A surgeon is using material from a donated heart to repair a patient's damaged aorta and needs to know the elastic characteristics of this aortal material. Tests performed on a 16.0 cm strip of the donated aorta reveal that it stretches 3.16 cm when a 1.60 N pull is exerted on it.
What is the force constant of this strip of aortal material?
If the maximum distance it will be able to stretch when it replaces the aorta in the damaged heart is 1.42 cm , what is the greatest force it will be able to exert there?
Well, let's have some fun with this!
The force constant of the strip of aortal material can be calculated using Hooke's Law, which states that the force applied is directly proportional to the amount of stretching. So, we can use the formula:
Force constant = Force applied / Amount of stretching
Plugging in the given values, the force applied is 1.60 N and the amount of stretching is 3.16 cm, we can calculate the force constant:
Force constant = 1.60 N / 3.16 cm
But wait, we have a bit of a mismatch here. We need to convert the distance from centimeters to meters for this calculation. So, let's do that:
Force constant = 1.60 N / (3.16 cm / 100 cm/m)
Simplifying this:
Force constant = 1.60 N / (3.16 cm * (1/100 cm/m))
Force constant = 1.60 N / (3.16 / 100) N/m
Force constant ≈ 50.6 N/m
So, the force constant of this strip of aortal material is approximately 50.6 N/m. Now, let's move on to the next question.
If the maximum distance the strip will be able to stretch in the damaged heart is 1.42 cm, we can use Hooke's Law again to find the greatest force it will be able to exert:
Force = Force constant * Amount of stretching
Plugging in the values, the force constant is 50.6 N/m and the amount of stretching is 1.42 cm:
Force = 50.6 N/m * 1.42 cm
Hold on, we need to convert the distance to meters for this calculation. Let's do that:
Force = 50.6 N/m * (1.42 cm / 100 cm/m)
Simplifying this:
Force = 50.6 N/m * (1.42 / 100) N/m
Force ≈ 0.719 N
So, the greatest force the strip will be able to exert in the damaged heart is approximately 0.719 N.
Hope that gave you a good stretch of laughter! However, please note that these calculations are based on idealized situations and actual results may vary.
To determine the force constant of the strip of aortal material, we can use Hooke's Law:
Hooke's Law: F = k * x
Where:
F is the applied force
k is the force constant
x is the amount of stretch
In this case, we have:
Applied force (F) = 1.60 N
Amount of stretch (x) = 3.16 cm = 0.0316 m
Plugging in the values into Hooke's Law, we have:
1.60 N = k * 0.0316 m
Dividing both sides of the equation by 0.0316 m:
k = 1.60 N / 0.0316 m
k = 50.63 N/m
Therefore, the force constant of the strip of aortal material is 50.63 N/m.
To calculate the greatest force the strip of aortal material will be able to exert when stretched to a maximum distance of 1.42 cm = 0.0142 m, we can rearrange Hooke's Law:
F = k * x
Plugging in the values, we have:
F = 50.63 N/m * 0.0142 m
Calculating:
F = 0.719 N
Therefore, the greatest force the strip of aortal material will be able to exert in the damaged heart is 0.719 N.
To find the force constant of the strip of aortal material, we can use Hooke's Law, which states that the force applied to an elastic material is directly proportional to the amount it stretches. Mathematically, this can be expressed as:
F = k * x
Where:
F is the force applied to the material
k is the force constant (also known as the spring constant)
x is the displacement or stretch of the material
Given that the strip of aorta stretches 3.16 cm (0.0316 m) when a 1.60 N force is applied, we can rearrange the equation to solve for the force constant:
k = F / x
Substituting the values, we have:
k = 1.6 N / 0.0316 m
Calculating this, we find that the force constant (k) is approximately 50.63 N/m.
Now, let's move on to the second part of the question. We are asked to determine the greatest force the strip of aortal material will be able to exert when it replaces the damaged aorta in the heart, given that it can stretch a maximum distance of 1.42 cm (0.0142 m).
Using Hooke's Law again, we can rearrange the equation to solve for the force (F):
F = k * x
Substituting the values:
F = 50.63 N/m * 0.0142 m
Calculating this, we find that the greatest force the strip of aortal material will be able to exert is approximately 0.719 N.
The force constant of the aorta material is
K = 1.60/0.0316 = ____ N/m
That's about 50 N/m
If stretched 0.0142 m, it will exert a force equal to
0.0142*K. N