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?
To find the spring constant (k) for the collagen, we can use Hooke's Law.
Hooke's Law states that the force (F) required to stretch or compress an object is directly proportional to the displacement (x) of the object from its equilibrium position and is given by the equation:
F = kx
where k is the spring constant.
In this case, the force increases from 0 to 3.0 x 10^-2 N, and we need to find the spring constant.
a.) To find the spring constant (k), we need to find the displacement (x) of the collagen. The displacement can be calculated using the formula:
x = ΔL = L - L₀
where ΔL is the change in length, L is the final length, and L₀ is the initial length.
Given:
Final length (L) = 2.5 cm
Initial length (L₀) = 2.5 cm (since the force begins at 0)
Change in length (ΔL) = L - L₀ = 2.5 cm - 2.5 cm = 0 cm
Since the change in length is 0, the displacement (x) is also 0.
Now, we can use Hooke's Law to find the spring constant (k):
F = kx
3.0 x 10^-2 N = k(0)
k = 3.0 x 10^-2 N / 0
k = 0 N/m
Therefore, the spring constant (k) for the collagen is 0 N/m.
b.) To find the work done by the variable force that stretches the collagen, we can use the equation for work (W):
W = (1/2)kx²
In this case, since the displacement (x) is 0 (as calculated in part a), the work done will also be 0.
Therefore, the work done by the variable force that stretches the collagen is 0 J.