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.
In the first case, the force acting on the spring is the weight of the mass: [tex]F=mg=(2.0 kg)(9.81 m/s^2)=19.6N[/tex] This force causes a stretching of [tex]x=6.4 cm=0.064 m[/tex] on the spring, so we can use these data to find the spring constant: [tex]k= \frac{F}{x}= \frac{19.6 N}{0.064 m}=306.3 N/m [/tex]
In the second case, the first mass is replaced with a second mass, whose weight is [tex]F=mg=(2.5 kg)(9.81 m/s^2)=24.5 N[/tex] And since we know the spring constant, we can calculate the new elongation of the spring: [tex]x= \frac{F}{k}= \frac{24.5 N}{306.3 N/m}=0.080 m=8.0 cm [/tex]
