Why use 1/3 inches instead of 4 inches in this spring problem?

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Homework Help Overview

The discussion revolves around a spring problem involving a mass weighing 24 pounds that stretches a spring 4 inches. The original poster questions why the equation uses 1/3 inches instead of 4 inches, particularly in the context of the spring constant and the units involved.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of unit conversions between inches and feet, questioning the units for the spring constant. There is also a discussion about the initial conditions provided in the answer key and their relevance to the problem statement.

Discussion Status

Some participants have offered insights regarding the units of measurement and their importance in physics problems. There is an ongoing exploration of how to interpret the problem statement and the answer key, with no explicit consensus reached on the correct approach yet.

Contextual Notes

Participants note the potential confusion arising from unspecified units in the problem statement and the importance of clarifying these units in the context of the course. The original poster expresses uncertainty about the units used by their teacher, which adds to the complexity of the discussion.

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Homework Statement


a mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 3 inches above the equilibrium position

Homework Equations


F = ks

The Attempt at a Solution


24 = k*(1/3)

why is it 1/3 inches and not 4 inches?
 
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It isn't "1/3 inches"- it is 1/3 feet. What are the units for k? Since k times a distance is equal to a force, k must have units of "force divided by distance". If k has units of "pounds per inch" then "4 inches" would be correct. But if k has units of "pounds per foot" then 4/12= 1/3 feet is correct..
 
HallsofIvy said:
It isn't "1/3 inches"- it is 1/3 feet. What are the units for k? Since k times a distance is equal to a force, k must have units of "force divided by distance". If k has units of "pounds per inch" then "4 inches" would be correct. But if k has units of "pounds per foot" then 4/12= 1/3 feet is correct..

Ahh, I see. I'm not sure what the units are, these are just some questions assigned by my differential equations teacher. I'll have to ask and verify, but what you said makes sense.
 
Also, the answer key has initial conditions. x(0) = -1/4 , x'(0) = 0.

Where does that come from?
 
24 = k*(1/3)
why is it 1/3 inches and not 4 inches?
... yah: I was going to say: why did you write down 1/3 if you don't know what it means?
If you want to use 4", why not use 4"? It's allowed.
If it's part of an answer key, then it's someone elses work: why are you writing down someone elses work? Do it the way you'd do it, and then compare results.

I see this is a math course - in physics, as you just found out, numbers have units to give them physical meaning. You'll have seen this in other examples by now - but it does take a bit of practice to get used to.

Also, the answer key has initial conditions. x(0) = -1/4 , x'(0) = 0.
Comes from the problem statement ... translate those math expressions into English sentences and see. x(0) is x(t) at t=0, and x'(0) is the 1st derivative of x(t) at t=0.
 
Simon Bridge said:
... yah: I was going to say: why did you write down 1/3 if you don't know what it means?
If you want to use 4", why not use 4"? It's allowed.
If it's part of an answer key, then it's someone elses work: why are you writing down someone elses work? Do it the way you'd do it, and then compare results.

I see this is a math course - in physics, as you just found out, numbers have units to give them physical meaning. You'll have seen this in other examples by now - but it does take a bit of practice to get used to.

Comes from the problem statement ... translate those math expressions into English sentences and see. x(0) is x(t) at t=0, and x'(0) is the 1st derivative of x(t) at t=0.
Yes, I just realized that 1/4 foot is 3 inches. These units are confusing me. I actually did have my work right but the answer key made me think otherwise.

I've taken physics courses, in fact completed all my requirements for physics. My differential equation teacher likes to use random units without telling me what she is using. I get very confused.
 
The initial problem statement did not include a question - what was the wording of the task?
If you are not told the units, you can use what you like as long you specify at the start.
If you are self-marking it can be tricky, but your teacher appears to have been consistent in the question to use the f-p-s standard units.
http://en.wikipedia.org/wiki/Foot–pound–second_system
Is it possible that your teacher expects you to use this standard when units are not specified - just like others prefer SI (m-k-s) units by default?
 
Simon Bridge said:
The initial problem statement did not include a question - what was the wording of the task?
If you are not told the units, you can use what you like as long you specify at the start.
If you are self-marking it can be tricky, but your teacher appears to have been consistent in the question to use the f-p-s standard units.
http://en.wikipedia.org/wiki/Foot–pound–second_system
Is it possible that your teacher expects you to use this standard when units are not specified - just like others prefer SI (m-k-s) units by default?

I am going to verify, but it appears to be true.
 

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