JEE Advanced · Physics · 5. Laws of Motion
A block is moving on an inclined plane making an angle \(45^{\circ}\) with the horizontal and the coefficient of friction is \(\mu\). 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 \(N=10 \mu\), then \(N\) is
- A 10
- B 15
- C 5
- D 20
Answer & Solution
Correct Answer
(C) 5
Step-by-step Solution
Detailed explanation


\(
\begin{aligned}
& F_1=m g \sin \theta+\mu m g \cos \theta \\
& F_2=m g \sin \theta-\mu m g \cos \theta
\end{aligned}
\)
Given that, \(F_1=3 F_2\)
or \(\left(\sin 45^{\circ}+\mu \cos 45^{\circ}\right)\)
\(
=3\left(\sin 45^{\circ}-\mu \cos 45^{\circ}\right)
\)
On solving, we get \(\mu=0.5\)
\(
\therefore \quad N=10 \mu=5
\)
\(\therefore\) 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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