Finding the Maximum Compression Distance of a Spring: A Physics Problem

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Homework Help Overview

The problem involves a 2kg block dropped from a height of 40 cm onto a spring with a force constant of 1960 N/m. The goal is to determine the maximum compression distance of the spring, which is stated to be 10 cm.

Discussion Character

  • Exploratory, Conceptual clarification, Energy considerations

Approaches and Questions Raised

  • Participants discuss using force and energy equations to find the maximum compression of the spring. There are attempts to apply the formula F = ky and the energy conservation principle mgh = 1/2 k x^2. Some participants express confusion over their calculations and the expected answer.

Discussion Status

The discussion includes various approaches to the problem, with some participants questioning their calculations and the assumptions made. There is acknowledgment that the energy method is typically effective, but discrepancies in expected results are noted. No consensus has been reached on the correct approach or interpretation of the problem.

Contextual Notes

Participants mention that the expected answer is 10 cm, leading to discussions about the height from which the block is dropped and its impact on the spring's compression. There is a suggestion that a different drop height might yield the expected result.

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Homework Statement


A 2kg block is dropped from a height of 40 cm onto a spring of force constant k=1960 N/m. Find the maximum distance the spring is compressed.

(The answer must be 10 cm)

The Attempt at a Solution



Well, I know that if a spring is stretched "y", it will be compressed "y". The problem is that I can't find out how long will the spring stretch once it is dropped from that height. Therefore I tried to use the formula

F = ky

Where F= mg = 2 \times 9.81 and k =1960 N/m which is 196000 N/cm

2 \times 9.81 = (196000) k

y= \frac{2 \times 9.81}{196000}

But this gives me the wrong answer. Why is that?? :confused:

By the way, I know that the tension is T= 2 \pi \sqrt{\frac{m}{k}} = 2 \pi \sqrt{\frac{2}{196000}}= 0.02 N.
 
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It is best to consider energy for this problem, graviational pe = potential pe of springAlso that T is period time and not the tension.
 
rock.freak667 said:
It is best to consider energy for this problem, graviational pe = potential pe of spring

I tried that, it doesn't seem to work:

mgh= \frac{1}{2} k x^2

2 (9.81) 40 = 1/2 (196000) x2

x=0.089 cm

Because the correct answer should be 10 cm! Is there anything wrong with my calculations?

Also that T is period time and not the tension.

Yep, my mistake. :blushing:
 
Not sure, the only way to get 10cm exact is if the mass was dropped from 50cm.

Normally, the energy method should work.
 

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