Question

1. a spring hangs at rest from a fixed support. if you suspend a 370 g mass from the spring with a spring constant of 100.0 n/m, determine the stretch of the spring.

a. 3.6 m
b. 27 cm
c. 3.6 x 10^-3 m
d. 3.6 cm

Explanation:

Step1: Convert mass to SI units

The mass \(m = 370g=370\times10^{- 3}kg = 0.37kg\).

Step2: Use Hooke's law \(F = kx\) and equilibrium condition \(F = mg\)

Since \(F = mg\) (force due to gravity) and \(F = kx\) (Hooke's law), we can equate them: \(mg=kx\).
We need to solve for \(x\), so \(x=\frac{mg}{k}\).
Substitute \(m = 0.37kg\), \(g = 9.8m/s^{2}\), and \(k = 100.0N/m\) into the formula:
\(x=\frac{0.37\times9.8}{100.0}\)
\(x=\frac{3.626}{100.0}=0.03626m\approx0.036m = 3.6cm\)

Answer:

D. \(3.6\ cm\)