# Why is my solution to finding the spring constant incorrect?

• LCSphysicist
In summary: The solution you found on the net interprets the "eventually comes to rest" as meaning after oscillations have died down (due to damping). This makes the initial fall of the mass irrelevant; it could have been placed on the pan gently.Of course this will give a different answer to the one you found.
LCSphysicist
Homework Statement
A mass of 10 kg falls 50 cm onto the platform of a spring scale,
and sticks. The platform eventually comes to rest 10 cm below its initial
position. The mass of the platform is 2 kg. Find the spring constant.
Relevant Equations
All below

This is my scope of the question, i could think to solve it by two steps, but before, let's give name to the things.

X is positive down direction.
X = 0 at the initial position o the platform
Mass of the falling block is m1
Mass of the platform, m2
Spring constant k
Δx is the initial stretched length of the spring
h1 is the height of the block
h2 is the final distance (0,1m) of the origin
vb is the block speed
vc is the both bodies speed

vb² = 2gh1
m1*vb = (m1+m2)*vc
(m1+m2)*vc²/2 + kΔx²/2 = -(m1+m2)*g*|h2| + k(Δx + h2)²/2
Δx = m2*g/k

That is a system possible and determined, but why this is wrong?
I see we could apply kx before and in the final scenario, but why the first attempt is wrong?

The question is not entirely clear. It says "eventually " comes to rest. That suggests it is at the new equilibrium position, so the initial drop is irrelevant.
You have interpreted it as "first" comes to rest. Under that view, your work looks correct to me. How do you know it is considered wrong? Do you get a numerical answer different from a given one? If so, what are the two numbers?

Merlin3189
haruspex said:
The question is not entirely clear. It says "eventually " comes to rest. That suggests it is at the new equilibrium position, so the initial drop is irrelevant.
You have interpreted it as "first" comes to rest. Under that view, your work looks correct to me. How do you know it is considered wrong? Do you get a numerical answer different from a given one? If so, what are the two numbers?
HI :D I have no a numerical answer, but there is this solution on the internet

I think this is correct too, since the question says that in both cases the spring is in rest. If this is right, i agree that my answer would need too, but solving leads to differing values.

LCSphysicist said:
i agree that my answer would need too, but solving leads to differing values.

The solution you found on the net interprets the "eventually comes to rest" as meaning after oscillations have died down (due to damping). This makes the initial fall of the mass irrelevant; it could have been placed on the pan gently.
Of course this will give a different answer to the one you found.

Are you sure the answer on the net is the intended interpretation? Was it provided by the person who set the question?

## 1. What is the spring constant k?

The spring constant, denoted by k, is a measure of the stiffness of a spring. It represents the force required to stretch or compress a spring by a certain distance.

## 2. How do you calculate the spring constant k?

The spring constant k can be calculated by dividing the force applied to the spring by the displacement of the spring from its equilibrium position.

## 3. What units is the spring constant k measured in?

The spring constant k is measured in units of force per unit length, such as newtons per meter (N/m) in the SI system or pounds per inch (lb/in) in the US customary system.

## 4. How does the spring constant k affect the behavior of a spring?

The spring constant k determines the stiffness of a spring and how much it will stretch or compress when a force is applied. A higher spring constant means the spring is stiffer and requires more force to stretch or compress, while a lower spring constant means the spring is more flexible and requires less force.

## 5. Can the spring constant k change?

Yes, the spring constant k can change depending on factors such as the material and shape of the spring, as well as the temperature and other environmental conditions. It can also change if the spring is stretched or compressed beyond its elastic limit.

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