TS EAMCET 11-Sep-2020 Shift 1 Solved Paper

In addition to the forces shown in figure, there will be another upcoming normal force FN exerted by the floor on the block. Using Newton's second law in horizontal and vertical directions, we getIn horizontal direction,
F1−fs=ma...(i)
and in vertical direction,
F2+FN−mg=0...(ii)
Given, F2=10N,m=2kg and g=10ms−2
⇒‌‌10+FN−mg=0
or ‌‌FN=(2×10)−10=10N...(iii)
Hence, static friction, fs=µsFN
=0.4×10=4N...(iv)
Clearly, it is lesser than static force F1(=6N), so block will move, i.e. we are dealing with kinetic friction now.
Hence, friction force (kinetic) acting on the body is
fk=µkFN=0.25×10=2.5N