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.