If a spring has a constant of #1 (kg)/s^2#, how much work will it take to extend the spring by #35 cm #?

1 Answer
Dec 13, 2015

Answer:

#61.25mJ#

Explanation:

Note that the spring stiffness constant comes from Hooke's Law #F=kx=>k=F/x# and hence has units #N//m#.

I am going to assume its value is hence #1N//m# (which is a very small value and hence a very elastic spring, very easy to compress and stretch).

From the formula for work done to compress or elongate a spring, we get : (Let me know if you require the derivation of this formula)

#W=1/2kx^2#

#=1/2xx1xx0.35^2#

#=61.25mJ#