/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 A balky cow is leaving the barn ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 32

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from \(x=0\) to \(x=6.9 \mathrm{m}\) as you apply a force with \(x\) -component \(F_{x}=-[20.0 \mathrm{N}+(3.0 \mathrm{N} / \mathrm{m}) x] .\) How much work does the force you apply do on the cow during this displacement?

Short Answer

Expert verified
The work done by the applied force is approximately -209.415 J.

Step by step solution

01

Understanding the Work Done Formula

The work done by a force is given by the formula \( W = \int F(x) \, dx \), where \( F(x) \) is the force as a function of position \( x \). Here, the force applied is \( F_x = -[20.0 \text{ N} + (3.0 \text{ N/m})x] \).
02

Setup the Integral for Work Done

We need to calculate \( W = \int_{0}^{6.9} (-20.0 - 3.0x) \, dx \). This integral will give us the work done as the cow moves from \( x = 0 \) to \( x = 6.9 \text{ m} \).
03

Evaluate the Integral

Calculate the integral. We have:\[W = \int_{0}^{6.9} (-20.0 - 3.0x) \, dx\]This can be split as two integrals:\[W = \int_{0}^{6.9} (-20.0) \, dx - \int_{0}^{6.9} 3.0x \, dx\]Evaluate each part:\[= -20.0[x]_{0}^{6.9} - 3.0\left[\frac{x^2}{2}\right]_{0}^{6.9}\]\[= -20.0[6.9 - 0] - 3.0\left[\frac{6.9^2}{2}-\frac{0^2}{2}\right]\]\[= -20.0 \times 6.9 - 3.0 \times \frac{6.9^2}{2}\]Now calculate each term separately.
04

Calculations

Calculate the term \(-20.0 \times 6.9\): \[- 20.0 \times 6.9 = -138 \text{ J}\]Calculate \(6.9^2\):\[6.9^2 = 47.61\]Then calculate \(-3.0 \times \frac{47.61}{2}\):\[-3.0 \times 23.805 = -71.415 \text{ J}\].
05

Add Up the Results

Sum the results from each part of the integral:\[-138 \text{ J} - 71.415 \text{ J} = -209.415 \text{ J}\].The negative sign indicates the direction of the work done by the force as the cow moves away from the barn.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Calculus in Physics
Integral calculus is a powerful tool in physics, especially when it comes to calculating work done by a force over a certain path. In essence, it helps us understand how quantities like force, pressure, or even electric fields change and accumulate over a path or area. When dealing with forces that vary along a distance, as in the case of the balky cow moving away from the barn, integral calculus allows us to sum up the small bits of work done at each point of the displacement.
The integral for calculating work is given by:
  • \( W = \int F(x) \, dx \)
  • Where \( F(x) \) is the force function, and you integrate over the path the object moves.
This approach simplifies calculating work in situations where the force is not constant. By breaking down the force into tiny intervals and summing them up, we capture the complete picture of work done over a path.
Net Work Done
The net work done is essentially the total work performed by a force acting over a distance. In physics, work is defined as the energy transferred to or from an object via the application of force along a displacement. When calculating net work, both the magnitude and direction of force and displacement must be considered.
In our cow scenario, the force applied attempts to push the cow back, while it moves forward. The mathematical representation shows us how to account for this negative force working against the cow鈥檚 direction of movement. After integrating the force expression given in the problem, we calculate:
  • \( W = -138 \text{ J} - 71.415 \text{ J} = -209.415 \text{ J} \)
The negative sign in the resulting work illustrates that the direction of the force is opposite to the direction of displacement, demonstrating that work done by force doesn't always mean pushing something forward successfully.
Force-Displacement Relationship
The relationship between force and displacement is critical when analyzing the work done on objects. Here, the force exerted on the cow is not constant; it varies linearly with the cow's position according to the formula:
  • \( F(x) = -[20.0 \, \text{N} + (3.0 \, \text{N/m})x] \)
This relationship shows that as the cow moves farther from the barn door, the force increases in magnitude. The negative sign indicates the direction of the force is towards the barn, opposing the cow鈥檚 displacement.
Understanding this relationship helps us break down and approach problems involving varying forces more systematically. While a constant force might be straightforward to calculate, varying forces require integration to determine the total net work done across the path of influence.

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Most popular questions from this chapter

A Spring with Mass. We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass \(M\) , equilibrium length \(L_{0}\) , and spring constant \(k\) . The work done to stretch or compress the spring by a distance \(L\) is \(\frac{1}{2} k X^{2}\) , where \(X=L-L_{0}\) (a) Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\) . Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. Calculate the kinetic energy of the spring in terms of \(M\) and \(v\) . (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each piece in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L\) . The result is not \(\frac{1}{2} M v^{2},\) since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 \(\mathrm{kg}\) and force constant 3200 \(\mathrm{N} / \mathrm{m}\) is compressed 2.50 \(\mathrm{cm}\) from its unstretched length. When the trigger is pulled, the spring pushes horizontally on a 0.053 -kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncompressed length (b) ignoring the mass of the spring and (c) including, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

A 75.0 kg painter climbs a ladder that is 2.75 \(\mathrm{m}\) long leaning against a vertical wall. The ladder makes an \(30.0^{\circ}\) angle with the wall. (a) How much work does gravity do on the painter? (b) Does the answer to part (a) depend on whether the painter climbs at constant speed or accelerates up the ladder?

You are a member of an Alpine Rescue Team. You must project a box of supphes up an incline of constant slope angle \(\alpha\) so that it reaches a stranded skier who is a vertical distance \(h\) above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \(\mu_{\mathbf{k}}\) . Use the work-energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of \(g, h, \mu_{k},\) and \(\alpha .\)

(a) How many joules of kinetic energy does a \(750-\mathrm{kg}\) auto- mobile traveling at a typical highway speed of 65 \(\mathrm{mi} / \mathrm{h}\) have? (b) By what factor would its kinetic energy decrease if the car traveled half as fast? (c) How fast (in mi/h) would the car have to travel to have half as much kinetic energy as in part (a)?

All birds, independent of their size, must maintain a power output of \(10-25\) watts per kilogram of body mass in order to fly by flapping their wings. (a) The Andean giant hummingbird (Patagona gigas) has mass 70 \(\mathrm{g}\) and flaps its wings 10 times per second while hovering. Estimate the amount of work done by such a hummingbird in each wingbeat. (b) A \(70-\mathrm{kg}\) athlete can maintain a power output of 1.4 \(\mathrm{kW}\) for no more than a few seconds; the steady power output of a typical athlete is only 500 \(\mathrm{W}\) or so. Is it possible for a human-powered aircraft to fly for extended periods by flapping its wings? Explain.
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