How do you find the spring constant of a trampoline? Say a mass of $35kg$ is placed in the middle of a trampoline and causes a vertical displacement of $22cm$.
Why is the spring constant $35g/0.22$? 
Why are we considering the applied force rather than the force within the material itself? Why does that work?
It certainly doesn't work in the case of a mass being a hung on a horizontal rope, so what logic is one meant to use in order to deduce that the spring constant is $35g/0.22$?
Here's where the question comes from, for your interest:
https://qualifications.pearson.com/content/dam/pdf/International%20Advanced%20Level/Physics/2013/Exam%20materials/WPH05_01_que_20150618.pdf 
 A: The issue here is that for a shape like a trampoline, you are interested in the "effective force" for a certain "effective displacement". You are right that there are much larger forces present in the springs/elastic bands that hold the trampoline in place - but the experiment you did cannot tell you about those forces without some more information about the geometry.
In the case of a string under tension, you can ask two different questions: (1) how hard are we currently pulling on the end of the string, or (2) what force is needed to displace the (center of the) string sideways by a certain amount?
Again - the answers are related, but they will be quite different. If you want to know, for example, what the resonant frequency of your string would be if you attached a mass at the midpoint and tried to excite lateral motion, you need the answer to (2) above. And that's basically the problem with the trampoline - you want to know what force gives rise to a certain vertical displacement, because that's the mode of operation of the trampoline.
A: This question is not rigorous and is open to other interpretations. When asking about Spring constant, we need to know the orientation. 
For example, consider an anisotropic elastic material, i.e. it stretches differently depending on the orientation. When applying a force $F$ horizontally it moves $\delta$. When applying the same force vertically $F$ it moves $2 \delta$.  
Say we want to know how much it stretches when it's applied a force $F$ at 45º. Doing the maths, $F_x = F_y = F/\sqrt 2$. Then the displacement $\Delta x=\delta/\sqrt 2$ and $\Delta y=2\delta/\sqrt 2$. Considering small displacements, the displacement along the diagonal axis can be approximated to $\Delta _ {45} = \delta\sqrt{5/2}$ (Pythagoras).
But now, what is the spring constant? Depending on the orientation, it can be:
$$K_x=F/\delta$$
$$K_y=\frac{F}{2\delta}$$
$$K_{45}=\sqrt{\frac{2}{5}}\frac{F}{\delta}$$
In this trampoline question, if you have enough data about the materials and do the maths, it can be verified that the $K$ of the rope (in this case, the fabric of the material), is related to the $K$ used to estimate the vertical displacement of the trampoline.
A: The correct formula is:
$$F= k\Delta l$$
where $k$ is in Newtons/meter, $l$ is in meters and $F$ is the applied force.
Therefore, 
$$F = 35 \times 9.8 = 343N$$
The spring constant is $343/0.22$ or $1559\space Nm^{-1}$
This is the basic formula for a constant $k$ but in most springs (or materials), it is not a constant so $k$ won't be a linear curve. It depends on materials elasticity. It's more an average.
A: Provided the springs supporting the canvas and the canvas itself remain essentially elastic, and that the deformations aren't large (questionable for the case of a trampoline - but we can ignore it for now), the force-displacement relationship of a mass in the centre of the trampoline would be expected to be basically linear.
Although there isn't an actual vertical spring in the centre of a trampoline, this linear force-displacement relationship may be considered as equivalent to a spring. And that 'spring' has a spring constant which you've calculated.
Another example is an elastic beam supported at two ends. If we subject it to a point load at its mid-span, it will deflect a certain amount that relates to the flexural stiffness of the beam and its end conditions. Clearly the beam is not a vertical spring, but for all intents and purposes we can conveniently represent its mid-point deflection using a spring constant. This is the basic approach for determining, for example, the beam's dynamic response - and this is done in practice in engineering.
Hope this helps somewhat.
A: When the trampoline reaches maximum extension, the force applied by the trampoline and the gravitational force on the object are equal because the object is rest (net force must be zero; $mg = F_{trampo}$). If you consider the trampoline as a spring, then the total extension is given as $0.22 m$ and the force applied is given to be $350N$ ($mg$).
Therefore, the spring constant is given by:
$$k = \frac{F}{\Delta x} = \frac{350}{0.22}$$
