Question

will answer brainliest pls solve fast
A block of mass 2 kg rests on a rough inclined plane making an angle 30 degree with the horizontal . the coefficient of static friction between the block and the plane is 0.7 . the frictional force on the block is ?
a) 9.8
b) 0.7*9.8√3
c) 9.8 *7 N
d) 0.8*9.8 N

Answer 1

Answer:

Mass of the block = 2 kg

Weight of the block = mg = 2 × 9.8 = 19.6 N

The component of the weight along normal = 19.6 Cos 30° = 16.97 N

Hence, the normal force (N) = 16. 97 N

Now, the component of the weight along the inclined plane = 19.6 Sin 30°

Along the inclined plane = 9.8 N

Now, the friction = μ N = 0.7 × 16.97 = 11.87 N

Since, 9.8 N is less than the maximum value of the friction (11.87 N)

Hence, the frictional force = 9.8 N