# If a spring has a constant of 4 (kg)/s^2, how much work will it take to extend the spring by 83 cm ?

Dec 11, 2015

$1.38 \text{J}$

#### Explanation:

Hooke's Law tells use that the extending force is proportional to the extension:

$F \propto x$

Which is expressed as:

$F = - k x$

Work done is equal to the force multiplied by the distance moved in the direction of the force.

Because the force changes as the extension increases we need to use some calculus:

$W = \int F . \mathrm{dx}$

$\therefore W = - \int k x . \mathrm{dx}$

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

$x = 83 \text{cm"=0.83"m}$

$\therefore W = 0.5 \times 4 \times {0.83}^{2} = 1.38 \text{J}$