Trampoline physics

I'll think of the trampoline as an object that can be analyzed by Hooke's Law, and that simply means that the force $F$ required to stretch it is proportional to the deflection $x$ . That translates into $F = kx$ , where $k$ is a constant.

I'll take the level where the maximum deflection of the trampoline is reached as my 0 gravitational potential energy level $U_g$ , and I'll call it ground.

I'll think of me as a point mass located at my feet when I'm straight, because along the analyzes, all what matters is the change of position, and that point would experience the same displacements as my center of mass.

First we must find the constant $k$ , because it is basically what describes the trampoline. To do that we use the fact that my weight $W = mg$ , where $m = 80 kg$ is my mass and $g$ is the acceleration of gravity, cause a deflection of $d = 10 cm = 0.1 m$ :

$W = mg = -kd$

$k = -\frac{mg}{d} = 7840 N/m$

The $-$ is because the trampoline's force is against my weight.

Now we can analyze the desired situation, and I'll call $x$ to the maximum deflection reached, which is also the minimum distance at which the trampoline must be placed to avoid hits. The energy of the system won't change, so, the initial total energy $E$ will be just equal to my potential energy at the top, $3 + x$ meters above the ground, because neither me nor the trampoline are moving and the trampoline is in it's initial equilibrium position.

That same energy $E$ at the state when the trampoline has reached it's maximum deflection, is completely stored as it's elastic potential energy, which is equal to $U_t = \frac{1}{2}kx^2$ given that Hooke's Law is applied. So we have:

$E = max(U_t) = max(U_g)$

$max(U_t) = \frac{1}{2}kx^2 = -\frac{1}{2} \frac{mg}{d}x^2$

$max(U_g) = 0 - mg(3 + x) = -mg(3 + x)$

$-\frac{1}{2} \frac{mg}{d}x^2 = -mg(3 + x)$

Working that we get and replacing:

$10x^2 - 2x - 6 = 0$

And so we find that our final answer x is:

$x = 0.881 m$

We can model this trampoline as a spring. Thus with the first diagram we can find the spring constant.

$mg = kx$

$k = mg/0.1 = 10mg$

Now, in the second diagram, the initial gravitational potential energy will be converted into elastic potential energy at the bottom.

At the top, $E = mg(h+x)$

The reason is that we consider the position where the box is at rest as the 0 GPE point.

At the bottom, $E = \frac {1}{2} kx^2$

Equate the energy using conservation of energy, $mg(h+x) = \frac {1}{2} kx^2$ Sub in value of k and simplify:

$mg (h+ x) = \frac {10}{2} mg x^2$

$5x^2 -x - h = 0$

Solve the quadratic,

$x = 0.881 or -0.681$

Since -0.681 is invalid, the answer is 0.881 m.

First, we have to find the spring constant $k$ of the trampoline. By balancing forces acting on the mass: $mg = kx$

We substitute in the values, and rearrange to get $k = (80*9.81)/(0.1) = 7848$

Let the height of the trampoline be h. Then by conservation of energy, Loss of Gravitational Potential Energy = Gain in Elastic Potential Energy.

So we have $mg(h+3) = 0.5kh^2$ rearrange to form a quadratic equation: $0.5kh^2 - mgh - 3mg = 0$

Solving for h and taking the positive root gives $h = 0.881 (to 3 sf)$

first we have to discover the constant K of the theorem Fel=K.X where X is the distance in meters that the trampoline flex itself.

in the first situation the Tensile Strenght is equal to weight strenght :

Fel=P KX=mg Kx0.1=80x9.8 K=7840N

second:

the Tensile Strenght Work :

KX^2/2=mgh

where mgh is the Weight strenght work;

solving we have:

x=0.7745 m

so we have to sum this result to 0.1 m , cause the guy that sitted down on the trampolim doesn't want to hit the ground.

$$ In this case, we can see that the potential energy will be converted to the kinetic energy and then elastic potential energy (the friction and heat energy during the process is neglected). Thus, by using the law of conservation of energy, we can obtain the minimum height of trampoline surface from the ground. Let $\,h$ be the minimum height of trampoline surface from the ground and $\,\Delta y$ be the displacement of trampoline surface when system reach equilibrium state, then $$ $\begin{aligned} mg(h+3)&=\frac{1}{2}kh^2\\ mg(h+3)&=\frac{1}{2}\left(\frac{mg}{\Delta y}\right)h^2\\ 2\Delta y(h+3)&=h^2\\ h^2-0.2h-0.6&=0. \end{aligned}$ $$ By solving the quadratic equation and taking $\,h>0$ , we obtain $\,h=\boxed{0.881\,\text{m}}$ . $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

first, we need to find the spring constant $k=\frac{F}{x}$ $k=\frac{mg}{x}$ $k=\frac{80 \textrm{ kg} \times 9.8 \textrm{ m}/\textrm{s}^2}{0.1 \textrm{ m}}$ $k=7840$

suppose the height of trampoline above the ground is y, and we use law of conservation formula $PE_{gravity}=PE_{trampoline}$ $mg(H+y)=\frac{1}{2} ky^{2}$ $80\times 9.8(3+y)=\frac{1}{2} \times 7840 \times y^{2}$ $y=\frac{\pm\sqrt{61}+1}{10}$ the answer must be positive, so the answer is $y=\frac{\sqrt{61}+1}{10}$ $y\approx 0.881$