A body of mass m dropped from a height h reaches the ground with a...

${K}_{1} + {U}_{1} = {K}_{2} + {U}_{2} + {W}_{f}$
${W}_{f}:$ work done against friction
$0 + mgh = \dfrac{1}{2}m{v}^{2} + 0 + {W}_{f}$
$mgh = \dfrac{1}{2}m\left(0 . 64\right)gh + {W}_{f}$
${W}_{f} = 0.68mgh$
Work done by friction ${W}_{f}^{'} = - {W}_{f} = - 0.68mgh$