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

$\frac{1}{2}kx_0^2 = \mu mgd$
We know that the block will stop when all of the spring energy becomes mechanical energy, meaning that it will be lost to friction. Thus, we find: $\frac{1}{2}kx_0^2 = F_f \times d$ $\frac{1}{2}kx_0^2 = \mu F_n \times d$ $\frac{1}{2}kx_0^2 = \mu mgd$