/*! 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 26 A spring of spring constant \(5.... [FREE SOLUTION] | 91Ó°ÊÓ

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A spring of spring constant \(5.0 \times 10^{3} \mathrm{~N} / \mathrm{m}\) is stretched initially by \(5.0 \mathrm{~cm}\) from the unstretched position. What is the work required to stretch it further by another \(5.0 \mathrm{~cm}\) ?

Short Answer

Expert verified
The work required is 18.75 J.

Step by step solution

01

Understand the Problem

We need to calculate the additional work required to further stretch a spring, which initially has been extended by \(5.0\, \text{cm}\). The spring constant \(k\) is given as \(5.0 \times 10^{3}\, \text{N/m}\).
02

Use the Work Formula for Springs

The work done on a spring is calculated using the formula \[ W = \frac{1}{2} k (x_2^2 - x_1^2) \]where \(x_1\) is the initial stretch and \(x_2\) is the final stretch.
03

Convert Units to Meters

Convert the initial and additional stretch from centimeters to meters, since the spring constant is in \(N/m\):- Initial stretch \(x_1 = 5.0 \text{ cm} = 0.05 \text{ m}\)- Additional stretch \(x_2 = 5.0 \text{ cm} = 0.05 \text{ m}\), so the total stretch \(x_2 = 0.05 + 0.05 = 0.10 \text{ m}\).
04

Substitute Values into the Formula

Substitute \(k = 5.0 \times 10^3 \text{ N/m}\), \(x_1 = 0.05 \text{ m}\), and \(x_2 = 0.10 \text{ m}\) into the work formula:\[ W = \frac{1}{2} \times 5.0 \times 10^3 \times (0.10^2 - 0.05^2) \]
05

Calculate the Work Done

Calculate the work:1. Find \(0.10^2 = 0.01\) and \(0.05^2 = 0.0025\)2. Compute \(0.01 - 0.0025 = 0.0075\)3. Plug into the equation: \[ W = \frac{1}{2} \times 5.0 \times 10^3 \times 0.0075 \]4. Calculate: \[ W = 0.5 \times 5000 \times 0.0075 = 18.75 \text{ J}\]
06

Conclusion

The work required to further stretch the spring by another 5.0 cm is \(18.75\, \text{J}\).

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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 symbolized as \( k \), is a measure of a spring's stiffness. It tells us how much force it takes to stretch or compress the spring by a certain amount. In essence, it's a way to quantify the spring's resistance to being deformed.

In our example, the spring constant is \( 5.0 \times 10^3 \ \mathrm{N/m} \). This means that it requires 5000 newtons of force to extend or compress the spring by one meter. Knowing the spring constant is crucial because it directly affects how much work is needed to stretch or compress the spring.

Here are some quick facts about spring constants:
  • Larger \( k \) values mean a stiffer spring, while smaller \( k \) values indicate a more flexible spring.
  • The unit of the spring constant is Newton per meter \((\mathrm{N/m})\).
  • Springs with higher spring constants require more force to deform.
Understanding the spring constant is key to solving problems involving springs, as it allows us to calculate the work and energy associated with them.
Elastic Potential Energy
Elastic potential energy is the energy stored in a spring (or any elastic material) when it is stretched or compressed. This form of potential energy is directly linked to the spring constant and the amount of deformation (stretch or compression) of the spring.

The potential energy stored in a spring is given by the formula:\[ U = \frac{1}{2} k x^2 \]where \( U \) is the elastic potential energy, \( k \) is the spring constant, and \( x \) is the distance the spring is stretched or compressed from its natural length.

Some important notes about elastic potential energy include:
  • It is zero when the spring is at its natural length, i.e., not stretched or compressed.
  • It increases as the spring is deformed further from its natural position.
  • This energy can be converted into other forms, such as kinetic energy, when the spring returns to its natural position.
In our exercise, calculating changes in stretch allows us to determine how much additional elastic potential energy is stored in the spring after further stretching.
Work Done on a Spring
The work done on a spring refers to the energy required to stretch or compress it. This work changes the elastic potential energy stored in the spring.

The mathematical expression for the work \( W \) done when a spring is stretched or compressed from an initial position \( x_1 \) to a final position \( x_2 \) is given by: \[ W = \frac{1}{2} k (x_2^2 - x_1^2) \]This equation accounts for the change in potential energy as the spring changes state.

Here are some key takeaways about work done on a spring:
  • Positive work means work is done to stretch or compress the spring, resulting in an increase in potential energy.
  • More work is required for larger stretches due to increased resistance from the spring.
  • Using the work formula allows us to calculate the energy needed for specific stretches, combining knowledge of both spring constants and deformation distances.
In the exercise, using the given values of spring constant, initial, and final stretches, we calculated that the work needed to further stretch the spring was 18.75 J, showcasing the practical application of these principles.

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Most popular questions from this chapter

(a) At a certain instant, a particle-like object is acted on by a force \(\vec{F}=(4.0 \mathrm{~N}) \hat{\mathrm{i}}-(2.0 \mathrm{~N}) \hat{\mathrm{j}}+(9.0 \mathrm{~N}) \hat{\mathrm{k}}\) while the object's velocity is \(\vec{v}=-(2.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(4.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\). What is the instantaneous rate at which the force does work on the object? (b) At some other time, the velocity consists of only a \(y\) component. If the force is unchanged and the instantaneous power is \(-15 \mathrm{~W}\), what is the velocity of the object?

A luge and its rider, with a total mass of \(85 \mathrm{~kg}\), emerge from a downhill track onto a horizontal straight track with an initial speed of \(37 \mathrm{~m} / \mathrm{s}\). If a force slows them to a stop at a constant rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\), (a) what magnitude \(F\) is required for the force, (b) what distance \(d\) do they travel while slowing, and (c) what work \(W\) is done on them by the force? What are (d) \(F\), (e) \(d\), and (f) \(W\) if they, instead, slow at \(4.0 \mathrm{~m} / \mathrm{s}^{2}\) ?

Across a horizontal floor, a \(102 \mathrm{~kg}\) block is pulled at a constant speed of \(5.5 \mathrm{~m} / \mathrm{s}\) by an applied force of \(125 \mathrm{~N}\) directed \(38^{\circ}\) above the horizontal. Calculate the rate at which the force does work on the block.

A can of sardines is made to move along an \(x\) axis from \(x=0.25 \mathrm{~m}\) to \(x=2.25 \mathrm{~m}\) by a force with a magnitude given by \(F=\exp \left(-4 x^{2}\right)\), with \(x\) in meters and \(F\) in newtons. (Here exp is the exponential function.) How much work is done on the can by the force?

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of \(12^{\circ}\) with the horizontal. The rope moves parallel to the slope with a constant speed of \(1.0 \mathrm{~m} / \mathrm{s}\). The force of the rope does \(880 \mathrm{~J}\) of work on the skier as the skier moves a distance of \(7.0\) \(\mathrm{m}\) up the incline. (a) If the rope moved with a constant speed of \(2.0\) \(\mathrm{m} / \mathrm{s}\), how much work would the force of the rope do on the skier as the skier moved a distance of \(8.0 \mathrm{~m}\) up the incline? At what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) \(1.0 \mathrm{~m} / \mathrm{s}\) and (c) \(2.0 \mathrm{~m} / \mathrm{s}\) ?

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