A2.0-kg block is on a perfectly smooth (frictionless) ramp that makes an angle of 30^\circ30 ∘ with the horizontal. what is the force of the ramp on the block?

Explanation:

Since the surface is frictionless therefore there will be no friction force on block but there will be weight of block which we can divide in to two components i.e. mgcosθ &mgsinθ which is perpendicular and parallel to the surface respectively.

In response to mgcosθ ramp will apply a normal force to the block which will be of equal magnitude to that of mgcosθ.

Therefore Ramp will apply a Force of mgcosθ on block where m is the mass of block.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west