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JEE Advanced 2011Paper 1
Exam
JEE Advanced
Year
2011
Paper
Paper 1
Subject
Physics
Chapter
Laws of Motion

JEE Advanced 2011 · Paper 1 · Physics · Laws of Motion

A block is moving on an inclined plane making an angle 45^ with the...

A block is moving on an inclined plane making an angle 45∘45^{\circ} with the horizontal and the coefficient of friction is μ\mu. 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=10μN=10 \mu, then NN isA block is moving on an inclined plane making an angle 45^ with the horizontal and the coefficient of friction is μ .A block is moving on an inclined plane making an angle 45^ with the horizontal and the coefficient of friction is μ .A block is moving on an inclined plane making an angle 45^ with the horizontal and the coefficient of friction is μ .A block is moving on an inclined plane making an angle 45^ with the horizontal and the coefficient of friction is μ .

Correct answer

5

Step-by-step solution

A block is moving on an inclined plane making an angle 45^ with the horizontal and the coefficient of friction is μ .
A block is moving on an inclined plane making an angle 45^ with the horizontal and the coefficient of friction is μ .
F1=mgsin⁡θ+\mumgcos⁡θF2=mgsin⁡θ−\mumgcos⁡θ\begin{aligned}& F_1=m g \sin \theta+\mum g \cos \theta\\ & F_2=m g \sin \theta-\mum g \cos \theta \end{aligned}Given that, F_1=3 F_2 or (sin⁡45∘+μcos⁡45∘)\left(\sin 45^{{}^{\circ}}+\mu \cos 45^{{}^{\circ}}\right) =3\left(\sin45^{{}^{\circ}}-\mu \cos 45^{{}^{\circ}}\right) On solving, we get μ=0.5\mu=0.5 \therefore \quad N=10 \mu=5 ∴\therefore Answer is 5 . Analysis of Question Question is simple. Only one has to take care of direction and magnitude of friction.
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