Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The plane's kinetic energy when it lands will be equal to the energy $U_s$ stored in the spring. We can find the landing speed of the plane. Therefore; $KE = U_s$ $\frac{1}{2}mv^2=\frac{1}{2}kx^2$ $v^2=\frac{kx^2}{m}$ $v=\sqrt{\frac{kx^2}{m}}$ $v=\sqrt{\frac{(60,000~N/m)(30~m)^2}{15,000~kg}}$ $v = 60~m/s$ The plane's landing speed was 60 m/s.