Over the smooth surface, the block has
• net force parallel to the surface
∑ F[para] = 0
• net force perpendicular to the surface
∑ F[perp] = F[normal] - (5 kg) g = 0
so that F[normal] = (5 kg) g = 49 N
Over the rough patch,
• net force parallel to surface
∑ F[para] = - F[friction] = (5 kg) a
where a is the acceleration of the block
• net force perpendicular to surface
∑ F[perp] = F[normal] - (5 kg) g = 0
so that, again, F[normal] = 49 N, and it follows that
F[friction] = µ F[normal] = 0.2 (49 N) = 9.8 N
As the block slides over this rough patch, friction performs
(-9.8 N) (2.0 m) = -19.6 J
of work on the block. No other forces act to speed up or slow down the block, so this is the total work done on it. By the work-energy theorem, this work is equal to the change in the block's kinetic energy, so the answer is -19.6 J. In other words, the block's kinetic energy is reduced by 19.6 J.
