Question

A spring scale hung from the ceiling stretches by 6.4 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and replaced with a 2.5 kg mass.

Answer 1

In the first case, the force acting on the spring is the weight of the mass:
$F=mg=(2.0 kg)(9.81 m/s^2)=19.6N$
This force causes a stretching of $x=6.4 cm=0.064 m$ on the spring, so we can use these data to find the spring constant:
$k= \frac{F}{x}= \frac{19.6 N}{0.064 m}=306.3 N/m$

In the second case, the first mass is replaced with a second mass, whose weight is
$F=mg=(2.5 kg)(9.81 m/s^2)=24.5 N$
And since we know the spring constant, we can calculate the new elongation of the spring:
$x= \frac{F}{k}= \frac{24.5 N}{306.3 N/m}=0.080 m=8.0 cm$