

Block will be at rest
mgsinθ+3N=P+f
10×10×21+3=P+μmgcosθ
( ∵ To get P minimum friction must be minimum)
Pmin=502+3−0.6×10×10×21
=202+3=32N
JEE Main 2019 — Physics Mechanics
A block of mass 10kg is kept on a rough inclined plane as shown in the figure. A force of 3N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P, such that the block does not move downward? (take g=10ms−2 )

Held on 9 Jan 2019 · Verified 6 Jul 2026.
23N
25N
18N
32N
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