## Essential University Physics: Volume 1 (4th Edition) Clone

$5.49 \ km/h$
We use conservation of energy to find: $\frac{1}{2}mv^2 = \frac{1}{2}kx^2$ $v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(1.3\times10^6N)(.05m)^2}{1400}}=1.52 \ m/s = 5.49 \ km/h$