AP EAMCET Engineering 21 Sep 2020 Shift 1 Paper

Static friction force between 10kg and 40kg block,Fs‌‌=µsR=0.6×mg
‌‌=0.6×10×9.8=58.8N
Here, we see that the applied force (F=100N) is greater than friction force, hence 10kg block will start motion due to application of 100N force.
Due to motion, kinetic friction force
fk=µkR=0.4mg
=0.4×10×9.8=39.2N
40kg body experiences a force of fk=39.2N
∴ Acceleration of 40kg slab
a=‌

fk

40

=‌

39.2

40

=0.98ms−2