JEE Main 2019 — Laws Of Motion Question with Solution
From: JEE Main 2019 (Online) 9th January Morning Slot
Question
A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N 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 = 10 ms2)
This question includes a diagram. The text above accompanies the figure.
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Show full solutionCorrect option: A
Correct answer
A32 N
Step-by-step explanation
At equilibrium,
N = mgcos45o . . . . . . (1)
3 + mgsin45o = P + N
3 + = P + 0.6
3 + = P + 0.6
P = 3 +
P = 3 + 20
P = 31.28 32 N
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