# If a spring has a constant of 7 (kg)/s^2, how much work will it take to extend the spring by 21 cm ?

Jul 23, 2017

$0.31 J$

#### Explanation:

$F = k \Delta x$, and $W = F \Delta s$, where:
$\Delta x = \Delta s$ = extension/distance travelled in the direction of the force ($m$)
$F$ = force ($N$)
$W$ = Work done ($N m$)
$k$ = spring constant ($N$ ${m}^{- 1}$)

Therefore, $W = k \left(\Delta x\right) \left(\Delta x\right) = k {\left(\Delta x\right)}^{2} = 7 \cdot {0.21}^{2} = 0.3087 \approx 0.31 J$