A 5 kg block moves with a constant speed of 10 ms to the right on a smooth surface where frictional forces are considered to be negligible. It passes through a 2. 0 m rough section of the surface where friction is not negligible, and the coefficient of kinetic friction between the block and the rough section μk is 0. 2. What is the change in the kinetic energy of the block as it passes through the rough section?.

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.

the answer is a because there all in high school