# If a spring has a constant of 12 (kg)/s^2, how much work will it take to extend the spring by 76 cm ?

Dec 9, 2015

$4 , 56 J$

#### Explanation:

I assume the units of the spring constant you meant was $12 N / m$ since from Hooke's Law, $F = k x \implies k = \frac{F}{x}$ and so the dimensions of k is N/m.

In this case, we use the formula for energy stored in a compressed or elongated spring. $W = \frac{1}{2} k {x}^{2}$, (this may be derived from calculus - ask me if you require the derivation as well), to obtain

$W = \frac{1}{2} \times 12 N / m \times 0 , 76 m$

$= 4 , 56 J$