The spring-flex exercise system consists of a spring with one end fixed and a handle on the other end. The idea is that you exercise your muscles by stretching the spring from its natural length, which is 32 cm. If a 190 Newton force is required to keep the spring stretched to a length of 47 cm, how much work is required to stretch it from 43 cm to 68 cm? Your answer must include the correct units.

Question

anonymous · Answer

$$F(x)=-\frac{1}{2}kx^2$$
You're told that the force (F(x)) is 190 Newtons at 47cm ($x=0.47-0.32=0.15m$), so you can work out what k is.

The displacement at 43cm is 0.11m.
The displacement at 68cm is 0.36m.
Find F(0.11) and F(0.36) to get the total force required to stretch the exercise system.
$$W=F\times d$$Where W is the work done, and d is the displacement during this stretch (0.36 - 0.11)