A block of mass 100 g is lying on an inclined plane of angle 30°. The frictional force on this block …….N.

Options

A. `4.9xx10^(2)`
B. `4.9xx10^(-1)`
C. `4.9xx10^(-1)`
D. `4.9 xx10^(1)`

Answer

B

Explanation

Weight of the block W= mg
The component of this weight parallel to the surface of a slope W sin `theta`
gives the friction force
F = mg sin `theta`

`=mg sin 30^(@) = 0.1 xx 9.8 xx (1)/(2) = 0.49 N`
`f= 4.9 xx 10^(-1) N`