The base equation for position when undergoing simple harmonic motion is:

\(\displaystyle x(t) = Asin(\omega t+\phi)\)

\(\displaystyle \omega = \sqrt{\frac{k}{m}}\)

First, solve for the phase constant.

\(\displaystyle x(0) = Asin(\phi) = 5, \phi = \frac{\pi}{2}\)

Plug all the variables into the equation and solve.

\(\displaystyle x(3) = 5sin(\sqrt{\frac{35}{5}}3+\frac{\pi}{2}) = -0.42 cm\)

Recall Hooke's Law, which states:

\(\displaystyle F=-kX\)

Here, \(\displaystyle F\) is the force exerted by the spring, \(\displaystyle k\) is the spring constant, and \(\displaystyle X\) is the displacement from the spring's rest position. This equation tells us that the force exerted is directly proportional to the displacement. We don't need to solve for \(\displaystyle k\) to determine the magnitude of the force on the spring stretched 10m. We can instead come up with a proportionality such that:

\(\displaystyle \frac{F_1}{X_1}=\frac{F_2}{X_2}\)

Here, \(\displaystyle F_1\) and \(\displaystyle F_2\) are forces applied on the string and \(\displaystyle X_1\) and \(\displaystyle X_2\) are the displacements of the spring from its respect position respectively. We assume that a stretched spring will have a positive displacement, whereas a shrunken spring will have a negative displacement. However, since we're looking for the magnitude of the force, regardless of direction, the direction of the displacement doesn't matter. Therefore, we can write the proportion as:

\(\displaystyle \mid \frac{F_1}{ X_1}\mid=\mid \frac{F_2}{ X_2 }\mid\)

In our case:

\(\displaystyle \mid \frac{F_{unknown}}{10m}\mid =\mid \frac{20N}{-5m} \mid\)

\(\displaystyle \frac{F_{unknown}}{10m}=\frac{20N}{5m}\)

\(\displaystyle F_{unknown}=40N\)

Hooke's law states that the spring force is equal to the product of the spring constant and the displacement of the spring:

\(\displaystyle F_{s}=-kx\)

The force is negative because it acts in the direction opposite of the displacement from the equilibrium position (i.e. when we stretch we do so in the positive direction). We are given the force and the displacement, so we just solve for k:

\(\displaystyle \frac{-6300}{-3}=2100 \frac{N}{m}\)

This question is giving us the amount of force needed to displace an object attached to a spring, and is asking us to calculate the spring constant. Thus, we will need to make use of Hooke's law.

\(\displaystyle F=-kx\)

The above equation, Hooke's law, tells us that the restoring force of the spring is related to the displacement of the attached object. It's important to note that in the question stem, we are told that an external force of \(\displaystyle 6\:Newtons\) is needed to displace this object. Thus, the restoring force of the spring, due to Newton's third law, has the same magniture of the applied force but in the opposite direction. Thus, the restoring force of the spring is equal to \(\displaystyle -6\:Newtons\). Plugging this value into Hooke's law, as well as the displacement of \(\displaystyle 3\: meters\), yields:

\(\displaystyle -6\: Newtons=-k(3\: meters)\)

\(\displaystyle k=\frac{6\: Newtons}{3\: meters}=2\: \frac{N}{m}\)

\(\displaystyle F_{total}=F_1+F_2+....\)

\(\displaystyle F_{total}=0\)

\(\displaystyle 0=kx+mg\)

Where \(\displaystyle k\) is the spring constant

\(\displaystyle x\) is the compression of the spring

\(\displaystyle m\) is the mass of the object.

\(\displaystyle g\) is the gravity constant, which will be treated as a negative number.

Solve for \(\displaystyle k\):

\(\displaystyle k=-\frac{mg}{x}\)

Plug in values:

\(\displaystyle k=-\frac{1.5*-9.8}{.005}\)

\(\displaystyle k=2940\frac{N}{m}\)

\(\displaystyle F_{total}=F_1+F_2+...\)

\(\displaystyle F_{total}=0=F_{spring1}+F_{spring2}+F_{gravity}\)

\(\displaystyle 0=k_1x+k_2x+mg\)

Where \(\displaystyle k_1\) and \(\displaystyle k_2\) are the respective spring constants, \(\displaystyle x\) is the stretch length, and \(\displaystyle g\) is the gravity constant, which is a negative as the vector is pointing down.

Since

\(\displaystyle k_1=.5k_2\)

Substitute and plug in values:

\(\displaystyle 1.5k_2+15*(-9.8)=0\)

Solving for \(\displaystyle k_2\):

\(\displaystyle k_2=\frac{15*9.8}{1.5}\)

\(\displaystyle k_2=98\frac{N}{m}\)

Determine the net forces on the submarine:

\(\displaystyle F_{net}=ma\)

Plug in values:

\(\displaystyle F_{net}=7500*2\)

\(\displaystyle F_{net}=15000\frac{kg*m}{s^2}\)

Determine what forces are acting on the submarine:

\(\displaystyle F_{net}=F_{spring}+F_g\)

\(\displaystyle F_{net}=kx+m*g\)

Plug in values

\(\displaystyle 15000=k(1.125-1)+7500*-9.8\)

*Note: Acceleration due to gravity is going down so it is a negative

Solve for \(\displaystyle k\):

\(\displaystyle k=7.08*10^5 \frac{N}{m}\)

Each spring will be subject to the same force, and since they have the same spring constant, stretch the same amount. Thus:

Total stretch:

\(\displaystyle 80cm-70cm=10cm\)

Stretch of one spring:

\(\displaystyle \frac{10cm}{2}=5cm\)

\(\displaystyle 5cm=.05m\)

Use Hooke's law:

\(\displaystyle F=kx\)

The force will be equal to the force of gravity on the mass:

\(\displaystyle mg=kx\)

Solve for \(\displaystyle k\):

\(\displaystyle k=\frac{mg}{x}\)

\(\displaystyle k=\frac{5*9.8}{.05}\)

\(\displaystyle k=980\frac{N}{m}\)

The force of the spring in relationship to strain is independent of direction. Thus, the same force pulling on the spring would result in an equal amount of length change, albeit by stretching instead of compressing.

In this problem the "spring within a spring" can be treated as a single spring.

Use Hooke's law:

\(\displaystyle F=kx\)

Plug in values:

\(\displaystyle 9.8*100=k*2\)

Solve for \(\displaystyle k\).

\(\displaystyle 490\frac{N}{mm}\)

Again use Hooke's Law:

\(\displaystyle F=kx\)

Plug in values:

\(\displaystyle 200*9.8=490x\)

Solve for \(\displaystyle x\)

\(\displaystyle x=4mm\)