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A \(35-\mathrm{N}\) force is applied to a spring with spring constant \(k=220 \mathrm{N} / \mathrm{m} .\) How much does the spring stretch?

Short Answer

Expert verified
The spring stretches by \(0.159 \, m\) or approximately \(16 \, cm\).

Step by step solution

01

Identify the Given Values

The force applied to the spring is given as 35 N and the spring constant is 220 N/m.
02

Write Out Hooke's Law

According to Hooke's Law, the force (F) exerted by a spring is equal to the displacement (x) times the spring constant (k). Mathematically, this can be written as \( F = k*x \).
03

Solve for Displacement (x)

Rearrange Hooke's Law to solve for the displacement \( x \). You get \( x = F / k \).
04

Substitute the Known Values

Substitute the force \( F = 35 \, N \) and the spring constant \( k = 220 \, N/m \) into the formula. You have \( x = 35 \, N / 220 \, N/m \).
05

Calculate the Displacement

Perform the division to get the displacement, \( x \), which is the amount the spring stretches.

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