When the mass of 2.0 kg is attached to the spring, the force applied to the spring is the weight of the mass, so
[tex]F_1 = m_1 g =(2.0 kg)(9.81 m/s^2)=19.62 N[/tex]
This force stretches the spring by [tex]\Delta x_1 = 5.6 cm=0.056 m[/tex], so we can use Hook's law to find the spring constant:
[tex]k= \frac{F_1}{\Delta x_1}= \frac{19.62 N}{0.056 m}=350.4 N/m [/tex]
Then the mass of 2.0 kg is removed and replaced with another mass m2=2.7 kg. The force produced by this mass is equal to its weight:
[tex]F_2 = m_2 g =(2.7 kg)(9.81 m/s^2)=26.5 N[/tex]
And so, we can use again Hook's law to calculate the new stretch of the spring:
[tex]x_2= \frac{F_2}{k}= \frac{26.5 N}{350.4 N/m}=0.076 m=7.6 cm [/tex]
