Q58 / 84 · JEE Advanced 2010 (Paper 1) · Physics
A block of mass $m$ is on an inclined plane of angle $\theta$. The coefficient of friction between the block and the plane is $\mu$ and $\tan \theta>\mu$. The block is held stationary by applying a force $P$ parallel to the plane. The direction of force pointing up the plane is taken to be positive. As $P$ is varied from $P_1=m g(\sin \theta-\mu \cos \theta) \quad$ to $P_2=m g(\sin \theta+\mu \cos \theta)$, the frictional force $f$ versus $P$ graph will look like
- Correct answerA
- B
- C
- D
Solution
When
$$
\begin{aligned}
P & =m g(\sin \theta-\mu \cos \theta) \\
f & =\mu m g \cos \theta \quad \text { (upwards) } \\
\text { when } \quad P & =m g \sin \theta \\
f & =0
\end{aligned}
$$
and when $P=m g(\sin \theta+\mu \cos \theta)$
$$
f=\mu m g \cos \theta \text { (downwards) }
$$
Hence friction is first positive, then zero and then negative.
$\therefore$ correct option is (a).
Hence, the correct answer is option: 'A'.