# If a spring has a constant of 1 (kg)/s^2, how much work will it take to extend the spring by 15 cm ?

May 20, 2018

Approximately $0.01$ joules.

#### Explanation:

Work exerted on a spring is given by:

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

where:

• $k$ is the spring constant, usually in newtons per meter or kilograms per second squared.

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

Here: $x = 15 \setminus \text{cm"=0.15 \ "m}$.

:.W=1/2*1 \ "kg/s"^2*(0.15 \ "m")^2

$= \frac{1}{2} \setminus {\text{kg/s"^2*0.0225 \ "m}}^{2}$

$= 0.01125 \setminus {\text{kg m"^2"/s}}^{2}$

~~0.01 \ "J" \ (because 1 \ "J"=1 \ "kg m"^2"/s"^2)