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The length of the spring

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Turns of a uniform spring of relaxed length l = 1.00 m and force constant
k = 500 N/m almost touch each other. A light glue is applied evenly
between every adjacent turn. Breaking strength of the glue is $F_b = 100
N$. The spring is placed on a frictionless horizontal floor and pulled from
one of its ends. If the pulling force is gradually increased to a value F =
200 N. how much will the length of the spring become?

asked Oct 15 in Physics Problems by koolman (4,116 points)

1 Answer

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Best answer

It is not obvious how to deal with the glue. I shall discuss that later.

The main mistake you made is to assume that equal and opposite forces $F$ are applied at both ends of the spring. In the question there appears to be only one force $F$ applied to the RHS of the spring. Therefore the spring will accelerate. The tension in it will not be uniform, and the extension will not be the same as when there are equal forces at both ends.

How you have dealt with the glue is reasonable. You have assumed that the coils of the spring do not expand at all until the glue has broken. However, because the tension is not constant along the spring (as explained above), only those parts of the spring in which the tension is greater than $100N$ will extend.

Effectively the glue and spring are in parallel because the extension must be the same before the glue (or spring) breaks. The tension in the glue and spring need not be the same : the sum of tensions equals the applied force. If the glue and spring were in series then the tension in both would be the same but the extension could be different. Moreover, when the glue breaks then the spring would break also - the coils would come apart.

You have assumed that the glue has infinite stiffness (aka spring constant), so that it does not extend before it breaks. The question does not tell us what its stiffness is. It need not be infinite, but this is probably the only assumption we can make. If the stiffness is not infinite then we will get a different answer.


Comment : You might wonder why the question specifies that the force $F$ is gradually increased. This is because a force applied suddenly at one end causes the spring to oscillate as it accelerates. The maximum extension is then twice what it would be if a constant force is applied. See Total energy of the block and spring.


Solution :

The tension in the accelerated spring varies linearly by mass from $F=200N$ at the RHS to $0N$ at the LHS. The tension in the middle is $100N$. This is where the glue comes unstuck. So the left half remains at its initial length of $0.5m$; only the right half extends.

The right half is now a spring with half its original length so the spring constant is $2k=1000N/m$. The tension at the left end of this spring is $100N$ and that at the right end is $200N$ so the average tension is $150N$. Its extension is $\frac{150N}{1000N/m}=0.15m$.

The new length of the whole spring is $1.15m$.

answered Oct 15 by sammy gerbil (26,096 points)
selected Nov 4 by koolman
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