/*! 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 1 What is the spring constant of a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the spring constant of a spring that stores \(25 \mathrm{~J}\) of elastic potential energy when compressed by \(7.5 \mathrm{~cm} ?\)

Short Answer

Expert verified
The spring constant is approximately 8888.89 N/m.

Step by step solution

01

Understanding the Formula for Elastic Potential Energy

Elastic potential energy stored in a spring is given by the formula: \[ E = \frac{1}{2} k x^2 \] where \( E \) is the elastic potential energy, \( k \) is the spring constant, and \( x \) is the displacement (or compression in meters).
02

Convert Compression Distance to Meters

The given compression is \( 7.5 \mathrm{~cm} \). Convert this to meters by dividing by 100: \( x = \frac{7.5}{100} = 0.075 \mathrm{~m} \).
03

Substitute Known Values into the Formula

We have \( E = 25 \mathrm{~J} \) and \( x = 0.075 \mathrm{~m} \). Substitute these values into the elastic potential energy formula: \[ 25 = \frac{1}{2} k (0.075)^2 \].
04

Solve for the Spring Constant

First, calculate \((0.075)^2\): \(0.075^2 = 0.005625\). Now substitute back to get: \[ 25 = \frac{1}{2} k \times 0.005625 \].Multiply both sides by 2 to eliminate the fraction: \[ 50 = k \times 0.005625 \]. Finally, solve for \( k \) by dividing both sides by 0.005625: \[ k = \frac{50}{0.005625} \approx 8888.89 \mathrm{~N/m} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Elastic Potential Energy
Elastic potential energy is the energy stored in a spring when it is compressed or stretched. This type of energy is derived from the force needed to dislocate the spring from its equilibrium position. The amount of energy stored is determined by the stiffness of the spring and the amount of displacement from its natural position. The formula for elastic potential energy is given by:\[ E = \frac{1}{2} k x^2 \]- **E** stands for the elastic potential energy in joules (J).- **k** is the spring constant in newtons per meter (N/m).- **x** is the displacement from equilibrium, often referred to as compression or stretching distance, in meters.Understanding how elastic potential energy works helps clarify why the energy changes with different amounts of compression or stretching. It shows how energy is conserved and transformed in mechanical systems.
Compression Distance
Compression distance refers to how much a spring is pushed together or compressed from its original length. This is a critical measurement when calculating elastic potential energy since it directly influences the amount of energy stored. When dealing with exercises or actual measurements, the compression distance must always be converted into meters. This conversion is necessary because the international system of units (SI units) requires meters for calculations involving joules and newtons. For example, if a spring is compressed by 7.5 cm, it needs to be converted by dividing by 100 to obtain 0.075 meters. Accurate conversions are key to ensuring calculations remain consistent and correct across various scientific contexts. Using the right units makes mathematical equations easier to handle and avoids common mistakes that may arise from miscalculations or incorrect conversions.
Physics Formula
Physics formulas allow us to mathematically analyze and predict natural phenomena. In the case of elastic potential energy, the key formula \( E = \frac{1}{2} k x^2 \) helps us determine how much energy is stored in a spring based on its compression or stretch and the spring constant.To solve problems using this formula:1. **Identify Variables**: Find values for the elastic potential energy (E), spring constant (k), and displacement (x).2. **Convert Units**: Ensure all measurements are in SI units, such as converting compression distance to meters.3. **Substitute and Solve**: Input known values into the formula and solve for the unknown variable.For instance, in our original exercise where the energy is 25 J and compression is 0.075 m, substituting these into the formula allows us to solve for the spring constant. This step-by-step approach provides clarity and ensures accurate results when working with physical calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(0.63 \mathrm{~kg}\) ball thrown directly upward with an initial speed of \(14 \mathrm{~m} / \mathrm{s}\) reaches a maximum height of \(8.1 \mathrm{~m} .\) What is the change in the mechanical energy of the ball-Earth system during the ascent of the ball to that maximum height?

An automobile with passengers has weight \(16400 \mathrm{~N}\) and is moving at \(113 \mathrm{~km} / \mathrm{h}\) when the driver brakes, sliding to a stop. The frictional force on the wheels from the road has a magnitude of \(8230 \mathrm{~N}\). Find the stopping distance.

A machine pulls a \(40 \mathrm{~kg}\) trunk \(2.0 \mathrm{~m}\) up a \(40^{\circ}\) ramp at constant velocity, with the machine's force on the trunk directed parallel to the ramp. The coefficient of kinetic friction between the trunk and the ramp is \(0.40 .\) What are (a) the work done on the trunk by the machine's force and (b) the increase in thermal energy of the trunk and the ramp?

A block with mass \(m=2.00 \mathrm{~kg}\) is placed against a spring on a frictionless incline with angle \(\theta=30.0^{\circ}\) (Fig. 8-44). (The block is not attached to the spring.) The spring, with spring constant \(k=19.6 \mathrm{~N} / \mathrm{cm},\) is compressed \(20.0 \mathrm{~cm}\) and then released. (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block- Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?

To form a pendulum, a \(0.092 \mathrm{~kg}\) ball is attached to one end of a rod of length \(0.62 \mathrm{~m}\) and negligible mass, and the other end of the rod is mounted on a pivot. The rod is rotated until it is straight up, and then it is released from rest so that it swings down around the pivot. When the ball reaches its lowest point, what are (a) its speed and (b) the tension in the rod? Next, the rod is rotated until it is horizontal, and then it is again released from rest. (c) At what angle from the vertical does the tension in the rod equal the weight of the ball? (d) If the mass of the ball is increased, does the answer to (c) increase, decrease, or remain the same?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.