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In a room that is \(2.44 \mathrm{m}\) high, a spring (unstrained length \(=0.30 \mathrm{m}\) ) hangs from the ceiling. A board whose length is \(1.98 \mathrm{m}\) is attached to the free end of the spring. The board hangs straight down, so that its \(1.98-\mathrm{m}\) length is perpendicular to the floor. The weight of the board \((104 \mathrm{N})\) stretches the spring so that the lower end of the board just extends to, but does not touch, the floor. What is the spring constant of the spring?

Short Answer

Expert verified
The spring constant is 650 N/m.

Step by step solution

01

Understand the Problem

We have a spring with an unstrained length of 0.30 m and a board attached, hanging in a room 2.44 m in height. The board's weight causes the spring to stretch until the board's end just reaches the floor. We need to find the spring constant, denoted as \( k \).
02

Calculate Total Spring Elongation

The height of the room is 2.44 m and the unstrained length of the spring plus the length of the board is \(0.30 \text{ m} + 1.98 \text{ m} = 2.28 \text{ m}\). Since the board's bottom just reaches the floor, the total elongation of the spring is \(2.44 \text{ m} - 2.28 \text{ m} = 0.16 \text{ m}\).
03

Apply Hooke's Law

Hooke's Law relates the force required to stretch or compress a spring to the distance it is stretched or compressed: \( F = kx \), where \( F \) is the force (104 N in this case), \( k \) is the spring constant, and \( x \) is the elongation (0.16 m here).
04

Solve for the Spring Constant

Re-arrange Hooke's Law to solve for \( k \): \( k = \frac{F}{x} = \frac{104 \text{ N}}{0.16 \text{ m}} \). Calculate \( k \):\[k = \frac{104}{0.16} = 650 \text{ N/m}\]

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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 a fundamental principle in physics that relates to how springs behave under load. It states that the force needed to extend or compress a spring is directly proportional to the distance it is stretched. This relationship can be expressed by the formula:
  • \( F = kx \)
Here, \( F \) is the force applied to the spring, \( x \) is the displacement or elongation of the spring from its original length, and \( k \) is the spring constant. The spring constant \( k \) is a measure of the stiffness of the spring. It tells us how much force is needed to stretch or compress the spring by one unit of length. The greater the spring constant, the stiffer the spring.
Hooke's Law applies to situations where materials can return to their original shape after being stretched or compressed, which is known as the elastic limit. If the force exceeds this limit, the material will deform permanently and the law no longer holds true.
This principle is widely used in various applications, such as designing mechanical systems, evaluating materials, and solving physics problems involving elasticity.
Spring Elongation
Spring elongation refers to the increase in length of a spring when a force is applied. In the context of Hooke’s Law, elongation is the change in length from the spring’s original, unstressed length to its stretched length.
In the problem given, we see that the natural length of the spring is 0.30 m. When a weight of 104 N is applied due to the board, the spring stretches, making the total length of the spring-board system reach 2.44 m, which is the height of the room. Therefore, the elongation of the spring is calculated by subtracting the combined unstrained lengths of the spring and board from the total distance to the floor:
  • The initial setup: Total unstrained length = Spring length + Board length = 0.30 m + 1.98 m = 2.28 m.
  • Total room height: 2.44 m.
  • Spring elongation = Total room height - Total unstrained length = 2.44 m - 2.28 m = 0.16 m.
So, the spring elongates by 0.16 m in this scenario. Accurately calculating elongation is crucial as it plays a pivotal role in applying Hooke's Law to determine the spring constant.
Mechanics Problem Solving
Mechanics problem solving involves applying basic principles of physics to analyze situations involving forces and motion. When dealing with problems like finding the spring constant in this exercise, it's important to break the problem down into understandable steps.
To solve mechanics problems effectively:
  • **Understand the Problem**: Clearly know what is being asked. Identify all known values such as forces, distances, and mass involved in the problem.
  • **Calculate Needed Quantities**: Determine what the problem requires you to calculate, such as the spring elongation in our example.
  • **Apply Relevant Physics Principles**: Use principles like Hooke's Law where applicable. In this exercise, acknowledge that the stretch or compression of the spring is linearly related to the force applied up to the elastic limit.
  • **Calculate and Verify**: Perform the necessary calculation, like solving for the spring constant by rearranging and computing from the Hooke’s Law formula.
Each step should be approached methodically to ensure thorough understanding and correct solutions. This systematic way of tackling mechanics problems builds a solid foundation for solving more complex physics challenges.

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Most popular questions from this chapter

A 70.0 -kg circus performer is fired from a cannon that is elevated at an angle of \(40.0^{\circ}\) above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by \(3.00 \mathrm{m}\) from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as the height of the net into which he is shot. He takes 2.14 s to travel the horizontal distance of \(26.8 \mathrm{m}\) between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of \(7.0 \mathrm{rad} / \mathrm{s} .\) The drawing indicates the position of the block when the spring is unstrained. This position is labeled " \(x=\) \(0 \mathrm{m} .\) "The drawing also shows a small bottle located \(0.080 \mathrm{m}\) to the right of this position. The block is pulled to the right, stretching the spring by \(0.050 \mathrm{m},\) and is then thrown to the left. In order for the block to knock over the bottle, it must be thrown with a speed exceeding \(v_{0} .\) Ignoring the width of the block, find \(v_{0}\).

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is \(1.25 \mathrm{m} / \mathrm{s},\) and its maximum acceleration is \(6.89 \mathrm{m} / \mathrm{s}^{2} .\) How much time elapses between an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum?

A heavy-duty stapling gun uses a 0.140 -kg metal rod that rams against the staple to eject it. The rod is attached to and pushed by a stiff spring called a "ram spring" \((k=32000 \mathrm{N} / \mathrm{m})\). The mass of this spring may be ignored. The ram spring is compressed by \(3.0 \times 10^{-2} \mathrm{m}\) from its unstrained length and then released from rest. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{m}\) when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

The femur is a bone in the leg whose minimum cross-sectional area is about \(4.0 \times 10^{-4} \mathrm{m}^{2} .\) A compressional force in excess of \(6.8 \times 10^{4} \mathrm{N}\) will fracture this bone. (a) Find the maximum stress that this bone can with- stand. (b) What is the strain that exists under a maximum-stress condition?

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