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

Dec 13, 2015

$61.25 m J$

#### Explanation:

Note that the spring stiffness constant comes from Hooke's Law $F = k x \implies k = \frac{F}{x}$ and hence has units $N / m$.

I am going to assume its value is hence $1 N / m$ (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)

$W = \frac{1}{2} k {x}^{2}$

$= \frac{1}{2} \times 1 \times {0.35}^{2}$

$= 61.25 m J$