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Chapter 9: Q.57 (page 230)

Two identical horizontal springs are attached to opposite sides of a box that sits on a frictionless table. The outer ends of the springs are clamped while the springs are at their equilibrium lengths. Then a 2.0N force applied to the box, parallel to the springs, compresses one spring by 3.0cm while stretching the other by the same amount. What is the spring constant of the springs?

Short Answer

Expert verified

The value of spring constant is33N/m.

Step by step solution

01

Content Introduction

Hooke's law says the force applied on spring is directly proportional to deformation in the spring.

Fsp=kx

Here,Fspis force applied on spring,kis spring constant,xis deformation in the spring.

02

Content Explanation

As we know two springs are identical therefore, spring constant for both springs is same.

Total deformation in spring is (3.0cm+3.0cm)=6cm

Use Hooke's law

Fsp=kx2.0N=k(6cm)(1m100cm)k=2.0N6.0脳10-2mk=33N/m

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