Question

Work of 2 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (in Newtons) that holds the spring stretched at the same distance (19 cm)?

Answer 1

Work done ,
Wd = 1/2*kx^2 2joules = 1/2*k*(0.19)^2
find k , spring constant Force that holds the spring stretched = k*0.19 N