If a spring has a constant of 4 (kg)/s^24kgs2, how much work will it take to extend the spring by 65 cm 65cm?

1 Answer
Apr 11, 2018

The work done is 0.8450.845 joules.

Explanation:

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

W=1/2kx^2W=12kx2

  • kk is the spring constant in "N/m"N/m

  • xx is the extension of the spring in meters

Note that:

1 \ "N/m"=("kg m""/s"^2)/("m")=1 \ "kg/s"^2

:.4 \ "kg/s"^2=4 \ "N/m"

So, we have:

65 \ "cm"=0.65 \ "m"

And so, the work done is:

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

=2 \ "N/m"*0.4225 \ "m"^2

=0.845 \ "N m"

=0.845 \ "J"