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Chapter 8: Problem 1

(1) A spring has a spring constant \(k\) of 82.0 \(\mathrm{N} / \mathrm{m} .\) How much must this spring be compressed to store 35.0 \(\mathrm{J}\) of potential energy?

Short Answer

Expert verified
The spring must be compressed approximately 0.924 meters.

Step by step solution

01

Understand the Formula for Potential Energy in a Spring

The potential energy (PE) stored in a compressed or stretched spring can be calculated using the formula: \[ PE = \frac{1}{2} k x^2 \]where \(k\) is the spring constant, and \(x\) is the displacement or compression of the spring. In this exercise, the potential energy \(PE\) is given as 35.0 \(\mathrm{J}\), and the spring constant \(k\) is 82.0 \(\mathrm{N/m}\). We need to find \(x\).
02

Rearrange the Formula to Solve for Compression

To find \(x\), we need to rearrange the potential energy formula:\[ 35.0 = \frac{1}{2} \times 82.0 \times x^2 \]To isolate \(x^2\), we first multiply both sides by 2 to get rid of the fraction:\[ 70.0 = 82.0 \times x^2 \]Then, divide both sides by 82.0 to solve for \(x^2\):\[ x^2 = \frac{70.0}{82.0} \]
03

Calculate the Compression Distance

Now calculate \(x^2\):\[ x^2 = \frac{70.0}{82.0} \approx 0.85366 \]Next, take the square root of both sides to find \(x\):\[ x = \sqrt{0.85366} \approx 0.924 \]Thus, the spring must be compressed approximately 0.924 meters.

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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 to understanding how springs behave under forces. It states that the force required to compress or extend a spring is directly proportional to the amount of displacement it experiences. This is expressed as:
  • \( F = kx \)
where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement.Hooke's Law only applies within the elastic limit of the spring鈥攖his means the spring will return to its original shape and length once the force is removed, provided this limit is not exceeded.
Understanding and applying this law is crucial when dealing with problems involving potential energy in springs. It helps us identify how much force a spring can handle before it becomes permanently deformed.
Spring Constant
The spring constant, denoted by \( k \), is a measure of a spring's stiffness. It tells us how much force is needed to extend or compress the spring by a unit length. The spring constant is either given in scenarios like our problem or can be calculated through experimentation.Think of it this way:
  • A high spring constant means the spring is stiff, requiring more force to change its length.
  • A low spring constant indicates a more flexible spring.
In the given exercise, the spring constant \( k \) is 82.0 \( \mathrm{N/m} \). This means for every meter we wish to compress the spring, 82 Newtons of force would be required.
Knowing the spring constant is essential for calculating potential energy and dealing with mechanical energy in systems involving springs.
Mechanical Energy
Mechanical energy is the sum of kinetic and potential energy in any given system. When dealing with springs, we often focus on potential energy, especially when they are compressed or stretched. In the spring potential energy formula:
  • \( PE = \frac{1}{2} k x^2 \)
the stored mechanical energy is largely due to the position and state of the spring.
This energy can be converted from potential to kinetic when the spring is released, propelling objects attached to it or returning to its original form.
Understanding this energy transformation is crucial. It allows us to grasp how mechanical energy can be conserved and manipulated in systems. It also helps us solve real-world problems, such as those requiring calculations of spring compression or energy storage.

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Most popular questions from this chapter

Electric energy units are often expressed in the form of "kilowatt-hours." (a) Show that one kilowatt-hour (kWh) is equal to \(3.6 \times 10^{6} \mathrm{~J} .\) (b) If a typical family of four uses electric energy at an average rate of \(580 \mathrm{~W}\), how many \(\mathrm{kWh}\) would their electric bill show for one month, and (c) how many joules would this be? ( \(d\) ) At a cost of \(\$ 0.12\) per \(\mathrm{kWh}\), what would their monthly bill be in dollars? Does the monthly bill depend on the rate at which they use the electric energy?

An elevator cable breaks when a \(920-\mathrm{kg}\) elevator is \(24 \mathrm{~m}\) above a huge spring \(\left(k=2.2 \times 10^{5} \mathrm{~N} / \mathrm{m}\right)\) at the bottom of the shaft. Calculate ( \(a\) ) the work done by gravity on the elevator before it hits the spring, \((b)\) the speed of the elevator just before striking the spring, and (c) the amount the spring compresses (note that work is done by both the spring and gravity in this part).

(II) A 1400 -kg sports car accelerates from rest to 95 \(\mathrm{km} / \mathrm{h}\) in 7.4 \(\mathrm{s} .\) What is the average power delivered by the engine?

(II) How much work would be required to move a satellite of mass \(m\) from a circular orbit of radius \(r_{1}=2 r_{\mathrm{E}}\) about the Earth to another circular orbit of radius \(r_{2}=3 r_{\mathrm{E}} ?\) \(\left(r_{\mathrm{E}}\right.\) is the radius of the Earth.)

(III) The position of a 280 -g object is given (in meters) by \(x=5.0 t^{3}-8.0 t^{2}-44 t,\) where \(t\) is in seconds. Determine the net rate of work done on this object \((a)\) at \(t=2.0 \mathrm{~s}\) and (b) at \(t=4.0 \mathrm{~s}\). (c) What is the average net power input during the interval from \(t=0 \mathrm{~s}\) to \(t=2.0 \mathrm{~s},\) and in the interval from \(t=2.0 \mathrm{~s}\) to \(4.0 \mathrm{~s} ?\)
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