If a spring has a constant of #4 (kg)/s^2#, how much work will it take to extend the spring by #19 cm #?

1 Answer
Jun 28, 2017

Answer:

3.04J of work.

Explanation:

19cm=0.19m

Hooke's law says that #FpropDeltax#, or #F=kDeltax#. In this question we know both the extension, and spring constant, so we just have to do 0.19x4=0.76N.

The question however asks for work done, which is represented by #W=FDeltas#, #Deltas=Deltax#. We just need to do 0.76x4=3.04J of work.