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Chapter 8: Q1P (page 202)

What is the spring constant of a spring that stores25 Jof elastic potential energy when compressed by7.5 cm?

Short Answer

Expert verified

Spring constant is,k=8.9×103 N/m .

Step by step solution

01

Given

Elastic potential energyU=25 J

Compressed lengthx=7.5 cm=7.5×10−2 m .

02

To understand the concept 

The problem is based on the concept of elastic potential energy. It is energy stored as a result of applying a force to deform an elastic object. By using the concept of elastic potential energy, we can find the spring constant.

Formula:

U=12kx2

03

Calculate the spring constant of the spring 

Elastic potential energy can be written as

U=12kx2k=2Ux2

Substitute all the value in the above equation.

k=2×25 J(7.5×10−2 m)2k=8.9×103 N/m

Hence the spring constant is, k=8.9×103 N/m.

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