## College Physics (4th Edition)

The amount of work that must be done to stretch the spring is $1.60~J$
We can find the elastic potential energy stored in the spring when it is pulled back: $U_s = \frac{1}{2}kx^2$ $U_s = \frac{1}{2}(20.0~N/m)(0.40~m)^2$ $U_s = 1.60~J$ The amount of work that must be done to stretch the spring is $1.60~J$