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 `1.5` N upwards
C. Contact force between the block and belt is `10.5 N`
D. Contact force between the block and the belt is `5 sqrt3N`

Answer

A, C

Explanation

Block is moving upwards due to friction `f_(r) - mg sin 30^(@) = ma`
` (##GRB_AM_PHY_C07_E01_104_S01.png" width="80%">
`implies f_(r) - 1 xx 10 xx (1)/(2) = 1 xx l implies f_(r) = 6N`
Contact force is the resultant of N and `f_(r)`
`= sqrt(N^(2) + f_(r)^(2)) = sqrt((mg cos 30)^(2) + (6)^(2))`
`=10.5 N`