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 1.5nN (1.5×10−9N) pulls on the other end, causing the molecule to stretch by 5.0nm (5.0×10−9m). What is the spring constant of that DNA molecule?

Answer 1

Answer:

Spring constant, k = 0.3 N/m

Explanation:

It is given that,

Force acting on DNA molecule, $F=1.5\ nN=1.5\times 10^{-9}\ N$

The molecule got stretched by 5 nm, $x=5\times 10^{-9}\ m$

Let k is the spring constant of that DNA molecule. It can be calculated using the Hooke's law. It says that the force acting on the spring is directly proportional to the distance as :

$F=-kx$

$k=\dfrac{F}{x}$

$k=\dfrac{1.5\times 10^{-9}\ N}{5\times 10^{-9}\ m}$

k = 0.3 N/m

So, the spring constant of the DNA molecule is 0.3 N/m. Hence, this is the required solution.

Answer 2

The larger the spring constant, the stiffer the spring, and the more difficult it is to stretch.

The value of the spring constant for DNA molecules is 0.3 N/m.

What is the spring constant?

The spring constant is defined as a measure of the stiffness of the spring.

Given that the force F on the DNA constant is 1.5nN (1.5×10−9N). The stretched length l of the molecule is 5.0nm (5.0×10−9m).

Let's consider that k is the spring constant of that DNA molecule. It can be calculated using Hooke's law.

Hooke's law states that the force needed to extend or compress a spring by some distance scales linearly with respect to that distance.

$F = kl$

Substituting the values, we get the value of the spring constant.

$1.5\times 10^{−9} = k ( 5\times 10^{-9})$

$k = 0.3 \;\rm N/m$

Hence we can conclude that the value of the spring constant for DNA molecule is 0.3 N/m.

To know more about the spring constant, follow the link given below.

https://brainly.com/question/4291098.