STANDARD Physics Friction on Inclined Planes MCQ 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 m s⁻²)
25 N
23 N
18 N
32 N
Correct Answer
Detailed Explanation
The limiting friction is calculated as fₛ = μ mg cos 45°. Substituting the values, fₛ = 0.6 × 10 × 10 × 1/√2 = 42.43 N. The downward force when the block starts to slide is F = 3 + mg sin 45° = 3 + 10 × 10 × 1/√2 = 73.71 N. The block will not move if P = F - fₛ = 73.71 - 42.43 = 31.28 N, which is approximately 32 N.
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