Question

A spring scale hung from the ceiling stretches by 5.6 cm when a 2.0 kg mass is hung from it. the 2.0 kg mass is removed and replaced with a 2.7 kg mass.what is the stretch of the spring

Answer 1

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
$F_1 = m_1 g =(2.0 kg)(9.81 m/s^2)=19.62 N$
This force stretches the spring by $\Delta x_1 = 5.6 cm=0.056 m$, so we can use Hook's law to find the spring constant:
$k= \frac{F_1}{\Delta x_1}= \frac{19.62 N}{0.056 m}=350.4 N/m$

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:
$F_2 = m_2 g =(2.7 kg)(9.81 m/s^2)=26.5 N$
And so, we can use again Hook's law to calculate the new stretch of the spring:
$x_2= \frac{F_2}{k}= \frac{26.5 N}{350.4 N/m}=0.076 m=7.6 cm$