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. (b) If the maximum distance it will be able to stretch when it replaces the aorta in the damaged heart is 1.14 cm, what is the greatest force it will be able to exert there?

Video transcript

Hey everyone. So today we're dealing with the problem, We'll need to use Hook's law. So we're being told that a technician wishes to perform a repair by riveting an elastic material onto a gap in an object. She cuts 18 cm off the material of the material and applies a force of newtons on it. When applying this force, she observes that the material increases in length by four cm. So the material is therefore applied onto a patch with an elongation of 2.80 cm. With this information determined that the maximum, or we're being asked to determine the maximum force applied onto the object. So, as I mentioned earlier, since we're mentioning a force and stretching, we should automatically come to the conclusion that we're going to need Hook's law in some way, shape or form due to the fact that those are the only two real values that are given were not given any masses or any other extra forces or anything like that, aside from the force being applied in the first place. So Hook's law states that the force or the absolute value of the force, the magnitude of the force will be equal to a constant. In this case, a force constant which is specific to the elastic material in question, multiplied by the distance through which it is compressed. So we are being told that the elongation that we're looking for the change in X is 2.80. And we're being asked to find f the force from that, however, to do so, we need to first determine K. But K. Is also unknown. And that's why we have the earlier information. The 10 newtons and the Change in length by four cm. So let's write that out in blue. We are forced physical two K. X. And we know that our forces 10 newtons, newtons, This K&X is four cm. So therefore K. is equal to 10 newtons per four cm or 2.5 newtons per centimeter. So, substituting that back into our first equation, We get that the uh force applied on the object with an elongation of 2.8 cm Will be equal to the force constant 2. newtons per centimeter. Multiplied by the elongation Of 2.8 cm. Giving us a final answer. Because we are sending, you will cancel out of 7.00 newtons newton's. Therefore the maximum force the spring applies to the object Through an elongation of 2.8 cm is option B seven Newtons. I hope this helps. And I look forward to seeing you all in the next one