Page 2:
The force of a spring is calculated by -kx. (k) is the constant of elasticity and x is the deformation of the spring. The negative is to show how the spring force is opposite in the direction opposite of the displacement. If there are 2 springs working in parallel, they will have the same amount of force that adds up to the magnitude of the applied force. This happens because of twice the spring constant is against the applied. In series each force has the same magnitude as that of the applied.

The spring constant determines the elasticity of a spring, whether it is stiff or soft and is expressed in N/m, derived from the applied force needed to move the spring by a certain distance.
A higher value of N/m, will result to a stiffer spring, while a lower value result to a softer spring.
A stiffer spring has a higher spring constant (k), since more force is required to move it by a meter. A softer spring, on the other hand, has a lower k, since less force is required to move it by a meter.

Page 3: 
As for the parallel question where the springs are parallel I had noticed that the displacement would be less so the more parallel springs the more restoring force there is and there is less displacement.

When I went through the series simulation the restoring force was less than the parallel and there was greater displacement which shows that the series springs stretch more than parallel springs when put under the same force.

The spring constant determines the elasticity of a spring, whether it is stiff or soft and is expressed in N/m, derived from the applied force needed to move the spring by a certain distance.
A higher value of N/m, will result to a stiffer spring, while a lower value result to a softer spring.
A stiffer spring has a higher spring constant (k), since more force is required to move it by a meter. A softer spring, on the other hand, has a lower k, since less force is required to move it by a meter.

Page 2::
Hooke's law states that spring force (Fs), is equal to the product of minus the spring constant (-k) and the displacement of the object (x). Fs = -kx
Since the spring force is the opposite of the applied force, the minus sign is placed in the equation.

Page 3:

Page (1)
-The applied force and spring force are equal in size, but opposite in directions.
-Constant of Elasticity or K is measured in N/m (Newtons/meter)
-The higher the value for K, the more stiff and less stretchy the spring is.
-The higher the value for k, the lower the value for the displacement
-The relationship between the applied force and the green displacement vector for a constant k is a linear relationship

Page (2)
-F(s)=-kx
-The negative sign is just to show the opposite direction
-I predict that when the springs are parallel to each other, they will be stretchy but when they are working in series, they won't be stretchy.

Page 2:

Page 2: Robert Hook introduced the formula of Fs= -kx. Fs is known as a spring force which could be described by the unit of N. K describes the stiffness or stretchiness of a material. and x represents the elongation of a spring by the length of the spring when its stretched. The negative sign represents the opposite direction of the displacement. When the two springs are parallel, we would know that the applied force wouldn't make too much of a difference since there are two spring forces to pull or push.  it would lead to less displacement, since they are separate. However if two springs are attached to each other, the two springs work together to move and would lead to a higher displacement rather than the parallel springs

page 3: When mass changes, we could say the elongation would change as well. So if there is more mass, there will be more elongation

Conclusion 1:
Spring force and applied force are opposite in direction but equal. The unit of k is newtons/meters, which is dividing the force with displacement. As the k's values increases or decreases the spring and applied force aren't affected. The relationship between applied force and displacement is linear. The k and displacement are indirectly proportional.

Conclusion 2:
The force of the spring is calculated by the equation 
Fs=-kx when Fs is the spring force, k is the constant of elasticity and x is the elongation or deformation of the spring. The k is negative because the spring force and constant of elasticity are in opposite direction. The spring force is called the restoring force it restores the force it its equilibrium.

conclusion 3:
When the two springs are parallel The k is doubled because there's only one applied force acting on both of them and the components are in the same direction. When the springs are end to end the components are in opposite directions so the k is divided by two. We also relate the equation to math where you can find the k by getting the slope of the graph and after finding the k, you can use it to find the mass of an object.

Page 1:

The applied force and the spring force are equal in size but opposite in direction.  The constant of elasticity(K) is measured in N/m . K is inversely proportional to the displacement.  The more the K, the lesser the displacement and vice versa.  As the K increases, the spring becomes more rigid/stiffer. This is true because when K increases, the displacement is decreased and the applied force is pulled more towards the starting point. The applied force and the displacement are directly proportional and it's a linear relationship since the change is constant. When force increases by 1 newton, the displacement increases by 0.005 m.

Page 2:

The formula for the spring force is equal to Fs=-kx where the negative sign is just used to indicate the direction. Th direction of the spring force is opposite of the applied force. If the applied force is to the right, the spring force would be to the left. I predict that if a two parallel  springs that have 400 N/m elasticity each and are being pulled with 75 N they will have half the displacement of one spring with K=400N/m and Fapp=75N. I predict in the two series spring that have 400 N/m elasticity each and are being pulled with 75 N they will have half the displacement of one spring with K=400N/m and Fapp=75N

Page 3:

My prediction for the "parallel" prediction for Q7 was correct. The phenomenon states that when two parallel springs that have a K=400N/m each and are pulled to the right with 75 N, the displacement would be half of when one spring with K=400N/m is pulled with 75 N.  For the "series"  springs my prediction was wrong.  I predicted that when two series springs that have a K=400 N/m each and are pulled to the right with 75 N, the displacement would be half of when one spring with K=400 N/m is pulled with 75 N. In reality however, when two series springs that have k=400 N/m each and are pulled by 75 N, the displacement would be double of what one spring with k=400 N/m and 75 N applied to it to the right would have. For question 11 we need to get the constant of elasticity. The spring force formula is Fs=-kx but if you need to get the K, the formula would be K=Fs/x. Fs would be equal to Fg since the only applied force is the force of gravity and Fs is equal to the applied force.  X is the displacement of the spring and spring after the stretch due to the added weight. Then we just graph the spring force on the y axis and the displacement or x on the x-axis. The graph should be linear and it should pass through the origin since there is no shift. The slope of the graph is the constant of elasticity or K and it should be the same for all the values.  As for Q13, we need to find the mass of unknown boxes. Since Fs=-kx and Fs is equal to mass times gravity or mg. We can equate mg to -kx. Since we want to find the mass we can divide the kx by g and therefore we will get the mass of each unknown box. 
Final formula= 
m = -kx/g

The applied force is equal to the spring force . their forces though are in different directions. when applied force is to the right the spring force is to the left. As for the units for k with k being the spring constant force it is measured by newtons per meter .this shows the amount of force that is applied and the displacement the spring had undergone . When k is high the spring is stiffer and less stretchy but when k is low the spring is is less stiff and more stretchy .  I had created a table with 4 situations in which I changed the applied force in all 4 situations the results I had gotten was for 100N there was a 0.500m displacement . AS for -100N there was also a 0.500 meter displacement .When the force was zero the displacement was zero and finally when the applied force was 50N the displacement was 0.250. This show that the forces are equal. When the applied force is 100 and the spring force is 1000 the displacement is 0.100 m yet when both are zero the displacement is 1 meter.

Page 2 Notes :

Robert Hook the known physicist came up with a claim being " As the extension , so the force" This means that the more the spring is  extended the more the applied force and restoring force. The equation we could use to calculate it would be Fs=-kx. As for the equation the Fs would be the spring force and this is also called a restoring force as it tries to restore the spring to normal. The K would be the constant of elasticity and this shows  the measure of the material could be stretchy or elastic of some sort. Finally X being how deformed or displaced the spring would be for example . Spring force would be negative if its not the same side as the displacement.  They gave us a question to solve in where the values were 400 N/m and 75 N where force is applied to the right . The answer would be 5.3 m . Since the applied force is towards the right the the restoring force would be to the left. The 2 questions given for us to predict was a situation in where there are two springs that are parallel and there are two springs that are working in series. When the springs are parallel they require more force than when the springs are in series. The displacement of the series was greater than the displacement of the parallel.

Page 3: 

As for the parallel question where the springs are parallel I had noticed that the displacement would be less so the more parallel springs the more restoring force there is and there is less displacement.

When I went through the series simulation the restoring force was less than the parallel and there was greater displacement which shows that the series springs stretch more than parallel springs when put under the same force. I then had created a table testing our different weights on spring 1. Through this experiment I had concluded that as the weight increased ,the spring force increased and the displacement increased .

k= applied force/displacement 

Firstly we can see this from the unit of k itself which is newtons per meters. but using values from the simulations you can also prove this. i kept my elasticity as 200N/m and changed my applied force. first I made it 20N which caused my displacement to be 0.100m. if you divide 20N/0.100m which is 200N/m (e.g. 100N/0.500m=200N/m). if you try this for any other value you will get the same result. since k is a constant it can't be negative. the higher the value of k the more stiff the spring. k and displacement are indirectly proportional because as we increase k the displacement decreases. this makes sense as the stiffer the spring gets, the less it allows an object to move.

Page 2:

the force of spring is calculated by -kx. where k is the constant of elasticity and x is the elongation or deformation of the spring. the negative is to show ha spring force is opposite in the direction opposite of the displacement. if you have two springs working in parallel, they will have the same force that adds up to the magnitude applied force this is because of two times the spring constant is against the applied. in series however each force has the same magnitude as that of the applied.

Page 3:
Two springs working in parallel half their force becuase they work separately against one applied force meaning they should add up to the magnitude of the applied force. The two springs in a series however both have an individual force equivalent to that of the applied force because they work together against one applied force. The first spring pulls opposite of the other spring which pulls opposite to applied force. The equation f=Kx can be used and compared to normal linear algebraic equations (y=mx+b). And if you graphed a elongation against spring force graph the slope would be the spring constant. You could calculate the slope graphical and algebraically. Also using pervious knowledge about f=ma we can use masses and gravity to calculate the force. Elongation and spring force are directly proportional as one increases so does the other.