/*! 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} Q25P If a chain of 50 identical shor... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If a chain of 50identical short springs linked end to end has a stiffness of 270N/m, what is the stiffness of one short spring?

Short Answer

Expert verified

The stiffness of the one short spring is 1.35×104N/m.

Step by step solution

01

Identification of given data

The given is listed as follows:

  • Number of springs is 50
  • The stiffness of the spring is,kt=270N/m
02

Significance of the stiffness of spring

The stiffness of a spring is described as the ability of a particular material for resisting the deformation of a particular object.

The concept of the stiffness gives the stiffness of one short spring.

03

Calculation for the stiffness of one short spring

When the springs are calculated in the series form then the equation of the stiffness is expressed as follows:

1kt=1k+..............+50k

Here, ktis the total stiffness of the spring and k is the stiffness of one spring.

Substitute all the values in the above equation.

1270N/m=50kk=1.35×104N/m

Thus, the stiffness on one spring is 1.35×104N/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!

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

Steel is very stiff, and Young’s modulus for steel is unusually large, 2×1011N/m. A cube of steel 28 cm on a side supports a load of 85 kg that has the same horizontal cross section as the steel cube. (a) What is the magnitude of the normal force that the steel cube exerts on the load? (b) What is the compression of the steel cube? That is, what is the small change in height of the steel cube due to the load it supports? Give your answer as a positive number. The compression of a wide, stiff support can be extremely small.

The diameter of a copper atom is approximately2.28×10-10m. The mass of one mole of copper is 64g. Assume that the atoms are arranged in a simple cubic array. Remember to convert to SI units. (a) What is the mass of one copper atom, in kg? (b) How many copper atoms are there in a cubical block of copper that is 4.6 cmon each side? (c) What is the mass of the cubical block of copper, in kg?

A chain of length L, and mass M is suspended vertically by one end with the bottom end just above a table. The chain is released and falls, and the links do not rebound off the table, but they spread out so that the top link falls very nearly the full distance L. Just before the instant when the entire chain has fallen onto the table, how much force does the table exert on the chain? Assume that the chain links have negligible interaction with each other as the chain drops, and make the approximation that there is very large number of links. Hint: Consider the instantaneous rate of change momentum of the chain as the last link hits the table.

In a spring-mass oscillator, when is the magnitude of momentum of the mass largest: when the magnitude of the net force acting on the mass is largest, or when the magnitude of the net force acting on the mass is smallest?

A spring has stiffness ks. You cut the spring in half. What is the stiffness of the half-spring?

(a) 2ks,

(b) ks,

(c) ks/2

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.