JEE Main 16 March 2021 Shift 1 Solved Paper

The diagram and the required components of force on given block are shown belowFrom the above diagram,
N=mg−F‌sin‌θ...(i)
where, N=‌ normal force ‌
and F‌cos‌θ−fk=ma
⇒‌F‌cos‌θ−µkN=ma ...(ii)
where, fk= kinetic friction force.
From Eq. (i) and Eq. (ii), we get
F‌cos‌θ−µk(mg−F‌sin‌θ)=ma
⇒‌‌a=‌

F

m

‌cos‌θ−µk(g−‌

F

m

‌sin‌θ)
This is the required acceleration of the block.