# If a spring has a constant of 4 (kg)/s^2, how much work will it take to extend the spring by 65 cm ?

Apr 11, 2018

The work done is $0.845$ joules.

#### Explanation:

We use the work equation for a spring, which states that,

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

• $k$ is the spring constant in $\text{N/m}$

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

Note that:

$1 \setminus {\text{N/m"=("kg m""/s"^2)/("m")=1 \ "kg/s}}^{2}$

$\therefore 4 \setminus \text{kg/s"^2=4 \ "N/m}$

So, we have:

$65 \setminus \text{cm"=0.65 \ "m}$

And so, the work done is:

W=1/2*4 \ "N/m"*(0.65 \ "m")^2

$= 2 \setminus {\text{N/m"*0.4225 \ "m}}^{2}$

$= 0.845 \setminus \text{N m}$

$= 0.845 \setminus \text{J}$