A cylindrically shaped piece of collagen is being stretched by a force that increases from 0 to 3.0 x 10^-2 N. The length and radius of the collagen are 2.5 and 0.091 cm. Young's Modulus is 3.1 x 10^6 N/m^2. a.) If the stretching obeys Hooke's Law, what is the spring constant, k, for the collagen? b.) How much work is done by the variable force that stretches the collagen?
a.) To find the spring constant, k, we can use Hooke's Law equation:
F = k * x
Where F is the applied force, k is the spring constant, and x is the displacement.
Since the stretching obeys Hooke's Law, we can rewrite the equation as:
F = (Young's Modulus) * A * ΔL / L
Where A is the cross-sectional area of the collagen, ΔL is the change in length, and L is the original length.
Given:
Force, F = 3.0 x 10^-2 N
Length, L = 2.5 cm = 0.025 m
Radius, r = 0.091 cm = 0.00091 m
Young's Modulus, Y = 3.1 x 10^6 N/m^2
The cross-sectional area, A, of the collagen can be calculated using the formula:
A = π * r^2
The change in length, ΔL, can be calculated using the formula:
ΔL = F / [(Young's Modulus) * A / L]
Let's calculate the spring constant, k:
A = π * (0.00091 m)^2
ΔL = (3.0 x 10^-2 N) / [(3.1 x 10^6 N/m^2) * A / 0.025 m]
k = F / ΔL
First, calculate the cross-sectional area:
A = π * (0.00091 m)^2
A ≈ 2.617e-7 m²
Next, calculate the change in length:
ΔL = (3.0 x 10^-2 N) / [(3.1 x 10^6 N/m^2) * A / 0.025 m]
ΔL ≈ 0.000246 m
Now, calculate the spring constant:
k = F / ΔL
k ≈ (3.0 x 10^-2 N) / 0.000246 m
k ≈ 0.122 N/m
Therefore, the spring constant, k, for the collagen is approximately 0.122 N/m.
b.) To find the work done by the variable force that stretches the collagen, we can use the formula:
Work (W) = ∫ F * dx
Where W is the work, F is the force, and dx is the displacement. Since the force is changing, we need to integrate the force with respect to the displacement.
The force can be written as:
F = k * x
Since we already calculated the spring constant, k, in part a, we can use it in the equation.
Integrating the force from 0 to x gives us:
W = ∫ (k * x) dx
W = (1/2) * k * x^2
Now, we can calculate the work done by the variable force:
W = (1/2) * k * x^2
W = (1/2) * (0.122 N/m) * (0.025 m)^2
W ≈ 3.05 x 10^-4 N·m
Therefore, the work done by the variable force that stretches the collagen is approximately 3.05 x 10^-4 N·m.
To solve part (a) of the problem, we need to find the spring constant, k, for the collagen. We know that the force applied, F, is related to the displacement, x, by Hooke's Law, which states that F = -kx.
In this case, the force increases from 0 to 3.0 x 10^-2 N, and the displacement, x, corresponds to the change in length of the collagen. Since the force and displacement have opposite directions, we take F = -kx to represent the stretching process.
Given:
Force, F = 3.0 x 10^-2 N
Initial length, L0 = 2.5 cm = 0.025 m
Final length, L = L0 + x
Young's Modulus, Y = 3.1 x 10^6 N/m^2
Now, let's calculate the change in length, x:
x = L - L0 = (L0 + x) - L0 = L - L0
To relate stress (force per unit area) and strain (change in length divided by original length), we use the Young's modulus formula:
Y = stress / strain
From this formula, we can express the stress as the force divided by the cross-sectional area of the collagen, A:
stress = F / A
The cross-sectional area of a cylinder is given by:
A = πr^2
where r is the radius.
With these equations, we can find the spring constant k and then proceed to part (b) to calculate the work done by the variable force.
Let's start solving part (a) with finding the spring constant, k:
Step 1: Convert the radius from centimeters to meters:
radius, r = 0.091 cm = 0.091 x 10^-2 m
Step 2: Calculate the cross-sectional area:
A = πr^2
Step 3: Calculate the change in length, x:
x = L - L0 = 0.025 m - 0.025 m = 0
Step 4: Find the stress on the collagen:
stress = F / A
Step 5: Calculate the spring constant, k:
F = -kx (from Hooke's Law)
k = -F / x
Using these steps and the given values, we can now calculate the spring constant, k.