Hooke's law describes a certain light spring of unstretched length 33.6 cm. when one end is attached to the top of a doorframe and a 6.89 kg object is hung from the other end, the length of the spring is 43.2 cm.
Missing question: "What is the spring's constant?"
Solution: The object of mass m=6.89 kg exerts a force on the spring equal to its weight: [tex]F=mg=(6.89 kg)(9.81 m/s^2)=67.6 N[/tex] When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is [tex]\Delta x=43.2 cm-33.6 cm=9.6 cm=0.096 m[/tex] And by using Hook's law, we can find the constant of the spring: [tex]k= \frac{F}{\Delta x}= \frac{67.6 N}{0.096 m}=704.2 N/m [/tex]
