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A block of mass m is on an inclined plan...

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 `tanthetagtmu`. 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=mg(sintheta-mucostheta)` to `P_2=mg(sintheta+mucostheta)`, the frictional force f versus P graph will look like

A

B

C

D

Text Solution

Verified by Experts

The correct Answer is:
A

From FBD of block, `N=mg cos theta`
`p+f=mg sing theta or f=mg sin theta-P`
As P varies from `mg(sin theta-mu cost thea) "to" mg (sin theta+mu cos theta)`
`"f varies from"+mu m g cos theta "to"- mu mg cos theta`
This value of friction is always less than equal to `muN` is magnitude. Hence, option (a) is correct
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