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 3.75 cm when a 1.50-N pull is exerted on it. (a) What is the force constant of this strip of aortal material?

Video transcript

Hey everybody. So today we're being told that a specialist has purchased repair material to fix a torn canvas of theirs. However, they are unable to trace the documentation of the material how much it's moved. So using a technique to test the elastic properties of the material, they cut a 25 cm piece of said material and apply a force of 2.4 newtons on it. Through this, they observed that the material stretches by a distance of 2. centimeters. So with this we're being asked to calculate the force constant K. Of the material. Now, at the mention of stretching and force constants, we can actually use or deduce that we need to use hooks law, which states that if magnitude of force is equal to a specific constant, a force constant which is specific to a material multiplied by X. The change in distance or the change in distance via stretching more compression for that matter. So, with the materials cut out, we know a few things, we know that the force applied. The magnitude of the force at least is 2.40 newtons. We also know that the change in X is 2.20 cm. Which is why we don't really, while this is important to know if we didn't have the change in distance, we're already given the actual change in distance. So with that in mind, let's rearrange our formula to solve for the force constant what we're looking for. So we get that K. And I'll write this in blue is equal to the magnitude of the force divided by the change in distance due to stretching or compression. So substituting in our values, we get 2.4 newtons Divided by 2.20 cm. Now our answer choices are all in Newton's times meters or newton's divided newtons per meter. So we can use the conversion factor by recalling that. We have one m for every 10 to the second or 100 cm. So centimeters will cancel. That will be left with uh 0.022 meters. And finally our answer will be 109, 109 loops. 109 newtons per meter for answer choice. C. Therefore the force constant of the material is c. Newtons Per meter. I hope this helps. And I look forward to seeing you all in the next one.