Question

Experiments using "optical tweezers" measure the elasticity of individual DNA molecules. For small enough changes in length, the elasticity has the same form as that of a spring. A DNA molecule is anchored at one end, then a force of 1.5 nN (1.5 x 10-9 N) pulls on the other end, causing the molecule to stretch by 5.0 nm (5.0 x 10-9 m). What is the spring constant of that DNA molecule?

Answer 1

Answer:

The spring constant of DNA molecule is $0.3\frac{N}{m}$

Explanation:

As we know ,

Formula for spring constant

$F = -kX$

Where,

$F =$ Force exerted on the DNA molecule in Newton

$k =$ spring constant in $\frac{Newton}{Meter}$

$X =$ displacement in meters

Substituting the given values in above equation, we get -

$1.5 * 10^{-9} = -k * 5.0* 10^{-9}\\k = 0.3$$\frac{N}{m}$

Thus, the spring constant of DNA molecule is $0.3\frac{N}{m}$