KEAM 2021 Physics and Chemistry Solved Paper

FBD of the given block is as shown belowNet force on the block due to which it is moving in horizontal direction,
F‌net ‌=ma
=10×3=30N
Also, F‌net ‌=F‌applied ‌−f
where, f is frictional force.
⇒‌‌F‌applied ‌=F‌net ‌+f...(i)
As in equilibrium,
ΣFx=0
⇒ R=mg ....(ii)
Also, F=µrR=µrmg [from Eq. (ii)]
Now, Eq. (i) can be written as
F‌applied ‌=F‌net ‌+µrmg
Substituting the values in the above equation, we get
F‌applied ‌=30+0.3×10×10
=30+30
=60N