterns have been disrupted by stress

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this