A block of mass `1 kg` is stationary with respect to a conveyer belt that is accelerating with `1 m//s^(2)` upwards at an angle of `30^(@)` as shown in figure. Which of the following is`//`are correct?

Options

A. Force of friction on block is `6 N` upwards.
B. Force of friction on block is `1.5N` upwards.
C. Contract force between the block & belt is `10.5 N`.
D. Contact force between the block & belt is `5sqrt(3)N`.

Answer

A, C

Explanation

`(A,C)` Block is moving upwards due to friction
`f_(r )-mg "sing" 30 = ma`

`rArr f_(r ) - 1 xx 10 xx (1)/(2) = 1 xx 1 rArr f_( r) = 6N`
Constant forces is the resultant of `N` and `f_(r )`
`= sqrt(N^(2)+r_(r )^(2)) = sqrt((mg cos 30)^(2)+(6)^(2)) = 10.5 N`