# If a spring has a constant of 3 (kg)/s^2, how much work will it take to extend the spring by 37 cm ?

May 16, 2018

Approximately $0.07$ joules.

#### Explanation:

Work done on a spring is given by:

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

where:

• $k$ is the spring constant, usually in newtons per meter

• $x$ is the extension of the spring, usually in meters

So here: $x = 37 \setminus \text{cm"=0.37 \ "m}$.

And so,

W=1/2*3 \ "kg s"^-2*(0.37 \ "m")^2

$= \frac{1}{2} \cdot 3 \setminus {\text{kg s"^-2*0.1369 \ "m}}^{2}$

=1/2*3*0.1369 \ "J" \ (because 1 \ "J"-=1 \ "kg m"^2 \ "s"^-2)

$= 0.06845 \setminus \text{J}$

$\approx 0.07 \setminus \text{J}$