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A spiral spring of 8cm extended to 9.2cm when a load of 1.6N is applied. what is the force constant of the spring, provided the elastic is not exceeded.​
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A spiral spring of 8cm extended to 9.2cm when a load of 1.6N is applied. what is the force constant of the spring, provided the elastic is not exceeded.​

User Durandal
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1 Answer

4 votes

Step-by-step explanation:

By Hooke's Law, Fe = kx.

Since Fe = 1.6N and x = 9.2cm - 8cm = 1.2cm,

k = Fe/x = 1.6N/1.2cm = 1.33N/cm.

User Deantawonezvi
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