Elastic potential energy of a spring?

Question

A spring used in an introductory physics laboratory stores 14 J of elastic potential energy when it is compressed 0.20 m. Suppose the spring is cut in half. When one of the halves is compressed by 0.20 m, how much potential energy is stored in it?

Answer 1

(elastic force exerted by spring) = F
= kx = (elastic constant of spring)(deformation)
= dU/dx = (derivative of elastic potential energy)

Therefore,
U = integral of (kx dx)
U = 1/2*kx²

14 = 1/2*k(0.20)²
k = 14*2/(0.20)² = 700 N/m
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The original spring can be seen as two halves of a spring coupled in series,

-/\/\/\/\/\/\- (long spring, K) = -/\/\/\ (one half, k) + /\/\/\- (another half, k)

1/K = 1/k+ 1/k = 2/k
K = k/2
k = 2K

The K for the original spring is 700N/m, so the k for one half is 2x 700 = 1400N/m.
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The potential energy stored in one half of the original spring is then

U = 1/2*kx² = 1/2 (1400)(0.20)² = 28 J.

The energy doubles because the half spring will undergo more deformation strain than the original spring, and elastic potential energy has to do directly with the deformation of the spring. For instance, a 1 m-long spring undergoes less deformation strain when compressed 0.20 m than a 0.50 m-long spring when compressed by the same amount.

Answer 2

The answer is 28 J. The elastic potential energy of a spring is 1/2*k*x^2 where k is the spring constant and x is the compression of the spring. The spring constant is equal to the material stiffness (which is constant) divided by the length of the spring, so cutting the spring in half doubles the spring constant. Thus the elastic potential energy is doubled when compressed by an equal amount.

Answer 3

When you cut a spring in half, each part has constant K'=2K, where K is the elasticity constant of the original spring

[the formulas for connecting strings in series (in parallel), are the same as the ones we use in order to calculate the total a) capacitance of connected capacitors in series (in parallel), or b)resistance of connected resistors in parallel (in series) respectively.]

In your case K=700N/m so K'=1400N/m.

Therefore, calculating the energy for the half spring we get (1/2)*1400*(0.2)^2 J=28J.

If you expected to get an answer of 7 J (thinking that since the whole spring has an elastic energy of 14 J, then each half ought to have 7 J), please remeber that if we compress the original spring by 0.20m, it means that each half is compressed by 0.10m. So in this case we would get for each part (1/2)*1400*(0.1)^2 J=7 J. So there's no inconsistency with logic (and, what's more, conservation of energy) there!

Answer 4

as the material of the spring is the same as before, the spring constant, which is a material specific constant does not change. the elastic potential energy is a function of the spring constant and the extension (or compression in this case). so for the same compression i.e. 0.20 m, the elastic potential energy remains the same. i.e. 14 J as u said.
Hats off to you Bhatia sir!

Answer 5

Potential Energy,

Pe=(kx^2)/2
=>k=2Pe/x^2
=>k=2*14/(0.2)^2 =700 N/m

case-2: Pe=(kx^2)/2={700*(0.2)^2}/2=14 J

in fact k varies on the element of spring mainly, if its unchanged, the calculation wont differ.

Answer 6

1/2k(0.2)^2=14
k=2800/4
k=700
after cutting k=1400
tehrefore
e=0.5*1400*0.04
28J