Question 103 of 210

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 $g=10m{s}^{-2}$ )

A$23N$B$25N$C$18N$D$32N$

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Answer: (D) $32N$

Block will be at rest

$mg\sin ⁡\theta +3N=P+f$

$10\times 10\times \dfrac{1}{\sqrt{2}}+3=P+\mu mg\cos ⁡\theta$

( $∵$ To get $P$ minimum friction must be minimum)

${P}_{min}=50\sqrt{2}+3-0.6\times 10\times 10\times \dfrac{1}{\sqrt{2}}$

$=20\sqrt{2}+3=32N$