Question

you use your hand to stretch a spring to a displacement x from its equilibrium position and then slowly bring it back to that position. What is the work done by hand?

Answer 1

F=Kx for a spring where K is the spring constant and x is displacement from equilibrium
\[W=\int\limits Kxdx=(1/2)Kx^2\]
but since the spring does an equal amount of work on your hand while the spring recoils, the net work done by your hand is 0
\[\sum W _{hand}=0\]