Question

a block of mass 2 kg is placed on tally on a plane in client at 30 degree with the horizontal the coefficient of friction between the block and the surface is 0.5 the frictional force act on the block is​

Answer 1

F=mgsin (theta)

F=mgsin (theta)F=2×9.8×1/2

$f = \: ms \times mg \times \cos(theta)$

=9.8N

Answer 2

Ans
9.8N
Explanat ion
There