Section 6.5 Reading Assignment: Work and Fluid Forces

Answer Only Exercise 1, 2, and 3 by using a screenshot provided Calculus Pearson textbook. Make sure you read these three questions very carefully and see on what it is asking for.

References: Thomas' Calculus: Early Transcendentals | Calculus | Calculus | Mathematics | Store | Pearson+

Be careful with Exercise #2 because I got the answer wrong for this second exercise and try to reread closely to that exercise to understand on what this problem is about.

Feedback: #2: The 0.08 interpretation is not correct here. It's best to reconsider this one closely. The 20 - x is technically true, but it might be better to consider this in terms of the rope instead. Especially in connection with the 0.08 factor. Wrong Answer: In Example 4 (p. 401), the integral factors represent the following in the physical situation:

0.08: This value represents the force due to the bucket, which is constant throughout the lifting process.

(20 - x): This expression represents the displacement of the bucket from its starting position. It changes as the bucket is lifted, and its value decreases as the bucket moves upward.

9.8: This value represents the acceleration due to gravity, which acts against the force exerted by the person lifting the bucket. It remains constant during the lifting process.

The integral is used to calculate the total work done in lifting the bucket by summing up the infinitesimal work done over each small displacement. The integral accounts for the varying force and displacement as the bucket is lifted.

Section 6.5 Reading Assignment: Work and Fluid Forces Instructions. Read through this assignment and complete the three exercises below by reading the appropriate passages of the textbook. The notion of work from physics gives a good example of how Riemann sums are typically used in physics and engineering. You may have learned that work is equal to force times displacement, but this formula assumes that the force is constant. How can we compute work when the force is changing? Note the definition boxed on p. 400. The textbook gives a more technical explanation, but the Riemann sum argument here is that for a small enough displacement, the force can be treated as constant. Over each small displacement, the work is F(c.) Ax. Adding these up give the Riemann sum and taking the limit as the displacement goes to zero gives the integral formula. This idea that over a small enough displacement or period of time, a quantity can be treated as though it were constant is a typical argument used when formulas involve integrals in physics. Exercise 1. Read the subsection "Hooke's Law for Springs: F=kx" (p. 400-401). Explain how the integral formula for work applies to springs. If you've seen Hooke's law before, you probably learned that the key issue is with the spring constant k. The same is true here, except with the addition of the integral formula. Exercise 2. Read Example 4 (p. 401). Describe what each factor in the integral, that is 0.08, (20- x), and 9.8 represent in the physical situation. Describe the meanings of all the terms separately. Note that the entire reason we have to consider an integral here is because of the rope. The force due to the bucket is constant. Since the example skips over some important steps, it's worth slowing down and thinking through how this problem is set up. We conclude our look into physical applications with the work used to pump liquids up. The justification for the formula for finding the work for pumping liquids is actually quite different from the formula we've considered thus far, so it's best to consider it from scratch. Exercise 3. Read Example 5 (p. 401-402). Summarize the Riemann sum argument in this example and how it gives the total work to pump the oil in the container. Do this by describing the physical meaning of AW and each factor of the expression for AW. Notice that the Ay, which becomes dy in the limit, in this example, is part of the units in the end result. We would not get ft-lb if not for multiplying that Ay. This is also true for the integral formula for work. Integrating Newtons (N) gives us Joules (J) because of the differential dx, which is in meters. So, force (Newton) times dx (meters) gives work (Joules). Thinking of that differential as a literal multiplication helps to track the units and makes it clearer how the physical units are affected by calculus operations. We will not be covering any material from the subsection "Fluid Pressures and Forces" due to time. That material will not appear for any assignments in this class. Another issue with this formula is that the force and the displacement must be in the same direction. We can deal with both these issues at once in Calculus III.