# If a spring has a constant of 4 (kg)/s^2, how much work will it take to extend the spring by 26 cm ?

##### 1 Answer
Jun 12, 2018

Approximately $0.14$ joules.

#### Explanation:

Work done on a spring is given by the equation:

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

where:

• $k$ is the spring constant

• $x$ is the extension in meters

$\therefore x = 26 \setminus \text{cm"=0.26 \ "m}$

So the work done is:

W=1/2*4 \ "kg/s"^2*(0.26 \ "m")^2

$= 2 \setminus {\text{kg/s"^2*0.0676 \ "m}}^{2}$

$\approx 0.14 \setminus {\text{kg m"^2"/s}}^{2}$

=0.14 \ "J" \ (because 1 \ "J"=1 \ "kg m"^2"/s"^2)