A molecule of DNA (deoxyribonucleic acid) is 2.33 µm long. The ends of the molecule become singly ionized: negative on one end, positive on the other. The helical molecule acts like a spring and compresses 1.01% upon becoming charged. Determine the effective spring constant of the molecule. I've seen a couple examples of this problem, including the one already solved on cramster. I tried that method but it didn't work. I got 8.114E-6, then 9.55 E-6, and webassign said that both answers were wrong. I need a fool proof method. One of my calculators said 9.544E-30 but I'm afraid to try it because I only get 10 chances to try an answer and I've tried about 5 times.

Question

Bennett Daugherty

Physics

26 April, 09:48

A molecule of DNA (deoxyribonucleic acid) is 2.33 µm long. The ends of the molecule become singly ionized: negative on one end, positive on the other. The helical molecule acts like a spring and compresses 1.01% upon becoming charged. Determine the effective spring constant of the molecule. I've seen a couple examples of this problem, including the one already solved on cramster. I tried that method but it didn't work. I got 8.114E-6, then 9.55 E-6, and webassign said that both answers were wrong. I need a fool proof method. One of my calculators said 9.544E-30 but I'm afraid to try it because I only get 10 chances to try an answer and I've tried about 5 times.

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Answers (1)

Know the Answer?

Isabelle Sexton · Answer 1

The effective spring constant is (4.25*10^7F) Newton per meter

Explanation:

From Hookes law of elasticity,

Force required to compress a spring (F) = force constant (k) * compression (c)

k = F/c

Length of the spring = 2.33 micrometers = 2.33*10^-6 meters

Compression (c) = 1.01% * 2.33*10^-6 = 0.0101 * 2.33*10^-6 = 2.3533*10^-8

k = F (Newton) / 2.3533*10^-8 meters = (4.25*10^7F) Newton per meter