A block rests on a rough inclined plane making an angle of `30 ^@` with the horizontal . The coefficient of static friction between the block and the plane is 0.8 If the frictional force on the block is 10 N the mass of the block is `" "(g=10 ms ^(-2))`

Options

A. 1 kg
B. 2 kg
C. 3 kg
D. 4 kg

Answer

B

Explanation

Here , angle of inclination , `theta = 30^@`
Coefficient of static friction between block and plane `mu_s=0.8`
Frictional force on the block , f = 10 N
Let the mass of the block be m.
The various forces acting on the block are shown in the figure given here.

As the block is resting on the plane,
`:. f = mgsintheta`
or `m=f/(g sintheta)=(10N)/((10 " m s"^(-2))sin30^@)=2kg`