Laws Of Motion Ques 41
- 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 $P_{1}=m g(\sin \theta-\mu \cos \theta)$ to $P_{2}=m g(\sin \theta+\mu \cos \theta)$, the frictional force $f$ versus $P$ graph will look like
(2010)
(a)
(b)
(c)
(d)
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Answer:
Correct Answer: 41.(a)
Solution:
Formula:
- When
$$ \begin{aligned} & P=m g(\sin \theta-\mu \cos \theta) \\ & f=\mu m g \cos \theta & (upwards)\\ & \text { when } \quad P=m g \sin \theta ; f=0 \\ & \text { and when } \quad P=m g(\sin \theta+\mu \cos \theta) \\ & f=\mu m g \cos \theta & (downwards) \end{aligned} $$
Hence, friction is first positive, then zero and then negative.
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