Question

a force 80N extends a natural length 8m by 0.4m,what will be the length of the spring when the applied force is 100N

Answer 1

Task. A force $F_{1}=80$ $N$ extends a natural length $L=8$ $m$ by $\Delta L_{1}=0.4$ $m$. What will be the length of the spring when the applied force is $F_{2}=100$ $N$?

Solution. By Hooke’s law the force $F$ needed to extend a spring $\Delta L$ is proportional to $\Delta L$:

$F=k\,\Delta L,$

where $k$ is a constant called stiffness of a spring.

In the first case

$\Delta L_{1}=0.4\ m,\qquad F_{1}=80\,N,$

whence

$F_{1}=k\,\Delta L_{1},$

and so

$k=\frac{F_{1}}{\Delta L_{1}}=\frac{80}{0.4}=200\ N/m.$

Now apply the force $F_{2}=100$ $N$ and let $\Delta L_{2}$ be the extension of the length in this case. Then similarly,

$F_{2}=k\Delta L_{2},$

whence

$\Delta L_{2}=\frac{F_{2}}{k}=\frac{100}{200}=0.5\ m.$

Hence the length of the spring in the second case will be

$L_{2}=L+\Delta L_{2}=8+0.5=8.5\ m.$

Answer. $8.5\ m.$