# Solve Spring Hooke's law: Stretch of each Spring

• zvee_y
In summary: N/m. A 44 kg person jumps from a 1.94m platform onto the innersprings. Assume the springs were initially unstretched and that they stretch equally. The potential energy of the person is 836.528J. The stretch displacement of the springs is 836.528J.
zvee_y
[SOLVED] Spring Hooke's law

## Homework Statement

The innerspring mattress is held up by 23 vertical springs, each having a spring constant of 5000N/m. A 44 kg person jumps from a 1.94m platform onto the innersprings. Assume the springs were initially unstretched and that they stretch equally. Determine the Stretch of each of the springs.

## The Attempt at a Solution

I use the equation mgh=(44)(9.8)(1.94) to find the potential energy. To find the stretch displacement, is it correct if I use Us=(1/2)kx^2 and substitute Us=mgh, k=5000 as given, and find x is the stretch of each springs?

It is if you multiply Us=(1/2)kx^2 by the number of springs and then equate to mgh.

Ok. So i did mgh=44*9.8*1.94=836.528J
Us=(1/2)*5000*x^2*23=836.528J. Find x from there to get stretch displacement. But the answer wasn't right. Did I do anything wrong in here?

As springs stretch(or compress), the person's potential energy also changes by mgx, where x is the stretch in the spring.

but they give h=1.94m. how is it a stretch?

The height is from spring when it is unstretched. When the person jumps on spring, the spring compresses and reduces in length, so the person's effective change in potential energy is mgh +mgx. (Hope I'm clear now)

oh ok i understand what you are explaining now. So to find x, we set it mgh+mgx=(1/2)kx^2*23 ?

can someone please help me with this questionn?? I am confused. :(

Didn't you get the correct answer using the formula in #7 ?

Well i got it right. But i just don't understand how is it mgh+mgx ??

Look, when the person is about to touch the bed (i.e springs), the change in PE is mgh; now as the springs compress, the person (along with spring) goes down (try to visualize the situation), and its PE further decreases.

Some other PF guy may give you a better explanation if my explanation is not clear enough.

There is nothing wrong with Sourabh N's explanation. When the springs are at their maximum compression, there is no kinetic energy. Only potential. So the change in the gravitational potential (over a distance of h+x) and potential in the springs (over a distance x) must be equal.

hehe Thank you. I think i got it. Just never think about that equation you know. Thanks a lots. Yes now i even understand more through DIck's explanation. mg(h+x)=(1/2)kx^2*23 hhehe...i totally got it now.

Last edited:

## 1. What is Hooke's Law?

Hooke's Law is a principle in physics that describes the relationship between the force applied to a spring and the resulting stretch or compression of the spring. It states that the force applied is directly proportional to the displacement of the spring.

## 2. How is Hooke's Law used to solve for the stretch of a spring?

To solve for the stretch of a spring using Hooke's Law, you need to know the spring constant (k) of the spring and the applied force (F). The formula for Hooke's Law is F = -kx, where x is the displacement of the spring. Rearranging the formula, we get x = -F/k, which allows us to calculate the stretch of the spring.

## 3. What is the spring constant and how is it determined?

The spring constant (k) is a measure of how stiff a spring is. It is determined by measuring the force applied to the spring and the resulting displacement, and then using the formula k = F/x. The unit for spring constant is Newtons per meter (N/m).

## 4. Can Hooke's Law be applied to all types of springs?

Yes, Hooke's Law can be applied to all types of springs, as long as they exhibit linear elasticity. This means that the force applied to the spring is directly proportional to the resulting stretch or compression, and the spring returns to its original length when the force is removed.

## 5. What are some real-life applications of Hooke's Law?

Hooke's Law has many practical applications, such as in the design of springs for mechanical devices, measuring the elasticity of materials, and studying the behavior of elastic materials under stress. It is also used in fields such as engineering, physics, and materials science to analyze and solve various problems.

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