# Compressed spring (trampolin)

• imontoya
In summary, a 60kg person at rest, compressed against a spring with a spring constant of 1000 N/m and a compressed distance of 0.2m from its equilibrium position, lands on a springy trampoline with a spring constant of 4000 N/m. The spring on the trampoline is maximally compressed a distance X2 below the equilibrium position. To solve for X2, the energy stored in the first spring must be equated to the energy in the second spring after the leap. The reference is taken below the trampoline (ground), so the distance will be positive.
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.

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.

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.

## 1. How does a compressed spring work?

A compressed spring works by storing potential energy when it is pushed or compressed. This energy is then released when the spring returns to its original shape, creating a bouncing or trampoline-like effect.

## 2. How is the amount of energy stored in a compressed spring determined?

The amount of energy stored in a compressed spring is determined by its spring constant, which is a measure of the stiffness of the spring. The higher the spring constant, the more energy the spring can store.

## 3. What factors affect the compression of a spring?

The amount of force applied to the spring, the stiffness of the spring, and the length of the spring all affect the compression of a spring. The more force applied, the greater the compression; the stiffer the spring, the less it will compress; and the longer the spring, the more it will compress.

## 4. Is there a limit to how much a spring can be compressed?

Yes, there is a limit to how much a spring can be compressed. This limit is called the elastic limit, and it is the point at which the spring will no longer return to its original shape and will instead permanently deform or break.

## 5. How can compressed springs be used in everyday life?

Compressed springs are used in a variety of everyday objects, such as trampolines, pogo sticks, shock absorbers in cars, and even pens. They are also used in various industrial and manufacturing processes, such as in machinery and tools.

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