Question

A spring is 16.2 cm long when it is lying on a table. One end is then attached to a hook and the other end is pulled by a force that increases to 28.0 N , causing the spring to stretch to a length of 19.0 cm .
Part A
What is the force constant of this spring?
Part B-How much work was required to stretch the spring from 16.2 cm to 19.0 cm ?
Part C- How long will the spring be if the 28.0 N force is replaced by a 56.0 N force?

Answer 1

Nvm guys, I got it !
A- k=F(delta x)=28N/(0.190-0.162)=1000N/m
B- w=1/2mv^2= 1/2(1000)(0.190-0.162)^2=0.392J
C- delta x = F/k=56N/1000=0.056 + 0.162= 0.218m= 21.8cm

Answer 2

B. Work = F*d = 28 * (0.19-0.162) = 32.8 Joules.

C. Length = 1m/1000N * 56N = 0.056 m.
or 56N/(1000N/m) = 0.056 m.