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Chapter 11: Problem 42

A load suspended by a massless spring produces an extension of \(x \mathrm{~cm}\), in equilibrium. When it is cut into two unequal parts, the same load produces an extension of \(7.5 \mathrm{~cm}\) when suspended by the larger part of length \(60 \mathrm{~cm} .\) When it is suspended by the smaller part, the extension is \(5.0 \mathrm{~cm}\). Then (a) \(x=12.5\) (b) \(x=3.0\) (c) the length of the original spring is \(90 \mathrm{~cm}\) (d) the length of the original spring is \(80 \mathrm{~cm}\)

Short Answer

Expert verified
The original spring length is 90 cm and \( x = 12.5 \), so answers are (a) and (c).

Step by step solution

01

Understand the Problem

We have a spring that extends by \( x \) cm with a load. It's cut into two parts: a larger part of 60 cm length and a smaller part. It produces different extensions when the same load is applied to each part. We are to find \( x \) and the original length of the spring.
02

Apply Hooke's Law

Hooke's Law states that the extension \( e \) is proportional to the force applied, given by \( e = \frac{F}{k} \), where \( k \) is the spring constant. If we divide the spring, its spring constant increases based on the division: \( k_1 = \frac{kL}{L_1} \) for parts of length \( L_1 \) and \( k_2 = \frac{kL}{L_2} \) for \( L_2 \).
03

Calculate Spring Constant Ratios

For the larger part, \( e_1 = 7.5 \) cm for a length \( L_1 = 60 \) cm, and for the smaller part, \( e_2 = 5 \) cm. The ratios of extensions give \( \frac{L_1}{e_1} = \frac{60}{7.5} \) and \( \frac{L_2}{e_2} = \frac{L-L_1}{5} \). We need equality of the product of extension and length in both parts to match.
04

Solve for Extensions and Original Length

Using \( 60 \) cm part, solving \( \frac{60}{7.5} = k \), gives \( k = 8 \), implying \( F = 7.5k = 60 \). Similarly, solve for the other part using \( L_2 = L - 60 \), getting \( \frac{L-60}{5} = k \). Equate both expressions for \( k \) to find \( L = 90 \) cm. Then \( x = \frac{L}{F} \times 60 = 12.5 \).
05

Verify Results

Check calculations by substituting back to ensure \( x = 12.5 \) and the original length is 90 cm. Both smaller and larger parts should satisfy provided extension equations.

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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, denoted as \( k \), is an essential parameter in Hooke's Law. It defines the stiffness of a spring. According to Hooke's Law, the larger the spring constant, the stiffer the spring, and consequently, the less it stretches under the same force.
In mathematical terms:
  • Hooke's Law is expressed as \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the extension.
  • A spring with a high \( k \) will extend less when a given force is applied.
Understanding the concept of spring constant is crucial when a spring is divided because each section of the spring will have a different \( k \) value depending on its length.
Spring Extension
Spring extension refers to how much a spring stretches when a force is applied. The amount of extension is directly proportional to the force applied, as described by Hooke's Law.
Here are some key points about spring extension:
  • When a load is applied, the extension \( x \) can be calculated using \( x = \frac{F}{k} \), where \( F \) is the applied force and \( k \) is the spring constant.
  • The original exercise problem involves a load suspending a spring creating specific extensions such as 7.5 cm and 5.0 cm, based on the part of the spring used.
  • The greater the spring extension, the larger the distance the spring stretches compared to its unstressed position.
Knowing the extension helps determine the force acting on the spring and its respective \( k \).
Equilibrium Position
The equilibrium position of a spring is its natural, unstressed state, where no net force acts on it. When a load is attached to a spring, it stretches to reach a new equilibrium position where the force from the spring counteracts the weight of the load.
Consider these points:
  • In the original exercise, the spring stretches by \( x \) cm to reach equilibrium with the load.
  • The actions of cutting and applying loads to sections of the spring change the equilibrium positions for each part, especially when cut into unequal divisions.
  • The equilibrium position is important for calculating the extension and changes when loads change or when the spring is altered.
In summary, determining the equilibrium position helps in finding how the spring's extensions change in different scenarios.
Spring Division
Dividing a spring alters its elastic properties significantly. When a spring is cut into two segments, each part behaves like a separate spring with its own spring constant different from the original.
Here's how division affects the spring:
  • Each divided part's spring constant is adjusted based on its length, given by equations like \( k_1 = \frac{kL}{L_1} \), where \( L_1 \) is the new length of one segment.
  • The exercise involves dividing a spring into lengths of 60 cm and another part, altering the extension properties for each segment.
  • By dividing the spring, each segment responds differently to the same load because \( k \) differs, making it crucial to consider the effects of division in calculations.
Understanding how division changes the spring's behavior is vital when solving problems involving parts of springs.
Original Spring Length
The original spring length is a foundational measure in problems involving springs. It's essential because all other calculations, like spring division and extensions, ultimately relate back to this original measure.
Here are some significant points involving the original spring length:
  • In the exercise, determining the original spring length (90 cm) was crucial to calculating the different extensions experienced by spring segments.
  • All segment lengths and extensions add up to derive the total original length, forming a baseline for solving the problem through backward calculation.
  • The known original length helps in determining the spring constant \( k \) when the spring is divided, as seen in deriving the exercise solution.
Understanding and calculating the original spring length provides the foundation for solving spring-related problems and applying Hooke's Law.

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Most popular questions from this chapter

A cube is compressed at \(0^{\circ} \mathrm{C}\) equally from all sides by an external pressure \(p\). By what amount should be temperature be raise to bring it back to the size it had before the external pressure was applied? (Given \(K\) is bulk modulus of elasticity of the material of the cube and \(\alpha\) is the coefficient of linear expansion.) (a) \(\frac{p}{K \alpha}\) (b) \(\frac{p}{3 K \alpha}\) (c) \(\frac{3 \pi \alpha}{p}\) (d) \(\frac{K}{3 p}\)

A substance breaks down by a stress of \(10^{6} \mathrm{Nm}^{-2}\). If the density of the material of the wire is \(3 \times 10^{3} \mathrm{kgm}^{-3}\), then the length of the wire of that substance which will break under its own weight when suspended vertically is nearly (a) \(3.4 \mathrm{~m}\) (b) \(34 \mathrm{~m}\) (c) \(340 \mathrm{~m}\) (d) \(3400 \mathrm{~m}\)

When a force is applied on a wire of uniform crosssectional area \(3 \times 10^{-6} \mathrm{~m}^{2}\) and length \(4 \mathrm{~m}\), the increase in length is \(1 \mathrm{~mm}\). Energy stored in it will be \(\left(Y=2 \times 10^{11} \mathrm{Nm}^{-2}\right)\) (a) \(6250 \mathrm{~J}\) (b) \(0.177 \mathrm{~J}\) (c) \(0.075 \mathrm{~J}\) (d) \(0.150 \mathrm{~J}\)

The length of a wire is increased by \(1 \mathrm{~mm}\) on the application of a given load. In a wire of the same material, but of length and radius twice that of the first, on the application of the same load, extension is (a) \(0.25 \mathrm{~mm}\) (b) \(0.5 \mathrm{~mm}\) (c) \(2 \mathrm{~mm}\) (d) \(4 \mathrm{~mm}\)

When a weight of \(5 \mathrm{~kg}\) is suspended from a copper wire of length \(30 \mathrm{~m}\) and diameter \(0.5 \mathrm{~mm}\), the length of the wire increases by \(2.4 \mathrm{~cm}\). If the diameter is doubled, the extension produced is (a) \(1.2 \mathrm{~cm}\) (b) \(0.6 \mathrm{~cm}\) (c) \(0.3 \mathrm{~cm}\) (d) \(0.15 \mathrm{~cm}\)
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