(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

When a 2.90-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.93 cm.

(a) What is the force constant of the spring?

(b) If the 2.90-kg object is removed, how far will the spring stretch if a 1.45-kg block is hung on it?

2. Relevant equations

Hooke's law: Fs=-kx

F=ma

3. The attempt at a solution

a. F=ma

F= (2.90 kg) (9.8 m/s^{2})

F=28.42N

Fs=-kx

27.42 N=-k (0.0293m)

k=970.0 N/m

b. F=ma

F= (1.45 kg) (9.8 m/s^{2})

F=14.21 N

Fs=-kx

14.21 N = (970 N/m) x

x=0.015 cm

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# Homework Help: Hooke's Law and the force constant

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