How Deep Does the Trampoline Spring Compress Under a 60kg Load?

[imontoya]

Homework Statement

A 60kg person is at rest with a vertical distance 1 m above the groud, compressed against the a spring. The spring is compressed a distance of .2m from its equilibrium position, with a spring constant of 1000 N/m. The person lands into a springy trampoline with a spring constant of 4000 N/m and the spring is maximally compressed a distance X2 below equilibrium position (ground). The reference is to be taken below the trampolin (ground), so the distance will have to be positive. How far below equilibrium will the spring be compressed?

Homework Equations

mgy + 1/2KX^2 = 1/2KX^2

The Attempt at a Solution

The fact that the reference is below the trampolin in confusing me, I tried soving the equation for X2, but I don't know how to acout for the vertical distance (1m + X) because we need to solve for X.

 

[Mentz114]

Is the person being thrown up by the first spring then landing on the second one ? It's not clear from your description.

The easiest way to do this problem is to work out how much energy is stored in the first spring, then equate that to the energy in the second spring after the leap.

 

[m2003]

I would first clarify the problem by asking for more information. Is the vertical distance of 1m above the ground the initial height of the person before they jump on the trampoline, or is it the distance between the equilibrium position of the spring and the ground? This clarification is important in order to accurately solve for the distance X2 below equilibrium position.

Assuming that the initial height of the person is 1m above the ground and that the distance between the equilibrium position of the spring and the ground is also 1m, the equation to solve for X2 would be:

mgy + 1/2KX^2 = 1/2KX^2

Where:
m = mass of the person (60kg)
g = acceleration due to gravity (9.8 m/s^2)
y = initial height (1m)
K = spring constant of trampoline (4000 N/m)
X = distance below equilibrium position (X2)

Solving for X2, we get:
X2 = (mgy + 1/2KX^2)/(1/2K)

Substituting the given values, we get:
X2 = (60kg x 9.8m/s^2 x 1m + 1/2 x 4000 N/m x 0.2m^2)/(1/2 x 4000 N/m)
X2 = 5.1m

Therefore, the spring will be compressed a distance of 5.1m below equilibrium position when the person lands on the trampoline. However, if the initial height of the person is not 1m above the ground, the equation and solution would be different. It is important to clarify this in order to provide an accurate response.