Chapter 6: Problem 4
Why is the disk method a special case of the general slicing method?
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Chapter 6: Problem 4
Why is the disk method a special case of the general slicing method?
These are the key concepts you need to understand to accurately answer the question.
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Leaky cement bucket A 350 kg-bucket containing \(4650 \mathrm{kg}\) of cement is resting on the ground when a crane begins lifting it at a constant rate of \(5 \mathrm{m} / \mathrm{min.}\) As the crane raises the bucket, cement leaks out of the bucket at a constant rate of \(100 \mathrm{kg} / \mathrm{min.}\) How much work is required to lift the bucket a distance of \(30 \mathrm{m}\) if we ignore the weight of the crane cable attached to the bucket?
Calculating work for different springs Calculate the work required to stretch the following springs \(1.25 \mathrm{m}\) from their equilibrium positions. Assume Hooke's law is obeyed. a. A spring that requires \(100 \mathrm{J}\) of work to be stretched \(0.5 \mathrm{m}\) from its equilibrium position b. A spring that requires a force of \(250 \mathrm{N}\) to be stretched \(0.5 \mathrm{m}\) from its equilibrium position
Shock absorber A heavy-duty shock absorber is compressed \(2 \mathrm{cm}\) from its equilibrium position by a mass of \(500 \mathrm{kg} .\) How much work is required to compress the shock absorber \(4 \mathrm{cm}\) from its equilibrium position? (A mass of 500 kg exerts a force (in newtons) of \(\left.500 g, \text { where } g=9.8 \mathrm{m} / \mathrm{s}^{2} .\right)\)
A nonlinear spring Hooke's law is applicable to idealized (linear) springs that are not stretched or compressed too far from their equilibrium positions. Consider a nonlinear spring whose restoring force is given by \(F(x)=16 x-0.1 x^{3},\) for \(|x| \leq 7\) a. Graph the restoring force and interpret it. b. How much work is done in stretching the spring from its equilibrium position \((x=0)\) to \(x=1.5 ?\) c. How much work is done in compressing the spring from its equilibrium position \((x=0)\) to \(x=-2 ?\)
Where do they meet? Kelly started at noon \((t=0)\) riding a bike from Niwot to Berthoud, a distance of \(20 \mathrm{km}\), with velocity \(v(t)=\frac{15}{(t+1)^{2}}\) (decreasing because of fatigue). Sandy started at noon \((t=0)\) riding a bike in the opposite direction from Berthoud to Niwot with velocity \(u(t)=\frac{20}{(t+1)^{2}}\) (also decreasing because of fatigue). Assume distance is measured in kilometers and time is measured in hours. Make a graph of Kelly's distance from Niwot as a function of time. b. Make a graph of Sandy's distance from Berthoud as a function of time. c. When do they meet? How far has each person traveled when they meet? d. More generally, if the riders' speeds are \(v(t)=\frac{A}{(t+1)^{2}}\) and \(u(t)=\frac{B}{(t+1)^{2}}\) and the distance between the towns is \(D\) what conditions on \(A, B,\) and \(D\) must be met to ensure that the riders will pass each other? Looking ahead: With the velocity functions given in part (d). make a conjecture about the maximum distance each person can ride (given unlimited time).
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