A block of mass `m=4kg` is placed oner a rough inclined plane as shown in figure. The coefficient of friction between the block and the plane is `mu=0.5` . A force `F=10N` is applied on the block at an angle of `30^(@)` .

Options

A. static in nature in the direction up the plane and have the value `30.2N`
B. static in nature in the direction down the plane and have the value `30.2N`
C. kinetic in nature in the direction up the plane and have the value `30.2N`
D. None of these

Answer

C

Explanation

Drawing free body diagram of block, `sumFy=0`
implies `N+Fsin30^(@)=mgcos37^(@)`
or, `N=mgcos37^(@)-Fsin30^(@)`
`=(4)(10)((4)/(5))-(10)((1)/(2))`
or, `N=27N`
`f_(max)=muN=0.5xx27=13.5N`
`mgsin37^(@)=(4)(10)((4)/(5))=32N`
and `Fcos30^(@)=(10)((sqrt(3))/(2))=8.66N`
Now since `mgsin37^(@)gtf_(max)+Fcos30^(@)`
Therefore block will slide down and friction will be kinetic.