A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 to 0.0436 N. The length and radius of the collagen are, respectively, 2.90 cm and 0.0815 cm, and Young's modulus is 3.1 x 106 N/m2. (a) If the stretching obeys Hooke's law, what is the spring constant k for collagen? (b) How much work is done by the variable force that stretches the collagen?
To answer part (a) of the question, we need to use Hooke's law, which states that the force needed to stretch or compress an elastic material is directly proportional to the change in length. The formula for Hooke's law is as follows:
F = k * ΔL
Where:
F is the force applied
k is the spring constant
ΔL is the change in length
In this case, we are given the final force, F = 0.0436 N, and the initial length, L = 2.90 cm. We need to calculate the spring constant, k.
First, we need to convert the length from centimeters to meters, as the units should be consistent in the equation.
L = 2.90 cm = 0.0290 m
Now we can rearrange Hooke's law equation to solve for k:
k = F / ΔL
Substituting the values we have:
k = 0.0436 N / 0.0290 m
k = 1.504 N/m
So, the spring constant for collagen is 1.504 N/m.
To answer part (b) of the question, we need to calculate the work done by the variable force to stretch the collagen. The formula for work is given by:
Work = Force * Distance
In this case, the force is not constant, so we need to calculate the work done for each infinitesimal displacement and integrate it over the range of displacement.
Let's assume the collagen is stretched from its initial length, L0, to a final length, L.
The work done can be calculated as follows:
Work = ∫ F(x) * dx from L0 to L
Since the force varies linearly as per Hooke's law, we can express it as:
F(x) = k * ΔL(x)
Where:
x is the displacement from the initial length L0
ΔL(x) is the change in length at displacement x
Now, we need to integrate the force equation over the displacement.
Work = ∫ k * ΔL(x) * dx from L0 to L
Since the force is directly proportional to the displacement, we can simplify the equation as follows:
Work = k * ∫ ΔL(x) * dx from L0 to L
Integrating ΔL(x) gives us the change in length, which is just the final length minus the initial length:
ΔL(x) = L - L0
Substituting this value back into equation:
Work = k * ∫ (L - L0) * dx from L0 to L
Integrating, we get:
Work = k * (L - L0) * ∫ dx from L0 to L
Integrating dx from L0 to L is simply the displacement itself, L - L0:
Work = k * (L - L0) * (L - L0)
Substituting the given values:
Work = 1.504 N/m * (L - 0.0290 m) * (L - 0.0290 m)
Now, you can calculate the work done by substituting the appropriate value for L.
To find the spring constant (k) for collagen and the work done by the variable force, we can use the formula for Hooke's law and the equation for work.
(a) To find the spring constant (k) for collagen, we can use the equation for Young's modulus (Y) and the formula for the spring constant:
k = Y × (π × r^2) / L
Where:
Y is Young's modulus
r is the radius of the collagen
L is the length of the collagen
Given values:
Y = 3.1 × 10^6 N/m^2
r = 0.0815 cm = 0.0815 × 10^(-2) m
L = 2.90 cm = 2.90 × 10^(-2) m
Substituting the values into the equation:
k = (3.1 × 10^6 N/m^2) × (π × (0.0815 × 10^(-2) m)^2) / (2.90 × 10^(-2) m)
Calculating this expression will give us the value of k.
(b) To calculate the work done by the variable force, we can use the equation:
Work = (1/2) × k × Δx^2
Where:
k is the spring constant
Δx is the change in length
In this case, the force increases from 0 to 0.0436 N. We can use Hooke's law to find Δx:
F = k × Δx
Given force F = 0.0436 N and k from part (a), we can solve for Δx:
Δx = F / k
Substituting the values into the equation for work:
Work = (1/2) × k × (F / k)^2
Calculate this expression to find the work done.