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 16.0 -cm strip of the donated aorta reveal that it stretches 3.75 3.75 cm when a 1.50 1.50 -N pull is exerted on it. What is the force constant of this strip of aortal material?

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.

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Work By Springs

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3:11 minutes

Problem 33a

Textbook Question

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. What is the force constant of this strip of aortal material?

Verified step by step guidance

Step 1: Understand the problem. The force constant (k) is a measure of the stiffness of the material and is defined by Hooke's Law: $ F = k \Delta x $, where $ F $ is the applied force, $ \Delta x $ is the extension, and $ k $ is the force constant. We need to calculate $ k $ using the given values.

Step 2: Identify the given values. The applied force $ F $ is 1.50 N, the original length of the strip is 16.0 cm, and the extension $ \Delta x $ is 3.75 cm. Note that $ \Delta x $ is the change in length, not the total length.

Step 3: Convert units if necessary. Since the force constant $ k $ is typically expressed in $ \text{N/m} $, convert the extension $ \Delta x $ from centimeters to meters. Use the conversion factor: $ 1 \, \text{cm} = 0.01 \, \text{m} $. Thus, $ \Delta x = 3.75 \, \text{cm} \times 0.01 \, \text{m/cm} = 0.0375 \, \text{m} $.

Step 4: Rearrange Hooke's Law to solve for $ k $. The formula becomes $ k = \frac{F}{\Delta x} $. Substitute the values of $ F $ and $ \Delta x $ into the equation.

Step 5: Perform the calculation. Divide the force $ F $ (1.50 N) by the extension $ \Delta x $ (0.0375 m) to find the force constant $ k $. The result will be in $ \text{N/m} $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hooke's Law

Hooke's Law states that the force exerted by a spring (or elastic material) is directly proportional to the amount it is stretched or compressed, as long as the elastic limit is not exceeded. Mathematically, it is expressed as F = kx, where F is the force applied, k is the spring constant (or force constant), and x is the displacement from the equilibrium position. This principle is fundamental in understanding how materials deform under stress.

Elasticity

Elasticity is the property of a material that enables it to return to its original shape after being deformed by an external force. It is quantified by the material's elastic modulus, which measures the stiffness of the material. In the context of the aorta, understanding its elasticity is crucial for determining how it will behave under physiological conditions and how it can withstand the forces exerted by blood flow.

Force Constant

The force constant, often denoted as k, is a measure of the stiffness of an elastic material. It is defined as the ratio of the force applied to the displacement produced, as described by Hooke's Law. A higher force constant indicates a stiffer material that requires more force to achieve the same amount of stretch, which is particularly important in medical applications where material properties can affect the success of surgical repairs.

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Textbook Question

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. 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?

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(II) A spring has k = 65 N/m . Draw a graph like that in Fig. 7–11 and use it to determine the work needed to stretch the spring from x = 3.0cm to x = 7.5cm, where x = 0 refers to the spring’s unstretched length.

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(II) If it requires 5.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be required to stretch it an additional 4.0 cm?

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