A spring gains 2.34 joules of elastic potential

energy as it is compressed 0.250 meter from its

equilibrium position. What is the spring constant

of this spring?

(1) 9.36 N/m (3) 37.4 N/m

(2) 18.7 N/m (4) 74.9 N/m

Elastic energy equals 1/2 k (spring constant) * Δx^2 (change in position) so you set the energy of the spring equal to the 1/2kΔx^2 and solve for K. 2.34=1/2k (.250) ^2 which gives you a spring constant of 74.9 N/m.