Spring Constant given mass, and equivalent mass in free fall from fixed height.

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Homework Help Overview

The problem involves two blocks of mass 54 kg, one resting on a spring and the other in free fall, both released from a height of 30 cm. The objective is to determine the spring constant based on the timing of their motions as one block hits the table while the other reaches an instantaneous rest position.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating the time for the free-falling block to reach the floor and equating it with the time for the spring-mounted block to reach its first instantaneous rest. There is mention of using the period of oscillation to find the spring constant.

Discussion Status

Some participants have provided guidance on the relationship between the time of free fall and the oscillation period of the spring-mounted block. There is an ongoing exploration of the definitions of instantaneous rest and the balance point in the oscillation, indicating a productive discussion without explicit consensus.

Contextual Notes

Participants are navigating potential misunderstandings regarding the definitions of motion terms and the setup of the problem, which may affect their calculations and interpretations.

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Homework Statement


Two 54kg blocks are held 30cm above a table. As shown in the figure, one of them is just touching a 30cm long spring. The blocks are released at the same time. The block on the left hits the table at exactly the same instant as the block on the right first comes to an instantaneous rest.

What is the spring constant?

Homework Equations


Delta L = mg/k


The Attempt at a Solution


I think the solution might start with finding the time it takes for the free fall block to hit the surface, which is the same as the block mounted on the spring. Then we can use to some how find the compression length. But that's as far as I have got.
 
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Yes, right. First of all, you can calculate the time that the block without a spring in its belly needs in order to reach the floor and equate this with the time that the other block needs in order to first come to instantaneous rest. Of course the second block's motion is going to be a part of a simple harmonic oscillation, so the time until first instantaneous rest is a quarter of the oscillation's period, meaning t = T/4. Use that equation to calculate the spring constant.
 
Awesome, so this is what I did from your suggestion:

d=vi * t + 0.5 * a * t^2
0.3m = 0 + 0.5 * 9.8 * t^2
t = 0.2473 seconds

0.2573 * 4 = 0.9897 = period

frequency = 1 / period
frequency = 1.01 hz

frequency = (1/2pie) * sqrt(k/mass)
1.01 = (1/2pie) * sqrt(k/0.054kg)
k = 2.176 N/m

But however, my professor told me this was wrong... please let me know where I made mistakes.
 
Oh yea, that's true, my bad. You see, the point of balance for the oscillation of the second block isn't the point that it initially is, where it touches the spring, but a little below that, so that the force from the spring is opposite to the block's weight. The second block first reaches that point and T/4 later it reaches it's first instantaneous rest point.There's a chance we're talking about different things here. When saying instantaneous rest point you mean i) the point where ΣF=0 for the second block or ii) the point where v=0 (velocity) for the second block?? Because I'm not perfectly familiar with the terms in english.
 
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