Question

If a spring has a constant of `1 (kg)/s^2`, how much work will it take to extend the spring by `35 cm`?

Answer 1

`61.25mJ`

Explanation:

Note that the spring stiffness constant comes from Hooke's Law `F=kx=>k=F/x` and hence has units `N//m`.

I am going to assume its value is hence `1N//m` (which is a very small value and hence a very elastic spring, very easy to compress and stretch).

From the formula for work done to compress or elongate a spring, we get : (Let me know if you require the derivation of this formula)

`W=1/2kx^2`

`=1/2xx1xx0.35^2`

`=61.25mJ`