If a spring has a constant of #12 (kg)/s^2#, how much work will it take to extend the spring by #76 cm #?

1 Answer
Dec 9, 2015

Answer:

#4,56J#

Explanation:

I assume the units of the spring constant you meant was #12N//m# since from Hooke's Law, #F=kx=>k=F/x# and so the dimensions of k is N/m.

In this case, we use the formula for energy stored in a compressed or elongated spring. #W=1/2kx^2#, (this may be derived from calculus - ask me if you require the derivation as well), to obtain

#W=1/2xx12N//mxx0,76m#

#=4,56J#