# If a spring has a constant of 2 (kg)/s^2, how much work will it take to extend the spring by 56 cm ?

May 27, 2018

Around $0.31$ joules.

#### Explanation:

Work on a spring is given by:

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

where:

• $W$ is the work done in joules

• $k$ is the spring constant

• $x$ is the extension, usually in meters

$\therefore x = 56 \setminus \text{cm"=0.56 \ "m}$

So, the work done is:

W=1/2*2 \ "kg/s"^2*(0.56 \ "m")^2

$\approx 0.31 \setminus {\text{kg m"^2"/s}}^{2}$

=0.31 \ "J" \ (because 1 \ "J"=1 \ "kg m"^2"/s"^2)