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Fundamentals Of Physics
Found in: Page 202

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Short Answer

What is the spring constant of a spring that stores 25 Jof elastic potential energy when compressed by 7.5 cm?

Spring constant is,k=8.9×103 N/m .

See the step by step solution

Step by Step Solution

Step 1: Given

Elastic potential energy U=25  J

Compressed lengthx =7.5 cm=7.5×102 m .

Step 2: 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

Step 3: 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(7.5×102 m)2k=8.9×103 N/m

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

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