Question

1. a spring is stretched 15 cm by a 45 - n force. how far would the spring be stretched if a 60 - n force were applied?

Explanation:

Step1: Find the spring constant

According to Hooke's law \(F = kx\), where \(F\) is the force, \(k\) is the spring constant, and \(x\) is the extension.
Given \(F_1=45\space N\) and \(x_1 = 15\space cm\), we can find \(k\) as \(k=\frac{F_1}{x_1}\).
Substituting the values: \(k=\frac{45}{15}= 3\space N/cm\)

Step2: Calculate the new extension

Now, when \(F_2 = 60\space N\) and \(k = 3\space N/cm\), using \(x=\frac{F}{k}\)
\(x_2=\frac{F_2}{k}=\frac{60}{3}=20\space cm\)

Answer:

The spring would be stretched \(20\space cm\)