Get the information you need with the help of IDNLearn.com's expert community. Our platform offers reliable and comprehensive answers to help you make informed decisions quickly and easily.
Sagot :
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.
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.