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Chapter 6: Problem 40

Half of a Spring. (a) Suppose you cut a massless ideal spring in half. If the full spring had a force constant \(k\) , what is the force constant of each half, in terms of \(k ?\) (Hint: Think of the original spring as two equal halves, each producing the same force as the entire spring. Do you see why the forces must be equal? (b) If you cut the spring into three equal segments instead, what is the force constant of each one, in terms of \(k ?\)

Short Answer

Expert verified
Half: spring constant is \( 2k \), third: spring constant is \( 3k \).

Step by step solution

01

Understand the Original Spring

The force constant, or spring constant \( k \), describes the stiffness of a spring. The spring follows Hooke's Law, \( F = kx \), where \( F \) is the force exerted by the spring and \( x \) is the displacement from equilibrium.
02

Cutting the Spring in Half

When the spring is cut in half, each half must support the same amount of force, implying each half acts like a spring with a doubled spring constant. This is because when you cut the spring in half, each piece must exert the same force over half the displacement. Thus, the spring constant of each half is \( 2k \).
03

Cutting the Spring into Three Segments

Similarly, when the spring is cut into three equal pieces, each piece behaves like a spring with a spring constant three times that of the original spring鈥檚 constant. This is because each piece must exert the same force over a third of the displacement. Thus, the spring constant for each segment is \( 3k \).
04

Conclude the Results

When you cut a spring into \( n \) equal parts, the spring constant for each segment becomes \( nk \), where \( n \) is the number of segments. For half, \( n = 2 \), so the spring constant is \( 2k \). For thirds, \( n = 3 \), so the spring constant is \( 3k \).

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Key Concepts

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

Spring Constant
The spring constant, often denoted as \( k \), is a crucial part of understanding how springs work. It is a measure of a spring's stiffness and is described in Hooke鈥檚 Law, represented by the equation \( F = kx \). Here, \( F \) stands for force, while \( x \) symbolizes the displacement from the spring's equilibrium position. A larger spring constant means that the spring is stiffer and requires more force to be stretched or compressed by the same amount.
  • A spring with a high value of \( k \) resists stretching, making it harder to change its shape.
  • Conversely, a smaller \( k \) indicates that the spring is more pliable and easier to stretch or compress.
With modifications such as cutting, as seen in exercises, the spring constant changes according to the number of segments.
Ideal Spring
An ideal spring is a theoretical concept within physics. It behaves perfectly according to Hooke's Law and does not experience energy loss from heat, friction, or deformation beyond its elastic limit. This means that, in an ideal spring, the force exerted by the spring is always proportional to the displacement, within the elastic range of the material.
  • No matter how much you stretch or compress an ideal spring, it will return to its original shape without any permanent deformation.
  • Ideal springs do not break or weaken over time or with repeated use.
Understanding the properties of ideal springs helps in visualizing scenarios where theoretical conditions are preserved, simplifying complex calculations in physics exercises.
Equilibrium Displacement
Equilibrium displacement refers to the distance a spring is stretched or compressed from its natural, resting position where no net force is acting on it. In Hooke's Law, this displacement is represented by \( x \). The concept of equilibrium displacement is essential because it determines the magnitude of the force that the spring exerts.
  • When a spring is displaced from its equilibrium, it stores potential energy, which is released when the spring returns to its original state.
  • Equilibrium displacement helps in calculating how different segments of a spring will react when cut, as each segment must provide the same force over a reduced displacement.
Analyzing the influences of equilibrium displacement allows for a deeper understanding of how springs respond to forces and how cutting a spring affects its properties.

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Most popular questions from this chapter

You and your bicycle have combined mass 80.0 \(\mathrm{kg}\) . When you reach the base of a bridge, you are traveling along the road at 5.00 \(\mathrm{m} / \mathrm{s}(\mathrm{Fig} .6 .35) .\) At the top of the bridge, you have climbed a vertical distance of 5.20 \(\mathrm{m}\) and have slowed to 1.50 \(\mathrm{m} / \mathrm{s}\) . You can ignore work done by friction and any inefficiency in the bike or your legs. (a) What is the total work done on you and your bicycle when you go from the base to the top of the bridge? (b) How much work have you done with the force you apply to the pedals?

Stopping Distance. A car is traveling on a level road with speed \(v_{0}\) at the instant when the brakes lock, so that the tires slide rather than roll. (a) Use the work-energy theorem to calculate the minimum stopping distance of the car in terms of \(v_{0}, g,\) and the coefficient of kinetic friction \(\mu_{x}\) between the tires and the road. (b) By what factor would the minimum stopping distance change if (i) the coefficient of kinetic friction were doubled, or (ii) the initial speed were doubled, or (iii) both the coefficient of kinetic friction and the initial speed were doubled?

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?

To stretch a spring 3.00 \(\mathrm{cm}\) from its unstretched length, 12.0 \(\mathrm{J}\) of work must be done. (a) What is the force constant of this spring? (b) What magnitude force is needed to stretch the spring 3.00 \(\mathrm{cm}\) from its unstretched length? (c) How much work must be done to compress this spring 4.00 \(\mathrm{cm}\) from its unstretched length, and what force is needed to streteh it this distance?

An elevator has mass 600 \(\mathrm{kg}\) , not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of 20.0 \(\mathrm{m}\) (five floors) in 16.0 \(\mathrm{s}\) , and it is driven by a motor that can provide up to 40 \(\mathrm{hp}\) to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass 65.0 \(\mathrm{kg}\).
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