A block is moving on an inclined plane making an angle 45° with the horizontal and the coefficient of friction is µ. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N = 10 µ, then N is

Engineering

Physics

Motion of Block on Inclined Surface

Static and Kinetic Friction

Question

A block is moving on an inclined plane making an angle 45° with the horizontal and the coefficient of friction is µ. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N = 10 µ, then N is

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Let F_up be the force to push up and F_down be the force to prevent sliding down on an inclined plane at angle θ=45°.

The forces are given by:

F_up = mg(sinθ + μcosθ)

F_down = mg(sinθ - μcosθ)

Given F_up = 3F_down. Substituting the expressions:

$m g (\sin θ + μ \cos θ) = 3 m g (\sin θ - μ \cos θ)$

Cancel mg and substitute sin45° = cos45° = 1/√2:

$\frac{1}{\sqrt{2}} + μ \frac{1}{\sqrt{2}} = 3 (\frac{1}{\sqrt{2}} - μ \frac{1}{\sqrt{2}})$

Multiply both sides by √2: 1 + μ = 3(1 - μ) → 1 + μ = 3 - 3μ → 4μ = 2 → μ = 0.5.

Therefore, N = 10μ = 10 × 0.5 = 5.

Final Answer: 5