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Chapter 3: Problem 29

A spring obeys Hooke's law. If an amount of work \(W\) is required to stretch the spring a length \(x\) beyond its unstretched length, how much work does it take to stretch it to \(3 x\) ? (A) \(W\) (B) \(3 W\) (C) \(6 W\) (D) \(9 \mathrm{~W}\) (E) \(27 W\)

Short Answer

Expert verified
Answer: (C) \(6W\)

Step by step solution

01

1. Recall Hooke's law and the definition of work

According to Hooke's law, the force exerted by the spring \(F\) is proportional to its displacement \(x\): \[F = -kx\] where \(k\) is the spring constant. The negative sign indicates that the force is in the opposite direction of the displacement. The work done while stretching the spring is given by the integral of force with respect to displacement: \[W = \int_{0}^{x} F dx\]
02

2. Express work in terms of spring constant and displacement

We integrate the force to find the work done in stretching the spring from its unstretched length to \(x\): \[W = \int_{0}^{x} (-kx) dx = -\frac{k}{2}x^2\] By solving for the spring constant, \(k\) we get: \[k = -\frac{2W}{x^2}\]
03

3. Calculate the work required to stretch the spring to 3x

Now we want to find the work required to stretch the spring to a new length, \(3x\). Let's call this new work \(W_{3x}\). \[W_{3x} = \int_{0}^{3x} F dx\] Using the expression for the force, we substitute \(k\) with the expression we found in step 2: \[W_{3x} = \int_{0}^{3x} \left(-\frac{2W}{x^2}x \right) dx = -\frac{2W}{x^2} \int_{0}^{3x} x dx = - \frac{2W}{x^2} \cdot \frac{9x^3}{3} = -6W\] However, we made an error in the signs because the force is negative. So, correcting this error, we get: \[W_{3x} = 6W\] Now we can find the correct answer from the given options: (C) \(6W\)

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Key Concepts

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

Hooke's Law
Hooke's law is fundamental in understanding how springs and other elastic materials behave when forces are applied. The law states that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, it is represented as:
\[F = -kx\]
where \(F\) is the force exerted by the spring, \(x\) is the displacement from the equilibrium position, and \(k\) is the spring constant. The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement, also known as the restoring force. It's important for students to grasp that the greater the displacement, the greater the force required to hold the spring stretched or compressed.
In our exercise, understanding Hooke's law allows us to set up the relationship between the work done to stretch the spring and how far it has been stretched. When a problem involves stretching a spring multiple times its original length, we can anticipate that the work done will increase significantly because the force required increases as the extension does. However, the relationship is not linear, which is a common point of confusion and an important concept to clarify.
Work-Energy Principle
The work-energy principle is a vital concept in physics that connects the dots between force, displacement, and energy. It states that the work done on an object is equal to the change in its energy. In the case of a spring, this relates to the potential energy stored in the spring as it is stretched or compressed.
In the given problem, when we apply the work-energy principle, we can express the work done on the spring in terms of the change in its potential energy. The calculation of work involves integrating the force over the displacement, which gives us:
\[W = \frac{k}{2}x^2\]
This formula illustrates that the work done on the spring, and hence the potential energy stored in it, increases with the square of the displacement. So, if the displacement is tripled, the work done (and energy stored) is increased by a factor of nine. This quadratic relationship is central to problems involving springs and energy, and helps explain why the answer to our exercise problem is not simply three times the original work for a displacement of \(3x\).
Physics Problem Solving
Physics problem solving requires a clear understanding of the principles involved and the ability to apply these principles to specific situations. This involves recalling relevant formulas, executing mathematical operations correctly, and understanding the physical meaning behind the calculations.
For the spring problem we are discussing, problem-solving starts with recalling Hooke's law to set up the force equation. Then, recognizing that work is the integral of force over displacement, we perform the integration to express work in terms of the spring constant and displacement. However, an equally important part of problem solving is the ability to find and correct mistakes, like the sign error encountered in the step-by-step solution. Developing these problem-solving skills involves practice, plenty of patience, and often, going back to the basics to ensure complete comprehension of the underlying concepts.
  • Recall and apply relevant physics laws and principles.
  • Perform the necessary mathematical calculations accurately.
  • Interpret the physical meaning of the mathematical results.
  • Review and correct any possible errors in the process.
These steps can guide students as they tackle a variety of physics problems, ranging from Hooke's law applications to more complex systems.

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Most popular questions from this chapter

Your dog Astro is nosing a dinner plate of mass \(M\) across a frozen pond at a constant velocity \(v\). There is a coefficient of friction \(\mu\) between the ice and the block. What is the rate of work Astro does on the plate? (A) \(\mu M v^2\) (B) \(\mu M g v\) (C) \(M g v^2 / \mu\) (D) \(\mu M v^2 / g\) (E) \(\mu M g / v\)

Two identical planets, Alpha and Beta, touch each other. Khan the Conqueror suddenly halves the density of Beta, and doubles the radius of Alpha. The gravitational force between them: (A) stays the same. (B) doubles. (C) halves. (D) goes up by a factor of \(16 / 9\). (E) goes down by a factor of \(9 / 16\).

A circus cannon fires an acrobatic dog, Astro, of mass \(m\), into a net. The speed of Astro as he leaves the cannon is \(v\). A cat, Beta, of mass \(0.5 \mathrm{~m}\) is then fired from the cannon. Assuming that the force exerted on each is constant throughout the cannon barrel, what is the speed of Beta as she leaves the mouth of the cannon? (A) \(2 v\) (B) \(\sqrt{2} v\) (C) \(v\) (D) \(v / \sqrt{2}\) (E) \(v / 2\)

An open railroad car is passing at a constant speed \(v_0\) under a water tower, which suddenly dumps a lot of water into the car. The speed of the car will (A) stay the same because of momentum conservation. (B) increase because of momentum conservation. (C) decrease because of momentum conservation. (D) increase because of energy conservation. (E) stay the same because the water is falling in the \(y\)-direction and the car is moving in the \(x\)-direction.

You find yourself stranded on planet Alpha, where the gravitational field is \(5 \mathrm{~N} / \mathrm{kg}\). Which of the following statements are true? The \(10 \mathrm{~kg}\) mass you hold in your hand i. will weigh about half as much on Alpha as on Earth. ii. will weigh about twice as much on Alpha as on Earth. iii. will have the same mass on Alpha as on Earth. iv. will fall with a greater acceleration than on Earth. v. will require a lesser force to push it across a horizontal surface with a small amount of friction. (A) i and iv (B) i, iii and \(v\) (C) i, iii, iv and \(v\) (D) ii, and iii (E) ii, iii and iv
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